Question 1015 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
AnswerHere, $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\cos^2\big(\frac{\text{vx}}{\text{x}}\big)}{\text{x}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\cos^2\text{v}$
$\frac{\text{dv}}{\cos^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\int\sec^2\text{vdv}=-\int\frac{\text{dx}}{\text{x}}$
$\tan\text{v}=-\log|\text{x}|+\log\text{C}$
$\tan\frac{\text{y}}{\text{x}}=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
View full question & answer→Question 1025 Marks
Solve the following differential equation:
$\text{y dx}+\Big\{\text{x}\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}-2\text{x dy}=0$
AnswerWe have, $\text{y dx}+\Big\{\text{x}\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}-2\text{x dy}=0$ $\Rightarrow\ \Big\{2\text{x}-\text{x}\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=\text{y dx}$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{x}-\text{x}\log\big(\frac{\text{y}}{\text{x}}\big)}$ This is a homogeneous differential equation. Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ we get$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{2\text{x}-\text{x}\log\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{2-\log\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-2\text{v + v}\log\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\log\text{v}-\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \frac{2-\log\text{v}}{\text{v}\log\text{v}-\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$ integrating both sides, we get $\int\frac{2-\log\text{v}}{\text{v}\log\text{v}-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\frac{1-(\log\text{v}-1)}{\text{v}(\log\text{v}-1)}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ Putting $\log\text{v}-1=\text{t}$ $\Rightarrow\ \frac{1}{\text{v}}\text{dv}=\text{dt}$ $\therefore\ \int\frac{1-\text{t}}{\text{t}}\text{dt}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\Big(\frac{1}{\text{t}}-1\Big)\text{dt}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \log|\text{t}|-\text{t}=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ \log|\log\text{v}-1|-(\log\text{v}-1)=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ \log|\log\text{v}-1|-\log\text{v}=\log|\text{x}|+\log\text{C}$ $\big($where, $\log\text{C}_1=\log\text{C}-1\big)$ $\Rightarrow\ \log\Big|\frac{\log\text{v}-1}{\text{v}}\Big|=\log|\text{C}_1\text{x}|$ $\Rightarrow\ \frac{\log\text{v}-1}{\text{v}}=\text{C}_1\text{x}$ $\Rightarrow\ \log\text{v}-1=\text{C}_1\text{xv}$ Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get $\log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{x}\times\frac{\text{y}}{\text{x}}$ $\Rightarrow\ \log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{y}$ Hence, $\log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{y}$ is the required solution.
View full question & answer→Question 1035 Marks
Solve the following differential equations:$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
AnswerWe have,
$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
$\Rightarrow\frac{\text{y}}{1+\text{y}^2}\text{dy}=\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Integrating both sides,
$\int\frac{\text{y}}{1+\text{y}^2}\text{dy}=\int\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Substituting $1+\text{y}^2=\text{t}$ and $1-\text{x}^2=\text{u}$
$2\text{ydy = dt}$ and $-2\text{x dx = du}$
$\therefore\frac{1}{2}\int\frac{1}{\text{t}}=\frac{-1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\frac{1}2{}\log|\text{t}|=-\frac{1}{2}\log|\text{u}|+\log\text{C}$
$\Rightarrow\frac{1}{2}|1+\text{y}^2|=-\frac{1}{2}\log|1-\text{x}^2|+\log\text{C}$
$\Rightarrow\frac{1}{2}\big[\log|1+\text{y}^2|+\log|1-\text{x}^2|\big]=\log\text{C}$
$\Rightarrow\log(|1+\text{y}^2||1-\text{x}^2|)=2\log\text{C}$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}^2$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1,$ where $\text{C}_1=\text{C}^2$
Hence, $(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1$ is the required solution.
View full question & answer→Question 1045 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\Big\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\Big\}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}\Big\{\log\Big(\frac{\text{vx}}{\text{x}}\Big)+1\Big\}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v + v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$
$\int\frac{1}{\text{v}\log\text{v}}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\log\log\text{v}=\log|\text{x}|+\log\text{C}$
$\log\text{v}=\text{xC}$
$\log\frac{\text{y}}{\text{x}}=\text{xC}$
$\frac{\text{y}}{\text{x}}=\text{e}^{\text{xC}}$
$\text{y}=\text{xe}^{\text{xC}}$
View full question & answer→Question 1055 Marks
Solve the following differential equation:$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$
AnswerHere, $\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$ $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{e}^{\text{x}}$ It is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{1}{\text{x}}, \text{Q}=\text{e}^{\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{\log\text{x}}$ $=\text{x}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}\times(\text{x})=\int\text{e}^{\text{x}}\times\text{xdx + C}$ $\text{xy}=\text{x}\int\text{e}^{\text{x}}\text{dx}-\int(1\times\int\text{e}^{\text{x}}\text{dx})\text{dx + C}$ Using integration by parts $=\text{x}\text{e}^{\text{x}}-\int\text{e}^{\text{x}}\text{dx}+\text{C}$ $=\text{x}\text{e}^{\text{x}}-\text{e}^{\text{x}}+\text{C}$ $\text{xy}=(\text{x}-1)\text{e}^{\text{x}}+\text{C}$ $\text{y}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}+\frac{\text{C}}{\text{x}},\text{x}>0$
View full question & answer→Question 1065 Marks
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation $\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}^{2}+\frac{\text{dy}}{\text{dx}}.$
Answer$\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}^{2}+\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}+\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{y}^{2}$
$(1-\text{x})\frac{\text{dy}}{\text{dx}}=\text{y}-\text{y}^{2}$
$\frac{\text{dy}}{\text{y}-\text{y}^{2}}=\frac{\text{dx}}{1+\text{x}}$
$\frac{\text{dy}}{\text{y}(1-\text{y})}=\frac{\text{dx}}{1+\text{x}}$
$\int\Big(\frac{1}{\text{y}}+\frac{1}{1-\text{y}}\Big)\text{dx}=\int\frac{\text{dx}}{1+\text{x}}$
$\log|\text{y}|-\log|1-\text{y}|=\log|1+\text{x}|+\log|\text{c}|$
$\frac{\text{y}}{1-\text{y}}=\text{c}(1+\text{x})$
$\text{y}=(1-\text{y})\text{c}(1+\text{x})\ ...(\text{i})$
It is passing through (2, 2) so,
$2=(1-2)\text{c}(1+2)$
$2=-3\text{c}$
$\text{c}=-\frac{2}{3}$
from eq.(i)
$\text{y}=-\frac{2}{3}(1-\text{y})(1+\text{x})$
$3\text{y}=-2(1+\text{x}-\text{y}-\text{xy})$
$3\text{y}+2+2\text{x}-2\text{y}-2\text{xy}=0$
$\text{y}+2\text{y}-2\text{xy}+2=0$
$2\text{xy}+2\text{x}-2-\text{y}=0$
Chapter 22 Differential eq.
It is passing through $\Big(1, \frac{\pi}{4}\Big)$,
$\tan\Big(\frac{\pi}{4}\Big)=-\log|1|+\text{C}$
$1-0+\text{C}$
$\text{C}=1$
Now, eq. (i) become
$\tan\Big(\frac{\text{y}}{\text{x}}\Big)=-\log|\text{x}|+\text{1}$
Therefore,
$\tan\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\frac{\text{e}}{\text{x}}|$
View full question & answer→Question 1075 Marks
Solve the following differential equations:
$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),\text{y}(1)=-1$
Answer$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),\text{y}(1)=-1$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{\text{x}+2}{\text{x}}\text{dx}$
$\Rightarrow\frac{\text{y}+2-2}{\text{y}+2}\text{dy}=\frac{\text{x}+2}{\text{x}}\text{dx}$
$\Rightarrow\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
Integrating both sides, we get
$\int\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\int\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|+\text{C}...(1)$
We know that at $\text{x}=1,\text{y}=-1$
Substituting the valuse of x and y in (1), we get
$-1-2\log|1|=1+2\log|1|+\text{C}$
$\Rightarrow-1=1+\text{C}$
$\Rightarrow\text{C}=-2$
Substituting the value of C in (1), we get
$\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$
Hence, $\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$ is the required solution.
View full question & answer→Question 1085 Marks
Solve the following differential equations:
$2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{y}^2$
AnswerHere, $2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{y}^2$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{y}^2}{2\text{xy}}$
It is homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2-2\text{v}^2}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}$
$\frac{2\text{v}}{1-\text{v}^2}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\int\frac{-2\text{v}}{1-\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|1-\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$
$(1-\text{v}^2)=\frac{\text{C}}{\text{x}}$
$\text{x}\Big(1-\frac{\text{y}^2}{\text{x}^2}\Big)=\text{C}$
$\frac{\text{x}(\text{x}^2-\text{y}^2)}{\text{x}^2}=\text{C}$
$\text{x}^2-\text{y}^2=\text{Cx}$
View full question & answer→Question 1095 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x},\text{ y}=0,\text{ when x}=\frac{\pi}{3}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=2\tan\text{x}$ and $\text{Q}=\sin\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{2\int\tan\text{x dx}}$
$=\text{e}^{2\log|\sec\text{x}|}$
$=\sec^2\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sec^2\text{x},$ we get
$\sec^2\text{x}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}\Big)=\sec^2\text{x}\times\sin\text{x}$
$\sec^2\text{x}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}\Big)=\tan\text{x }\sec\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sec^2\text{x}=\int\tan\text{x}\sec\text{x dx}+\text{C}$
$\text{y}\sec^2\text{x}=\sec{\text{x}}+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{3}\Big)=0$
$\therefore\ 0\Big(\sec\frac{\pi}{3}\Big)^2=\sec\frac{\pi}{3}+\text{C}$
$\Rightarrow\text{C}=-2$
Putting the value of C in (2), we get
$\text{y}\sec^2\text{x}=\sec\text{x}-2$
$\Rightarrow\text{y}=\cos\text{x}-2\cos^2\text{x}$
Hence, $\text{y}=\cos\text{x}-2\cos^2\text{x}$ is the required solution.
View full question & answer→Question 1105 Marks
Solve the following differential equations:$(1+\text{y}^2)\tan^{-1}\text{xdx}+2\text{y}(1+\text{x}^2)\text{dy}=0$
Answer$(1+\text{y}^2)\tan^{-1}\text{xdx}+2\text{y}(1+\text{x}^2)\text{dy}=0$
$(1+\text{y}^2)\tan^{-1}\text{xdx}=-2\text{y}(1+\text{x}^2)\text{dy}$
$-\frac{\tan^{-1}}{2(1+\text{x}^2)}\text{dx}=\frac{\text{y}}{(1+\text{y}^2)}\text{dy}$
Integrating on both the sides
$\int-\frac{\tan^{-1}\text{x}}{2(1+\text{x}^2)}\text{dx}=\int\frac{\text{y}}{(1+\text{y}^2)}\text{dy}$
$-\Big(\tan^{-1}\text{x}\Big(\frac{1}{2}\tan^{-1}\text{x}\Big)-\int\frac{1}{(1+\text{x}^2)}\Big(\frac{1}{2}\tan^{-1}\text{x}\Big)\text{dx}\Big)=\frac{1}{2}\ln(\text{y}^2+1)+\text{C}$
$-\frac{1}{4}(\tan^{-1}\text{x})^2=\frac{1}{2}\ln(\text{y}^2+1)+\text{C}_1$
$\frac{1}{2}(\tan^{-1}\text{x})^2+\ln(\text{y}^2+1)=\text{C}$
View full question & answer→Question 1115 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
Answer$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=2\text{e}^{2\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=2\int\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\frac{-1}{\text{y}}=\text{e}^{2\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=-1.$
Substituting the values of x and y in (1), we get
$1=1+\text{C}$
$\Rightarrow\text{C}=0$
Substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=\text{e}^{2\text{x}}$
$\Rightarrow\text{y}=-\text{e}^{-2\text{x}}$
Hence, $\text{y}=-\text{e}^{-2\text{x}}$ is the required soluton.
View full question & answer→Question 1125 Marks
Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\sin\text{x.}$
AnswerThe differential equation of the curve is:
$\text{y}'=\text{e}^{\text{x}}\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\sin\text{x}$
$\Rightarrow\text{dy}=\text{e}^{\text{x}}\sin\text{x}$
Integrating both sides, we get:
$\int\text{dy}=\int\text{e}^{\text{x}}\sin\text{x dx }...(1)$
Let $\text{I}=\int\text{e}^{\text{x}}\sin\text{x dx}.$
$\Rightarrow\text{I}=\sin\text{x}\int\text{e}^{\text{x}}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\sin\text{x}).\int\text{e}^{\text{x}}\text{dx}\Big)\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\int\cos\text{x}\cdot\text{e}^{\text{x}}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\Big[\cos\text{x}\cdot\int\text{e}^{\text{x}}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\cos\text{x})\cdot\int\text{e}^{\text{x}}\text{dx}\Big)\text{dx}\Big]$
$\Rightarrow\text{I}=\sin\text{x}\cdot\text{e}^{\text{x}}-\big[\cos\text{x}\cdot\text{e}^{\text{x}}-\int(-\sin\text{x})\cdot\text{e}^{\text{x}}\text{dx}\big]$
$\Rightarrow\text{I}=\text{e}^{\text{x}}\sin\text{x}-\text{e}^{\text{x}}\cos\text{x}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})}{2}$
View full question & answer→Question 1135 Marks
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
AnswerLet P(x, y) be the point on the curve y = f(x) such that tangent at P cuts the coordinate axes at A and B.
The quation of tangent is,
$\text{y}-\text{y}=\frac{\text{dy}}{\text{dx}}(\text{x}-\text{x})$
Put y = 0
$-\text{y}=\frac{\text{dy}}{\text{dx}}(\text{x}-\text{x})$
$-\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=\text{x}$
Coording of $\text{B}=\Big(-\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}, 0\Big)$
Here, x intercept of tangent = y
$-\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=\text{y}$
$\frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{\text{y}}=-1$
It is a differential equation on it with $\text{P}=\frac{1}{\text{y}}, \text{Q}=-1$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{y}}\text{dy}}$
$=\text{e}^{\log\text{y}}$
$=\frac{1}{\text{y}}$
Solution of the equation is given by,
$\text{x}(\text{I.F})=\int\text{Q}(\text{I.F})\text{dy}+\text{C}$
$\text{x}(\frac{1}{\text{y}})=\int\text{(-1)}(\frac{1}{\text{y}})\text{dy}+\text{C}$
$\text{x}(\frac{1}{\text{y}})=-\log\text{y}+\text{C}\ ...(\text{i})$
It is passing through (1, 1)
$\frac{1}{\text{1}}=-\log\text{1}+\text{C}$
$\text{C}=1$
Put C = 1 is equation (i)
$\frac{\text{x}}{\text{y}}=-\log\text{y}+\text{1}$
$\text{x}={\text{y}}-\text{y}\log\text{y}$
$\text{x}+\text{y}\log\text{y}=\text{y}$
View full question & answer→Question 1145 Marks
Solve the following differential equation:
$(\text{x}^2-2\text{xy})\text{dy}+(\text{x}^2-3\text{xy}+2\text{y}^2)\text{dx}=0$
AnswerHere, $(\text{x}^2-2\text{xy})\text{dy}+(\text{x}^2-3\text{xy}+2\text{y}^2)\text{dx}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-3\text{xy}+2\text{y}^2}{2\text{xy}-\text{x}^2}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-3\text{xvx}+2\text{v}^2\text{x}^2}{2\text{xvx}-\text{x}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2}{2\text{v}-1}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2-2\text{v}^2+\text{v}}{2\text{v}-1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}}{2\text{v}-1}$
$\frac{2\text{v}-1}{1-2\text{v}}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{1-2\text{v}}{1-2\text{v}}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\int\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\text{v}=-\log|\text{x}|+\text{C}$
$\frac{\text{y}}{\text{x}}+\log\text{x}=\text{C}$
View full question & answer→Question 1155 Marks
Solve the following differential equation:
$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
Answer$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$ $\Rightarrow\ \text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}=-\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}$$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\frac{-\big\{\text{y}^2-\text{x}^2\log\big(\frac{\text{y}}{\text{x}}\big)\big\}}{\text{xy}\log\big(\frac{\text{x}}{\text{y}}\big)}$
$=\frac{\text{x}^2\log\big(\frac{\text{x}}{\text{y}}\big)-\text{y}^2}{\text{xy}\log\big(\frac{\text{x}}{\text{y}}\big)}$
It is a homogeneous equation. We put x = vy $\frac{\text{dx}}{\text{dy}}=\text{v + y}\frac{\text{dv}}{\text{dy}}$ So, $\text{v + y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\text{y}^2\log(\text{v})-\text{y}^2}{\text{vy}^2\log(\text{v})}$ $\text{v + y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1}{\text{v}\log(\text{v})}-\text{v}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1-\text{v}^2\log(\text{v})}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{-1}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{v}\log(\text{v})\text{dv}=\frac{-1}{\text{y}}\text{dy}$ On integrating both sides we get, $\int\text{v}\log(\text{v})\text{dv}=-\int\frac{1}{\text{y}}\text{dy}$ $\Rightarrow\ \frac{\text{v}^2}2\log(\text{v})-\int\frac{\text{v}}2\text{dv}=-\log\text{y + C}$ $\Rightarrow\ \frac{\text{v}^2}2\log(\text{v})-\frac{\text{v}^2}4=-\log\text{y + C}$ $\Rightarrow\ \frac{\text{v}^2}2\Big[\log(\text{v})-\frac{1}2\Big]=-\log\text{y + C}$ $\Rightarrow\ \text{v}^2\Big[\log(\text{v})-\frac{1}2\Big]=-2\log\text{y + C}$ Now putting back the values of v as $\frac{\text{x}}{\text{y}}$ we get, $\frac{\text{x}^2}{\text{y}^2}\Big[\log(\text{v})-\frac{1}2\Big]+\log\text{y}^2=\text{C}$
View full question & answer→Question 1165 Marks
Solve the following initial value problems:
$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$
Answer$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$ $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{y}^2}{2\text{xy}}$ It is a homogeneous equation. Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ So,$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}-\text{v}$ $\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2-2\text{v}^2}{2\text{v}}$ $\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}$ $\int\frac{2\text{v}}{1-\text{v}^2}=\int\frac{\text{dx}}{\text{x}}$ $\log|1-\text{v}^2|=-\log|\text{x}|+\log|\text{C}|$ $\log|1-\text{v}^2|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$ $\Big|\frac{\text{x}^2-\text{y}^2}{\text{x}^2}\Big|=\Big|\frac{\text{C}}{\text{x}}\Big|$ $|\text{x}^2-\text{y}^2|=|\text{Cx}|\ \dots(\text{i})$ Put y = 0, x = 1 1 - 0 = C C = 1 Put the value of C in equation (i), $|\text{x}^2-\text{y}^2|=|\text{x}|$ $(\text{x}^2-\text{y}^2)^2=\text{x}^2$
View full question & answer→Question 1175 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$
$\Rightarrow\text{dy}=\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$
$\Rightarrow\text{dy}=\tan^2\frac{\text{x}}{2}$
$\Rightarrow\text{dy}=\Big(\tan^2\frac{\text{x}}{2}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\tan^2\frac{\text{x}}{2}\Big)\text{dx}$
$\Rightarrow\int\text{dy}=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$\Rightarrow\text{y}=2\tan\frac{\text{x}}{2}-\text{x}+\text{C}$
so, $\Rightarrow\text{y}=2\tan\frac{\text{x}}{2}-\text{x}+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$, where $\text{x}\in\text{R}$ is the solution o the given differential equation.
View full question & answer→Question 1185 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=4\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}4\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\int\text{e}^{2\text{x}}\text{dx}-4\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{2\text{x}}\text{dx}\Big]\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\frac{\text{e}^{2\text{x}}}2-4\times\frac{1}2\int\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-4\times\frac{1}4\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$ is the required solution.
View full question & answer→Question 1195 Marks
Solve the following differential equations:$(\text{xy}^2+2\text{x})\text{dx}+(\text{x}^2\text{y+2y})\text{dy}=0$
AnswerWe have,$(\text{xy}^2+2\text{x})\text{dx}+(\text{x}^2\text{y}+2\text{y})\text{dy}=0$
$\Rightarrow\text{x(y}^2+2)\text{dx+y}(\text{x}^2+2)\text{dy}=0$
$\Rightarrow\text{x(y}^2+2)\text{dx}=-\text{y}(\text{x}^2+2)\text{dy}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+2)}\text{dx}=-\frac{\text{y}}{(\text{y}^2+2)}\text{dy}$
Integration both sides, we get
$\int\frac{\text{x}}{\text{x}^2+2}\text{dx}=-\int\frac{\text{y}}{\text{y}^2+2}\text{dy}$
$\Rightarrow\frac{1}{2}\int\frac{2\text{x}}{\text{x}^2+2}\text{dx}=-\frac{1}{2}\frac{2\text{y}}{\text{y}^2+2}\text{dy}$
$\Rightarrow\frac{1}{2}\log|\text{x}^2+2|=-\frac{1}{2}\log|\text{y}^2+2|+\log\text{C}$
$\Rightarrow\frac{1}{2}\log|\text{x}^2+2|+\frac{1}{2}\log|\text{y}^2+2|=\log\text{C}$
$\Rightarrow\log|\text{x}^2+2|+\log|\text{y}^2+2|=2\log\text{C}$
$\Rightarrow\log\big(|\text{x}^2+2||\text{y}^2+2|\big)=\log\text{C}^2$
$\Rightarrow\big(|\text{x}^2+2||\text{y}^2+2|\big)=\text{C}^2$
$\Rightarrow(\text{x}^2+2)(\text{y}^2+2)=\text{K}$
$\Rightarrow\text{y}^2+2=\frac{\text{K}}{\text{x}^2+2}$
View full question & answer→Question 1205 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x + y}}$
Answer$\frac{\text{dy}}{\text{dx}}+1 = \text{e}^\text{x+y} .....(1)$
Let $\text{ x}+\text{y} = \text{t}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}}$
Substituting the value of $\text{x + y = t}$ and $1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}} (1),$ we get
$\frac{\text{dt}}{\text{dx}} = \text{e}^1$
$\Rightarrow \text{e}^{-1}\text{dt} = \text{dx}$
$\Rightarrow -\text{e}^{-1} = \text{x}+\text{C}$
$\Rightarrow -\text{e}^{-(\text{x+y})} = \text{x} +\text{C}$ $[\therefore \text{t} = \text{x} + \text{y}]$
View full question & answer→Question 1215 Marks
Form the differential equation corresponding to $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2$ by eliminating a and b.
View full question & answer→Question 1225 Marks
Solve the following differential equation:
$3\text{x}^2\text{dy}=(3\text{xy}+\text{y}^2)\text{dx}$
AnswerWe have,
$3\text{x}^2\text{dy}=(3\text{xy}+\text{y}^2)\text{dx}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3\text{xy}+\text{y}^2}{3\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{3\text{vx}^2+\text{v}^2\text{x}^2}{3\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{3\text{v}+\text{v}^2}{3}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2}3$
$\Rightarrow\ \frac{3}{\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$3\int\frac{1}{\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -3\times\frac{1}{\text{v}}=\log|\text{x}|+\text{C}$
$\Rightarrow\ -\frac{3}{\text{v}}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \frac{-3\text{x}}{\text{y}}=\log|\text{x}|+\text{C}$
Hence, $\frac{-3\text{x}}{\text{y}}=\log|\text{x}|+\text{C}$ is the required solution.
View full question & answer→Question 1235 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=6\text{e}^{\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=6\text{e}^{2\text{x}}\text{e}^\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=6\text{e}^{3\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=6\int\text{e}^{3\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=6\frac{\text{e}^{3\text{x}}}3+\text{C}$
$\Rightarrow\ \text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$
Hence, $\text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$ is the required solution.
View full question & answer→Question 1245 Marks
Solve the following differential equation:
$\Big(1+\text{e}^{\frac{\text{x}}{\text{y}}}\Big)\text{dx}+\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)\text{dy}=0$
AnswerWe have,
$\Big(1+\text{e}^{\frac{\text{x}}{\text{y}}}\Big)\text{dx}+\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)\text{dy}=0$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)}{1+\text{e}^{\frac{\text{x}}{\text{y}}}}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dx}}{\text{dy}}=\text{v + y}\frac{\text{dv}}{\text{dy}}$, we get
$\text{v + y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{e}^{\text{v}}(1-\text{v})}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{e}^{\text{v}}(1-\text{v})}{1+\text{e}^{\text{v}}}-\text{v}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{-\text{e}^{\text{v}}+\text{e}^{\text{v}}\text{v}-\text{v}-\text{v}\text{e}^{\text{v}}}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{v}+\text{e}^{\text{v}}}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \frac{1+\text{e}^{\text{v}}}{\text{v}+\text{e}^{\text{v}}}\text{dv}=-\frac{1}{\text{y}}\text{dy}$
Integrating both sides, we get
$\int\frac{1+\text{e}^{\text{v}}}{\text{v}+\text{e}^{\text{v}}}\text{dv}=-\int\frac{1}{\text{y}}\text{dy}$
$\Rightarrow\ \log|\text{v}+\text{e}^{\text{v}}|=-\log|\text{y}|+\log\text{C}$
$\Rightarrow\ |\text{v}+\text{e}^{\text{v}}|=\Big|\frac{\text{C}}{\text{y}}\Big|$
$\Rightarrow\ \text{v}+\text{e}^{\text{v}}=\frac{\text{C}}{\text{y}}$
Putting $\text{v}=\frac{\text{x}}{\text{y}}$, we get
$\frac{\text{x}}{\text{y}}+\text{e}^{\frac{\text{x}}{\text{y}}}=\frac{\text{C}}{\text{y}}$
$\Rightarrow\ \text{x}+\text{ye}^{\frac{\text{x}}{\text{y}}}=\text{C}$
Hence, $\text{x}+\text{ye}^{\frac{\text{x}}{\text{y}}}=\text{C}$ is the required solution.
View full question & answer→Question 1255 Marks
Show that $\text{y}=\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
AnswerWe have,
$\text{y}=\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})\ ...(1)$
$\frac{\text{dy}}{\text{dx}}\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})\ ...(1)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]\ ...(2)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]+\text{e}^\text{x}[-(\text{A}-\text{B})\cos\text{x}]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]$
$2\text{y}+\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence, $\text{y}=\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$ is the solution to the given differential equation.
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
View full question & answer→Question 1265 Marks
Find the differential equation of the family of curve $\text{y}=\text{Ae}^\text{2x}+\text{Be}^{-2\text{x}},$ where A and B are arbitrary constants.
AnswerThe equation of family of curves is
$\text{y}=\text{Ae}^\text{2x}+\text{Be}^{-2\text{x}}\ ...(1)$
where A and B is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of secound order.
Differentiating equation (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{Ae}^{2\text{x}}-2\text{Be}^{-2\text{x}}\ ...(2)$
Differentiating equation (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}}=4\text{Ae}^{2\text{x}}-2\text{Be}^{-2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4(\text{Ae}^{2\text{x}}+\text{Be}^{-2\text{x}})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{y}$
It is the required differential equation.
View full question & answer→Question 1275 Marks
Solve the following differential equations:
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
AnswerWe have,
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Integrating both sides, we get
$\int\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Putting $1+y^2=t^2$ and $1+x^2=u^2$, we get
2y dy = 2t dt and 2x dx = 2u du
$\Rightarrow\text{dy}=\frac{\text{t}}{\text{y}}\ \text{dt}\ \text{and}\ \text{dx}=\frac{\text{u}}{\text{x}}\ \text{du}$
$\therefore\int\frac{\text{t}^2}{\text{y}^2}\ \text{dt}=-\int\frac{\text{u}^2}{\text{x}^2}\ \text{dx}$
$\Rightarrow\int\frac{\text{t}^2}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\int\frac{\text{t}^2-1+1}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2-1+1}{\text{u}^2-1}\ \text{du}$
$\int\text{dt}+\int\frac{1}{\text{t}^2-1}\ \text{dt}=-\int\text{du}-\int\frac{1}{\text{u}^2-1}\ \text{du}$
Substituting t by $\sqrt{1+\text{y}^2}$ and u by $\sqrt{1+\text{x}^2}$
$\sqrt{1+\text{y}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=-\sqrt{1+\text{x}^2}\\-\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|+\text{C}$
$$$\Rightarrow\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$
Hence, $\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$ is the required solution.
View full question & answer→Question 1285 Marks
Solve the following differential equation
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
AnswerWe have,
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\text{dy}=\frac{1}{\text{x}^2+1}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\frac{1}{\text{x}^2+1}\Big)\text{dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$
so, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
View full question & answer→Question 1295 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$ This is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=1,\text{Q}=\text{e}^{-2\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int2\text{dx}}$ $=\text{e}^{\text{x}}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\times\text{e}^{\text{x}}=\int\text{e}^{-2\text{x}}\times\text{e}^{\text{x}}\text{dx + C}$ $=\int\text{e}^{-\text{x}}+\text{C}$ $\text{y}\text{e}^{\text{x}}=\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$ $\text{y}=-\text{e}^{-2\text{x}}+\text{C}\text{e}^{-\text{x}}$
View full question & answer→Question 1305 Marks
Solve the following differential equation:
$(\text{x}+\text{y}+1)\frac{\text{dy}}{\text{dx}} = 1$
AnswerWe have,
$(\text{x}+\text{y}+1)\frac{\text{dy}}{\text{dx}} = 1$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{1}{(\text{x}+\text{y}+1)}$
Let $\text{ x}+\text{y}+1 = \text{v}$
$\therefore1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}-1$
$\therefore\frac{\text{dv}}{\text{dx}}-1=\frac{1}{\text{v}}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{1}{\text{v}}+1$
$\Rightarrow \frac{\text{v}}{\text{v}+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}}{\text{v+1}}\text{dv} = \int \text{dx}$
$\Rightarrow \int \frac{\text{v}+1-1}{\text{v}+1}\text{dv}=\int\text{dx}$
$\Rightarrow\int\Big(1-\frac{1}{\text{v}+1}\Big)\text{dv} = \int\text{dx}$
$\Rightarrow\text{v}-\log|\text{v}+1|=\text{x + K}$
$\Rightarrow\text{x + y}+1-\log|\text{x + y}+1+1|=\text{x + K}$
$\Rightarrow\text{y}-\log|\text{x + y}+2|=\text{K}-1$
$\Rightarrow\text{y}-\log|\text{x + y}+2|=\text{C}_1$ $(\text{C}_1 = \text{K}-1)$
$\Rightarrow\text{y}-\text{C}_1=\log|\text{x + y}+2|$
$\Rightarrow\text{e}^{\text{y}-\text{C}_1}=\text{x + y}+2$
$\Rightarrow \frac{\text{e}^\text{y}}{\text{e}^{c_1}} = \text{x}+\text{y}+2$
$\Rightarrow \text{e}^{-\text{c}_1}\text{e}^{\text{y}} = \text{x}+\text{y}+2$
$\Rightarrow \text{Ce}^\text{y} = \text{x}+\text{y}+2$ $(\text{C} = \text{e}^{-\text{c}_1})$
$\Rightarrow \text{x} = \text{Ce}^\text{y} - \text{y}-2$
View full question & answer→Question 1315 Marks
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 year, and the present population is 100000, when will the city have a population of 500000?
AnswerLet the origional population be N and the population at any time t be P.
Given: $\frac{\text{dP}}{\text{dt}}\propto\text{P}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=\text{aP}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\text{a}\text{dt}$
$\Rightarrow\log|\text{P}|=\text{at}+\text{C}\ ...(\text{i})$
Now,
$\text{P}=\text{N}$ at $\text{t}=0$
Putting $\text{P}=\text{N}$ at $\text{t}=0$ in (i), we get
$\log|\text{N}|=\text{C}$
Putting $\text{C}=\log|\text{N}|$ in (i), we get
$\log|\text{P}|=\text{at}+\log|\text{N}|$
$\Rightarrow \text{log}|\frac{\text{P}}{\text{N}}|=\text{at}\ ...(\text{ii})$
According to the question,
$\log|\frac{2\text{N}}{\text{N}}|=25\text{a}$
$\Rightarrow\ \text{a}=\frac{1}{25}\log|2|$
$=\frac{1}{25}\times0.6931=0.0277$
Putting $\text{a}=0.0277$ in (ii), we get
$\log|\frac{\text{P}}{\text{N}}|=0.0277 \text{t}\ ...(\text{iii})$
For $\text{P}=500000$ and $\text{N}=100000$
$\log|\frac{500000}{100000}|=0.0277 \text{t}$
$\Rightarrow \text{t}=\frac{\log\ 5}{0.0277}=\frac{1.609}{0.0277}$
$=58.08\ \text{year}$
View full question & answer→Question 1325 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$
Answerwe have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$
$\Rightarrow(\sin\text{y+y}\cos\text{y})\text{dy = x}(2\log\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int(\sin\text{y+y}\cos\text{y})\text{dy}=\int\text{x}(2\log\text{x}+1)\text{dx}$
$\Rightarrow\int\sin\text{y dy}+\int\text{y}\cos\text{y dy }=2\int\text{x}\log\text{x dx}+\int\text{x dx}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\int\cos\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}\text{(y)}\int\cos\text{y dy}\Big\}\text{dy}\Big]\\=2\Big[\log\text{x}\int\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x dx}\Big\}\text{dx}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\sin\text{y}-\int\sin\text{y dy}\Big]=2\Big[\log\text{x}\times\frac{\text{x}^2}{2}-\int\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y+y}\sin\text{y}+\cos\text{y}=\text{x}^2\log\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow\text{y}\sin\text{y}=\text{x}^2\log\text{x + C}$
Hence, $\text{y}\sin\text{y = x}^2\log\text{x + C}$ is the required solution.
View full question & answer→Question 1335 Marks
Solve the following differential equations:$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
Answer$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\Big\{\frac{(\text{x + y})+(\text{x}-\text{y})}{2}\Big\}\cos\Big\{\frac{(\text{x + y})-(\text{x}-\text{y})}{2}\Big\}$
$=2\sin\Big(\frac{\text{x + y + x}-\text{y}}{2}\Big)\cos\Big(\frac{\text{x + y}-\text{ x}+\text{y}}{2}\Big)$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\text{x}\cos\text{y}$
$\frac{\tan\text{y}}{\cos\text{y}}\text{dy}=2\sin\text{x dx}$
$\int\sec\text{y}\tan\text{y dy}=2\int\sin\text{x dx}$
$\sec\text{y}=-2\cos\text{x + C}$
$\sec\text{y}+2\cos\text{x = C}$
View full question & answer→Question 1345 Marks
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
AnswerGiven,
Slope of tangent at $(x, y) = x^2$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{2}$
$\text{dy}=\text{x}^{2}\text{dx}$
$\int \text{dy}=\int\text{x}^{2}\text{dx}$
$\text{y}=\frac{\text{x}^{3}}{3}+\text{C}\ ...(\text{i})$
It is passing through (-1, 1)
$1=\frac{(-1)}{3}+\text{C}$
$1=-\frac{1}{3}+\text{C}$
$\text{C}=\frac{4}{3}$
Put is equation,
$\text{y}=\frac{\text{x}^{3}}{3}+\frac{4}{3}$
$3\text{y}=\text{x}^{3}+{4}$
View full question & answer→Question 1355 Marks
Solve the following differential equation:
$(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x + 2y}$
AnswerHere, $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x + 2y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2\text{y}}{\text{x}-\text{y}}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}+2\text{vx}}{\text{x}-\text{vx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1+\text{v}}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v + v}^2}{1-\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v + v}^2}{1-\text{v}}$
$\frac{1-\text{v}}{\text{v}^2+\text{v}+1}\text{dv}=\frac{\text{dx}}{\text{x}}$
$-\frac{\text{v}-1}{\text{v}^2+\text{v}+1}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{1}2\times\frac{2\text{v}-2}{\text{v}^2+\text{v}+1}\text{dv}=\frac{-\text{dx}}{\text{x}}$
$\int\frac{(2\text{v}+1)-3}{\text{v}^2+\text{v}+1}\text{dv}=-\int\frac{2\text{dx}}{\text{x}}$
$\int\frac{2\text{v}+1}{\text{v}^2+\text{v}+1}\text{dv}-\int\frac{3}{\text{v}^2+2\text{v}\big(\frac{1}2\big)+\big(\frac{1}2\big)^2-\big(\frac{1}2\big)^2+1}=-2\int\frac{\text{dx}}{\text{x}}$
$\int\frac{2\text{v}+1}{\text{v}^2+\text{v}+1}\text{dv}-\int\frac{3}{\big(\text{v}+\frac{1}2\big)^2+\big(\frac{\sqrt3}2\big)^2}\text{dv}=-2\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^2+\text{v}+1|-3\Big(\frac{2}{\sqrt3}\Big)\tan^{-1}\Bigg(\frac{\text{v}+\frac{1}2}{\frac{\sqrt3}{2}}\Bigg)=-2\log|\text{x}|+\text{C}$
$\log|\text{y}^2+\text{xy}+\text{x}^2|=2\sqrt3\tan^{-1}\Big(\frac{2\text{y + x}}{\text{x}\sqrt3}\Big)+\text{C}$
View full question & answer→Question 1365 Marks
If y(x) is a solution of the different equation $\Big(\frac{2+\sin\text{x}}{1+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=-\cos\text{x}$ and $\text{y}(0)=1,$ then find the value of $\text{y}\Big(\frac{\pi}{2}\Big).$
AnswerConsider the given equation
$\Big(\frac{2+\sin\text{x}}{1+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=-\cos\text{x}$
$\Rightarrow\frac{\text{dy}}{(1+\text{y})}=\frac{-\cos\text{x dx}}{(2+\sin\text{x})}$
Integrating both the sides,
$\Rightarrow\int\frac{\text{dy}}{(1+\text{y})}=\int\frac{-\cos\text{x dx}}{(2+\sin\text{x})}$
$\Rightarrow\log(1+\text{y})=-\log(2+\sin\text{x})+\log\text{C}$
$\Rightarrow\log(1+\text{y})+\log(2+\sin\text{x})=\log\text{C}$
$\Rightarrow\log(1+\text{y})(2+\sin)\text{x}=\log\text{C}$
$\Rightarrow(1+\text{y})(2+\sin\text{x})=\text{C}...(1)$
Given that $\text{y}(0)=1$
$\Rightarrow(1+1)(2+\sin0)=\text{C}$
$\Rightarrow\text{C}=4$
Substituting the value of C in equation (1) we have,
$\Rightarrow(1+\text{y})(2+\sin\text{x})=4$
$\Rightarrow(1+\text{y})=\frac{4}{(2+\sin\text{x})}$
$\Rightarrow\text{y}=\frac{4}{(2+\sin\text{x})}-1...(2)$
We need to find the value of $\text{y}\Big(\frac{\pi}{2}\Big)$
Substituting the value of $\text{x}=\frac{\pi}{2}$ in equation (2), we get,
$\text{y}=\frac{4}{\Big(2+\sin\frac{\pi}{2}\Big)}-1$
$\Rightarrow\text{y}=\frac{4}{(2+1)}-1$
$\Rightarrow\text{y}=\frac{4}{3}-1$
$\Rightarrow\text{y}=\frac{1}{3}$
Note: Answer given in the book is incorrect.
View full question & answer→Question 1375 Marks
Find the particular solution of $\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1,$ that $\text{y}=3,$ when $\text{x}=0.$
Answer$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$
$\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1),\text{y}=3$ at $\text{x}=0$
$\int\text{dy}=\int\log(\text{x}+1)\text{dx}$
$\text{y}=\log|\text{x}+1|\times\int1\times\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx}+\text{C}$
Using integration by parts
$\text{y = x}\log|\text{x}+1|-\int\frac{\text{x}}{\text{x}+1}\text{dx}+\text{C}$
$\text{y = x}\log|\text{x}+1|-\Big(\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}\Big)+\text{C}$
$=\text{x}\log|\text{x}+1|-(\text{x}-\log|\text{x}+1|)+\text{C}$
$\text{y = x}\log|\text{x}+1|-\text{x}+\log|\text{x}+1|+\text{C}$
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x + C}$
Put $\text{y}=3$ and $\text{x}=0$
$3=0-0+\text{C}$
$\text{C}=3$
Put $\text{C}=3$ in equation (1),
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x}+3$
View full question & answer→Question 1385 Marks
Solve the following differential equation
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
AnswerWe have
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$\Rightarrow(\text{x}-1)\text{dy}=2\text{xy dx}$
$\Rightarrow\frac{2\text{x}}{(\text{x}-1)}\ \text{dx}=\frac{1}{\text{y}}\text{ dy}$
Integrating both sides, we get
$2\int\frac{\text{x}}{(\text{x}-1)}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\int\frac{\text{x}-1+1}{\text{x}-1}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dx}$
$\Rightarrow2\int\text{dx}+2\int\frac{1}{\text{x}-1}\text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\text{x}+2\log|\text{x}-1|=\log|\text{y}|+\text{C}$
View full question & answer→Question 1395 Marks
Show that $\text{y}=\text{A}\cos2\text{x}+\text{B}\sin2\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
AnswerWe have,
$\text{y}=\text{A}\cos2\text{x}+\text{B}\sin2\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{A}\sin2\text{x}-\text{B}\cos2\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{A}\cos2\text{x}+4\text{B}\sin2\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4(\text{A}\cos2\text{x}+4\text{B}\sin2\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 1405 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
AnswerHere, $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$
It is a homogeneous equation
Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\cos^2\big(\frac{\text{vx}}{\text{x}}\big)}{\text{x}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\cos^2\text{v}$
$\frac{\text{dv}}{\cos^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\int\sec^2\text{vdv}=-\int\frac{\text{dx}}{\text{x}}$
$\tan\text{v}=-\log|\text{x}|+\log|\text{C}|$
$\tan\frac{\text{y}}{\text{x}}=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
View full question & answer→Question 1415 Marks
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$ given that $\text{y}=1.$ when $\text{x}=0.$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=-4\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=-4\int\text{x dx}$
$\Rightarrow-\frac{1}{\text{y}}=-4\times\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow-\frac{1}{\text{y}}=-2\text{x}^2+\text{C}...(1)$
It is given that at $\text{x}=0,\text{y}=1.$
Substituting the valuse of x and y in (1), we get
$\text{C}=-1$
Therefore, substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=-2\text{x}^2-1$
$\Rightarrow\text{y}=\frac{1}{2\text{x}^2+1}$
Hence, $\text{y}=\frac{1}{2\text{x}^2+1}$ is the required solution.
View full question & answer→Question 1425 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{e}^{-2\text{x}}\sin\text{x},\text{ y}(0)=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{e}^{-2\text{x}}\sin\text{x}\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=2$ and $\text{Q}=\text{e}^{-2\text{x}}\sin\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\text{e}^-{2\text{x}}\sin\text{x}$
$\Rightarrow\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\int\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+\text{C}\ ....(\text{ii})$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^0=-\cos0+\text{C}$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+1$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$
Hence, $\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$ is the required solution.
View full question & answer→Question 1435 Marks
Solve the differential equation $(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$
AnswerWe have,
$(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dy}}=\frac{\text{y}+3{\text{x}^{\text{2}}}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}=3{\text{x}}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=3\text{x}$
$\therefore \ \text{I}.\text{F}. = \text{e}^{\int{\text{P}\text{dx}}}$
$ =\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}\Big)=\frac{1}{\text{x}}3\text{x}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^{2}}\text{y}=3$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=3\int\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$
Hence, $\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$ is the required solution.
View full question & answer→Question 1445 Marks
Solve the following differential equations:$\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$
AnswerWe have,
$\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$
$\Rightarrow\text{y}\sqrt{1-\text{x}^2}\text{dy}=-\text{x}\sqrt{1-\text{y}^2}\text{dx}$
$\Rightarrow\frac{\text{y}}{\sqrt{1-\text{y}^2}}\text{dy}=-\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\sqrt{1-\text{y}^2}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Substituting $1-\text{y}^2=\text{t}$ and $1-\text{x}^2=\text{u},$ we get
$-2\text{y dy = dt}$ and $-2\text{x dy = du}$
$\therefore\frac{-1}{2}\int\frac{1}{\sqrt{\text{t}}}\text{dt}=\frac{1}{2}\int\frac{1}{\sqrt{\text{u}}}\text{du}$
$\Rightarrow-\text{t}^{\frac{1}{2}}=\text{u}^{\frac{1}{2}}+\text{K}$
$\Rightarrow\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=-\text{K}$
$\Rightarrow\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{C}$ (where, C = K)
Hence, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{C}$ is the required solution.
View full question & answer→Question 1455 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{e}^{-\text{y}}\sec^2\text{y dy}=\text{dx}+\text{x dy}$
AnswerWe have,
$\text{e}^{-\text{y}}\sec^2\text{y dy}=\text{dx}+\text{x dy}$
$\Rightarrow\text{dx}=\text{e}^{-\text{y}}\sec^2\text{y dy}-\text{x dy}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\text{e}^{-\text{y}}\sec^2\text{y}-\text{x}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\text{x}=\text{e}^{-\text{y}}\sec^2\text{y}\ ...(1)$
Clearly, it is a linear differential equation of tyhe form
$\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}$
Where
$\text{P}=1$
$\text{Q}=\text{e}^{-\text{y}}\sec^2\text{y}$
$\therefore\ \text{I.F.}=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\text{dy}}$
$=\text{e}^{\text{y}}$
Multiplying both sides of (1) by ey, we get
$\text{e}^{\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\text{x}\Big)=\text{e}^{\text{y}}\text{e}^{-\text{y}}\sec^2\text{y}$
$\Rightarrow\text{e}^{\text{y}}\frac{\text{dx}}{\text{dy}}+\text{e}^{\text{y}}\text{x}=\sec^2\text{y}$
Integrating both sides with respect to y, we get
$\text{e}^{\text{y}}\text{x}=\int\sec^2\text{y dy}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}\text{x}=\tan\text{y}+\text{C}$
$\Rightarrow\text{x}=(\tan\text{y}+\text{C})\text{e}^{-\text{y}}$
Hence, $\text{x}=(\tan\text{y}+\text{C})\text{e}^{-\text{y}}$ is the required solution.
View full question & answer→Question 1465 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0,\text{y}(0)=2,\text{y}'(0)=0$Function $\text{y}=\text{e}^\text{x}+\text{e}^{-\text{x}}$
AnswerWe have $\text{y}=\text{e}^{\text{x}}+\text{e}^{\text{-x}} ...(1)$ Differentiating both sides of (1) with respect to $x,$ we get $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}-\text{e}^{\text{x}} ...(2)$ Differentiating both sides of $(2)$ with respect to $x,$ we get $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{e}^{\text{x}}+\text{e}^{\text{-x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y}$ [Using (1)]$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
It is the given differential equation.Therefore, $y = e^x+ e^{-x} $ satisfies the given differential equation.
Also, when $x = 0; = e^0+ e^0= 1 + 1, i.e. y(0) = 2.$
And, when $x = 0; y_1= e^0- e^0= 1 - 1, i.e. y'(0) = 0$
Hence, $y = e^x+ e^{-x}$ ^is the solution to the given initial value problem.
View full question & answer→Question 1475 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{y}+\text{x}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{y}+\text{x}}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}}{2\text{vx + x}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1}{2\text{v}+1}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1}{2\text{v}+1}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}^2-\text{v}}{2\text{v}+1}$
$\Rightarrow\ \frac{2\text{v}+1}{1-2\text{v}^2-\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{2\text{v}+1}{1-2\text{v}^2-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{2\text{v}^2+\text{v}-1}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{2\text{v}(\text{v}+1)-1(\text{v}+1)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$
Solving left hand side integral of (1), we get
Using partial fraction,
Let $\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}=\frac{\text{A}}{(2\text{v}-1)}+\frac{\text{B}}{(\text{v}+1)}$
$\therefore\ \text{A}+2\text{B}=2\ \dots(2)$
And $\text{A}-\text{B}=1\ \dots(3)$
Solving (2) and (3), we get
$\text{A}=\frac{4}3$ and $\text{B}=\frac{1}3$
$\therefore\ \int\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}\text{dy}=\frac{4}3\int\frac{1}{2\text{v}-1}\text{dv}+\frac{1}3\int\frac{1}{\text{v}+1}\text{dv}$
$=\frac{4}{3\times2}\log|2\text{v}-1|+\frac{1}3\log|\text{v}+1|+\log\text{C}$
From (1), we get
$\frac{2}3\log|2\text{v}-1|+\frac{1}3|\text{v}+1|+\log\text{C}=-\log|\text{x}|+\log\text{C}_1$
$\Rightarrow\ \log\Big\{\big|(2\text{v}-1)^2\big||\text{v}+1|\Big\}=-3\log|\text{x}|+\log\text{C}_2$
$\Rightarrow\ \log\Big\{\big|(2\text{v}-1)^2\big||\text{v}+1|\Big\}=\log\Big|\frac{\text{C}_2^3}{\text{x}^3}\Big|$
$\Rightarrow\ (2\text{v}-1)^2(\text{v}+1)=\frac{\text{C}_2^3}{\text{x}^3}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \Big(\frac{2\text{y}-\text{x}}{\text{x}}\Big)^2\Big(\frac{\text{y}+\text{x}}{\text{x}}\Big)=\frac{\text{C}_2^3}{\text{x}^3}$
$\Rightarrow\ (\text{x + y})(2\text{y}-\text{x})^2=\text{K}$
View full question & answer→Question 1485 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{xy}^2$ when $\text{y}=0,\text{x}=0$
Answer$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{xy}^2$
$\frac{\text{dy}}{\text{dx}}=(1+\text{x})(1+\text{y}^2)$
$\frac{1}{(1+\text{y}^2)}\text{dy}=(1+\text{x})\text{dx}$
Integrating on both the sides we get
$\int\frac{1}{(1+\text{y}^2)}\text{dy}=\int(1+\text{x})\text{dx}$
$\tan^{-1}\text{y = x}+\frac{\text{x}^2}{2}+\text{C}...(1)$
Put $\text{y}=0,\text{x}=0$ then
$\tan^{-1}0=0+0+\text{C}$
$\text{C}=0$
From (1) we have
$\tan^{-1}\text{y = x}+\frac{\text{x}^2}{2}$
$\text{y}=\tan\Big(\text{x}+\frac{\text{x}^2}{2}\Big)$
View full question & answer→Question 1495 Marks
Solve the following differential equations:
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
AnswerWe have,
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
$\Rightarrow\ \text{x}^2\text{dy}=-\text{y}(\text{x + y})\text{dx}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}(\text{x + y})}{\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{vx}(\text{x + vx})}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=-\text{v}(1+\text{v})$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}-\text{v}-\text{v}^2$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-(\text{v}^2+2\text{v})$
$\Rightarrow\ \frac{\text{dv}}{(\text{v}^2+2\text{v})}=-\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{v}(\text{v}+2)}=-\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\int\frac{\text{dv}}{\text{v}(\text{v}+2)}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\int\Big[\frac{1}{\text{v}}-\frac{1}{\text{v}+2}\Big]\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\Big[\int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{\text{v}+2}\text{dv}\Big]=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\big[\log|\text{v}|-\log|\text{v}+2|\big]=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \frac{1}2\log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\Rightarrow\ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}^2}{\text{x}^2}\Big|$
$\Rightarrow\ \frac{\text{v}}{\text{v}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\frac{\text{y}}{\text{x}}}{\frac{\text{y}}{\text{x}}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\text{y}}{\text{y}+2\text{x}}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}^2\text{y}=\text{C}^2(\text{y +2x})$
$\Rightarrow\ \text{x}^2\text{y}=\text{K}(\text{y +2x})$ (Where, $K = C^2$)
View full question & answer→Question 1505 Marks
Solve the following differential equation:
$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$
Answer$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2\text{y}}{2\text{x}-\text{y}}$
This is a homogeneous differential equation. Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}+2\text{vx}}{2\text{x}-\text{vx}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{2-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2-\text{v}}$ $\Rightarrow\ \frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$ Integrating both sides, we get $\int\frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ 2\tan^{-1}\text{v}-\frac{1}2\log|1+\text{v}^2|=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log|\text{x}|+\log\text{C}+\log\Big|(1+\text{v}^2)^{\frac{1}2}\Big|$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log\Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|$ $\Rightarrow\ \Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|=\text{e}^{2\tan^{-1}\text{v}}$ Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get $\Rightarrow\ \Bigg|\text{Cx}\sqrt{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\Bigg|=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ $\Rightarrow\ \text{C}\sqrt{\text{x}^2+\text{y}^2}=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ Hence, $\sqrt{\text{x}^2+\text{y}^2}=\text{Ke}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ is the required solution.
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