Question 515 Marks
Solve the following differential equation
$\text{xy}(\text{y}+1)\text{dy}=(\text{x}^2+1)\text{dx}$
AnswerWe have $\text{xy}(\text{y}+1)\text{dy}=(\text{x}^2+1)\text{dx}$$\Rightarrow\{\text{y}(\text{y}+1)\}\text{dy}=\frac{\text{x}^2+1}{\text{x}}\ \text{dx}$
$\Rightarrow(\text{y}^2+\text{y})\text{dy}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\text{dx}$
Integrating both sides, we get
$\int(\text{y}^2+\text{y})\text{dy}=\int\Big(\text{x}+\frac{1}{\text{x}}\Big)\text{dx}$ $=\int\text{y}^2\text{dy}+\int\text{y dy}=\int\text{x dx}+\int\frac{1}{\text{x}}\text{ dx}$ $\Rightarrow\frac{\text{y}^3}{3}+\frac{\text{y}^3}{2}=\frac{\text{x}^2}{2}+\log|\text{x}|+\text{C}$ Hence, $\frac{\text{y}^3}{3}+\frac{\text{y}^3}{2}=\frac{\text{x}^2}{2}+\log|\text{x}|+\text{C}$ is the required solution.
View full question & answer→Question 525 Marks
Solve the following differential equation:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}-2\text{xy}=(\text{x}^2+2)(\text{x}^2+1)$
AnswerHere, $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}-2\text{xy}=(\text{x}^2+2)(\text{x}^2+1)$ $\frac{\text{dy}}{\text{dx}}-\frac{2\text{x}}{\text{x}^2+1}\text{y}=(\text{x}^2+2)$ It is a linear differential equation. Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=-\frac{2\text{x}}{\text{x}^2+1},\text{Q}=\text{x}^2+2$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{-\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}}$ $=\text{e}^{-\log|\text{x}^2+1|}$ $=\frac{1}{(\text{x}^2+1)}$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}\Big(\frac{1}{\text{x}^2+1}\Big)=\int\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx + C}$ $=\int\Big(1+\frac{1}{\text{x}^2+1}\Big)\text{dx}+\text{C}$ $\frac{\text{y}}{(\text{x}^2+1)}=\text{x}+\tan^{-1}\text{x + C}$ $\text{y}=(\text{x}^2+1)(\text{x}+\tan^{-1}\text{x + C})$
View full question & answer→Question 535 Marks
Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2$
AnswerThe equation of the family of curves is
$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2\ ...(1)$
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{x}-2\text{a}+4\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$
Now, from (1), we get
$2\text{a}=\frac{\text{x}^2-2\text{y}^2}{\text{x}}\ ...(3)$
From (2) and (3), we get
$2\text{x}-\frac{\text{x}^2+2\text{y}^2}{\text{x}}+4\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2\text{x}^2-\text{x}^2-2\text{y}^2+4\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}+\text{x}^2-2\text{y}^2=0$
$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}=2\text{y}^2-\text{x}^2$
It is the required differential equation.
View full question & answer→Question 545 Marks
A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.
AnswerLet $P_0$ be the intial amount and P be the amount at any time t. we have
$\frac{\text{dP}}{\text{dt}}=\frac{8\text{P}}{100}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=\frac{2\text{P}}{25}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\frac{2}{25}\text{dt}$
Intergrating both sides with respect to t, We get
$\log\text{P}=\frac{2}{25}\text{t}+\text{C}\ ...(\text{i})$
Now,
$\therefore \log\text{P}_{0}=0+\text{C}$
$\Rightarrow \text{C}=\log\text{P}_{0}$
Putting the value of C in (i), we get
$\log\text{P}=\frac{2}{25}\text{t}+\log\text{P}_{0}$
$\Rightarrow \log\frac{\text{P}}{\text{P}_{0}}=\frac{2}{25}\text{t}$
$\Rightarrow \text{e}^\frac{2}{25}\text{t}=\frac{\text{P}}{\text{P}_{0}}$
To find the amount after 1 year, we have
$ \text{e}^\frac{2}{25}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{e}^{0.08}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow 1.0833=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{P}=1.0833\text{P}_{0}$
Percentage increase $=\Big(\frac{\text{P}-\text{P}_{0}}{\text{P}_{0}}\Big)\times100\%$
$=\Big(\frac{1.0833 \text{P}_{0}-\text{P}_{0}}{\text{P}_{0}}\Big)\times100\%$
$=0.0833\times100\%$
$=8.33\%$
View full question & answer→Question 555 Marks
Solve the following initial value problems:
$(\text{xy}-\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0,\text{y}(1)=1$
Answer$(\text{xy}-\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0,\text{y}(1)=1$
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{xvx}-\text{v}^2\text{x}^2}{\text{x}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\text{v}^2-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}^2$
$-\int\frac{1}{\text{v}^2}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$-\Big(-\frac{1}{\text{v}}\Big)=\log|\text{x}|+\text{C}$
$\frac{\text{x}}{\text{y}}=\log|\text{x}|+\text{C}\ \dots(\text{i})$
Put y = 1, x = 1
1 = C
Using equation (1),
$\text{x}=\text{y}\big[\log|\text{x}|+1\big]$
$\text{y}=\frac{\text{x}}{\big[\log|\text{x}|+1\big]}$
View full question & answer→Question 565 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$
AnswerWe have
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{(1+\text{y}^2}{\text{y}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=-\frac{\text{y}}{1+\text{y}^2}$
$\Rightarrow\text{dx}=\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Integrating both sides, we get
$\int\text{dx}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
$\Rightarrow\text{x}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Putting $1 + y^2= t$ we get
2y dy dt
$\therefore\text{x}=-\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|\text{t}|+\text{C}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
Hence, $\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$ is the required solution.
View full question & answer→Question 575 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{ x, y}(0)=1$
Answer$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{ x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan\text{ x dx}$
Integrating both sides, we get
$\int \frac{1}{\text{y}}\text{dy}=\int\tan\text{ x dx}$
$\Rightarrow\log|\text{y}|=\log|\sec\text{x}|+\text{C}...(1)$
We know that at $\text{x}=0$ and $\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=\log|1|+\text{C}$
$\Rightarrow\text{C}=0$
substituting the value of C in (1), we get
$\log|\text{y}|=\log|\sec\text{x}|+0$
$\Rightarrow\text{y}=\sec\text{x}$
Hence, $\text{y}=\sec\text{x},$ where $\text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big),$ is the required solution.
View full question & answer→Question 585 Marks
Solve the following equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}^2-\text{y}$
$\Rightarrow\frac{1}{\text{y}^2-\text{y}}\ \text{dy}=\frac{1}{\text{x}}\ \text{dx}$
integrating both sides, we get
$\int\frac{1}{\text{y}^2-\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\frac{1}{\text{y}(\text{y}-1)}=\text{dy}=\int\frac{1}{\text{x}}\text{dx}\ ...(1)$
Let $\frac{1}{(\text{y}-1)}=\frac{\text{A}}{\text{y}}+\frac{\text{B}}{\text{y}-1}$
$\Rightarrow1=\text{A}(\text{y}-1)+\text{B}(\text{y})$
putting y = 0, we get
1 = -A
⇒ A = -1
putting y = 1, we get
1 = B
$\therefore\frac{1}{\text{y}(\text{y}-1)}=\frac{-1}{\text{y}}+\frac{1}{\text{y}-1}$
$\Rightarrow\int\frac{1}{\text{y}(\text{y}-1)}\text{dy}=\int\frac{-1}{\text{y}}\text{dy}+\int\frac{1}{\text{y}-1}\text{dy}\ ...(2)$
From (1) & (2) , we get
$\int\frac{-1}{\text{y}}\text{dy}+\int\frac{1}{\text{y}-1}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow-\log|\text{y}|+\log|\text{y}-1|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\Big|\frac{\text{y}-1}{\text{y}}\Big|-\log|\text{x}|=\log\text{C}$
$\Rightarrow\log\Big|\frac{\text{y}-1}{\text{xy}}\Big|=\log\text{C}$
$\Rightarrow\frac{\text{y}-1}{\text{xy}}=\text{C}$
$\Rightarrow\text{y}-1=\text{Cxy}$
hence, $\text{y}-1=\text{Cxy}$ is the required solution.
View full question & answer→Question 595 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\tan2\text{x dx}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\sec2\text{x}|+\frac{1}{2}\log\text{C}$
$\Rightarrow\text{y}^2=\text{C}\sec2\text{x}\dots(1)$
It is given that at $\text{x}=0,\text{y}=2.$
$\therefore\text{C}=4$
Substituting the value of C in (1), we get
$\therefore\text{y}^2=\frac{4}{\cos2\text{x}}$
$\Rightarrow\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$
Hence, $\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$ is the required solution.
View full question & answer→Question 605 Marks
verify that $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+\text{a}^2})^2$ is a solution of the differential equation $(\text{a}^2+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=0$
Answer$\text{y}=\log(\text{x}+\sqrt{\text{x}^2+\text{a}^2})^2$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}\frac{1}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})^2}\times2(\text{x}+\sqrt{\text{x}^2+\text{a}^2})\frac{\text{d}}{\text{dx}}(\text{x}+\sqrt{\text{x}^2+\text{a}^2})$
$=\frac{2}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})}\times\Big(1+\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}(2\text{x})\Big)$
$=\frac{2}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})}\Big(\frac{\sqrt{\text{x}2+\text{a}^2}+\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}$
$\sqrt{\text{a}^2+\text{x}^2}\frac{\text{dy}}{\text{dx}}=1$
Again differentiating it with respect to x,
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}^2}{\text{dx}^2}+\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})\frac{\text{dy}}{\text{dx}}=-\text{m}\frac{\text{dy}}{\text{dx}}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}^2}{\text{dx}^2}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}\frac{\text{dy}}{\text{dx}}-\Big(\frac{-\text{e}^{\text{m}^{\cos^{-1}}}\text{m}}{\sqrt{1-\text{x}^2}}\Big)=0$
Using equation (1)
$\sqrt{\text{a}^2+\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{2\text{x}}{2\sqrt{\text{a}^2+\text{x}^2}}\frac{\text{dy}}{\text{dx}}=0$
$(\text{a}^2+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 615 Marks
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
AnswerAccording to the quation,
$\frac{\text{dy}}{\text{dx}}=\text{x}+\text{xy}$
$\frac{\text{dy}}{\text{dx}}-\text{xy}=\text{x}$
Comparing it with $\text{P}=-\text{x}, \text{Q}=\text{x}$
$\text{I.F}=\text{e}^{\int\text{x}\text{dx}}$
$=\text{e}^{-\int\text{(x)}\text{dx}}$
$=\text{e}^{\frac{-\text{x}^{2}}{2}}$
Solution of equation is given by,
$\text{y}(\text{I.F})=\int\text{Q}(\text{I.F})\text{dx}+\text{C}$
$\text{y}\text{e}^{\frac{\text{x}^{2}}{2}}=\int\text{x}(\text{e}^{\frac{\text{x}^{2}}{2}})\text{dx}+\text{C}$
$\text{y}\text{e}^{\frac{\text{x}^{2}}{2}}=\text{I}+\text{C}$
Now,
$\text{I}=\int\text{xe}^{-\frac{\text{x}^{2}}{2}}\text{dx}$
Putting $\frac{-\text{x}^{2}}{2}=\text{t}$, we get
$-\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\text{e}^{\text{t}}\text{dt}$
$\Rightarrow \text{I}=-\text{e}^{\text{t}}$
$\Rightarrow \text{I}=\text{e}^{-\frac{\text{x}^{2}}{2}}$
Since the curve passes through the point (0, 1)
$1\text{e}_{0}=-\text{e}_{0}+\text{C}$
$\text{C}=2$
PUtting the value of C in the curve, we get
$\text{ye}^{\frac{-\text{x}}{2}}=-\text{e}^{\frac{-\text{x}^{2}}{2}}+2$
$\text{y}=-1+2\text{e}^{\frac{\text{x}^{2}}{2}}$
View full question & answer→Question 625 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
Answer$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
This is a homogeneous equation, Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}-\text{v + cosec v}=0$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{cosec v}$
$\frac{\text{dv}}{\text{cosec v}}=\frac{\text{dx}}{\text{x}}$
$\sin\text{v dv}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\int\sin\text{v dv}=\int\frac{\text{dx}}{\text{x}}$
$-\cos\text{v}=\log_{\text{e}}\text{x + C}$
$-\cos\text{v}+\log_{\text{e}}\text{x}=\text{C}$
$\cos\text{v}+\log_{\text{e}}\text{x}=-\text{C}$
$\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=-\text{C}$
As y(1) = 0
$\cos\Big(\frac{0}1\Big)=0+\log_{\text{e}}1=-\text{C}$
$1+0=-\text{C}$
$\Rightarrow\ \text{C}=-1$
$\Rightarrow\ \cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=1$
View full question & answer→Question 635 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
$\Rightarrow\frac{1}{\text{y}^3}\text{dy}=2\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^3}\text{dy}=\int2\text{e}^{\text{x}}\text{dx}$
$\Rightarrow-\frac{1}{2\text{y}^3}=2\text{e}^{\text{x}}+\text{C}...(1)$
Given: at $\text{x}=0,\text{y}=\frac{1}{2}$
Substituting the valuse of x and y in (1), we get
$-\frac{1}{2\times\frac{1}{4}}=2\text{e}^{0}+\text{C}$
$\Rightarrow\text{C}=-2-2$
$\Rightarrow\text{C}=-4$
Substituting the value of C in (1), we get
$\Rightarrow-\frac{1}{2\text{y}^2}=2\text{e}^{\text{x}}-4$
$\Rightarrow\text{y}^{2}(8-4\text{e}^{\text{x}})=1$
Hence, $\text{y}^{\text{x}}(8-4\text{e}^{\text{x}})=1$ is the required solution.
View full question & answer→Question 645 Marks
Solve the following initial value problems:
$\text{xe}^{\frac{\text{y}}{\text{x}}}-\text{y + x}\frac{\text{dy}}{\text{dx}}=0,\text{y(e)}=0$
Answer$\text{xe}^{\frac{\text{y}}{\text{x}}}-\text{y + x}\frac{\text{dy}}{\text{dx}}=0,\text{y(e)}=0$
This is also a homogeneous equation.
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\text{xe}^{\text{v}}-\text{vx + x}\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=0$
$\text{xe}^{\text{v}}-\text{vx + xv}+\text{x}^2\frac{\text{dv}}{\text{dx}}=0$
$\text{xe}^{\text{v}}+\text{x}^2\frac{\text{dv}}{\text{dx}}=0$
$\text{e}^{\text{v}}=-\text{x}\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dx}}{\text{x}}=-\frac{1}{\text{e}^{\text{v}}}\text{dv}$
On integrating both sides we get,
$\int\frac{\text{dx}}{\text{x}}=-\int\frac{1}{\text{e}^{\text{v}}}\text{dv}$
$\log_{\text{e}}\text{x}=-\int\text{e}^{-\text{v}}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\text{e}^{-\frac{\text{y}}{\text{x}}}+\text{C}$ $(\because\ \text{y}=\text{vx})$
As given y(e) = 0
$\log_{\text{e}}\text{e}=\text{e}^{-\frac{0}{\text{e}}}+\text{C}$
$1=1+\text{C}$
$\Rightarrow\ \text{C}=0$
$\therefore \log_{\text{e}}\text{x}=\text{e}^{-\frac{\text{y}}{\text{x}}}$
View full question & answer→Question 655 Marks
Find the differential equation of the family of curve $\text{x}=\text{A}\cos\text{nt}+\text{B}\sin\text{nt},$ where A and B are arbitrary constants.
AnswerThe equation of family of curves is
$\text{x}=\text{A}\cos\text{nt}+\text{B}\sin\text{nt}\ ...(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}}=-\text{An}\sin\text{nt}+\text{Bn}\cos\text{nt}\ ...(2)$
Differentiating equation (1) with respect to x, we get
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{An}^2\cos\text{nt}-\text{Bn}^2\sin\text{nt}$
$\Rightarrow\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{n}^2(\text{A}\cos\text{nt}+\text{B}\sin\text{nt})$
$\Rightarrow\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{n}^2\text{x}$
$\Rightarrow\frac{\text{d}^2\text{x}}{\text{dy}^2}+\text{n}^2\text{x}=0$
It is the required differential equation.
View full question & answer→Question 665 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=\log\text{x}$
AnswerHere, $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=\log\text{x}$ $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ It is a linear differential equation.Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$ $\text{P}=\frac{1}{\text{x}\log\text{x}},\text{Q}=\frac{1}{\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}\text{dx}}$ $=\text{e}^{\log|\log\text{x}|}$ $=\log\text{x}$Solution of the equaion is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}(\log\text{x})=\int\frac{1}{\text{x}}(\log\text{x})\text{dx + C}$ $\text{y}(\log\text{x})=\frac{(\log\text{x})^2}{2}+\text{C}$ $\text{y}=\frac{1}{2}\log\text{x}+\frac{\text{C}}{\log\text{x}},\text{x}>0,\text{x}\neq1$
View full question & answer→Question 675 Marks
Form the differential equation of the family of curve represented by $y^2 = (x - c)^3$
AnswerThe equation of the family of curves is
$y^2 = (x - c)^3 ...(1)$
where $\text{c}\in\text{R}$ is a parameter.
This equation contains only one parameter, so we shall obtain a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=3(\text{x}-\text{c})^2\ ...(2)$
Dividing equation (1) by equation (2), we get
$\frac{\text{y}^2}{2\text{y}\frac{\text{dy}}{\text{dx}}}=\frac{(\text{x}-\text{c})^3}{3(\text{x}-\text{c})^2}$
$\Rightarrow\frac{\text{y}}{2\frac{\text{dy}}{\text{dx}}}=\frac{(\text{x}-\text{c})}{3}$
$\Rightarrow\frac{3\text{y}}{2\frac{\text{dy}}{\text{dx}}}=\text{x}-\text{c}$
$\Rightarrow\text{c}=\text{x}-\frac{3\text{y}}{2\frac{\text{dy}}{\text{dx}}}$
Substituting the value of c in equation (1), we get
$\text{y}^2=\bigg(\text{x}-\text{x}+\frac{3\text{y}}{2\frac{\text{dy}}{\text{dx}}}\bigg)^3$
$\Rightarrow\text{y}^2=\frac{27\text{y}^3}{8\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}$
$\Rightarrow8\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=27\text{y}^3$
$\Rightarrow8\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^3-27\text{y}=0$
It is the required differential equation.
View full question & answer→Question 685 Marks
show that the differential equation of which $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^2}$ is a solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
Answer$\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^2}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=4-\text{ce}^{-\text{x}^{2}}2\text{x}$
$=2\text{x}[2\text{ce}^{-\text{x}^{2}}]$
$=-2\text{x}\big[2\text{x}^2-2+\text{ce}^{\text{x}^{2}}-2\text{x}\big]$
$=-2\text{x}\big[2(\text{x}^2-1)+\text{ce}^{\text{x}^{2}}-2\text{x}^2\big]$
$=2\text{x}[\text{y}-2\text{x}^2]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\text{xy}+4\text{x}^3$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 695 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=1,\text{y}(0)=2$
Function $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$
AnswerWe have,
$\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+\text{e}^{\text{x}}+\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+2\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{xe}^{\text{x}}+\text{e}^{\text{x}}+2\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{xe}^{\text{x}}+3\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2(\text{xe}^{\text{x}}+2\text{e}^{\text{x}})(\text{xe}^{\text{x}}+\text{e}^{\text{x}})$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2\frac{\text{dy}}{\text{dx}}-\text{y}$ [Using(1)and(2)]
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
It is the given differential equation.
Thus, $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$
satisfies the given differential equation.
Also, when $x = 0, y = 0 + 1 = 1, i.e. y(0) = 1$
And, when $x = 0, y' = 0 + 2 = 2, i.e. y'(0) = 2$
Hence, $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$ is the solution to the given initialvalue problem.
View full question & answer→Question 705 Marks
verify that $\text{y}=\text{ce}^{\tan^{-1}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}-1)\frac{\text{dy}}{\text{dx}}=0.$
AnswerWe have,
$\text{y}=\text{ce}^{\tan^{-1}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{ce}^{\tan^{1}\text{x}}\frac{1}{1+\text{x}^2}\ ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{(1+\text{x}^2)\text{e}^{\tan^{1}}\text{x}\frac{1}{1+\text{x}^2}-\text{e}^{\tan^{1}}\text{x}(2\text{x})}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{\text{e}^{\tan^{-1}}\text{x}-2\text{xe}^{\tan^{-1}}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{(1-2\text{x})\text{e}^{\tan^{-1}}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}(1-2\text{x})\frac{\text{e}^{\tan^{-1}}}{(1+\text{x}^2)}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=(1-2\text{x})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}-1)\frac{\text{dy}}{\text{dx}}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 715 Marks
Solve the following differential equations:$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$
Answer$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$
$\Rightarrow\text{cosec x }\log\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}^2\text{y}^2$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\frac{\text{x}^2}{\text{cosec x}}\text{dx}$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\text{x}^2\sin\text{x dx}$
$\Rightarrow\int\frac{1}{\text{y}^2}\log\text{ y dy}=-\int\text{x}^2\sin\text{x dx}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}+\int\frac{1}{\text{y}}\times\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+\int2\text{x}\cos\text{x dx}\Big]+\text{C}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}-\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}-2\int\sin\text{x dx}\Big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)=-\big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}+2\cos\text{x dx}\big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)-\text{x}^2\cos\text{x}+2(\text{x}\sin\text{x}+\cos\text{x})=\text{C}$
View full question & answer→Question 725 Marks
Form the differential equation corresponding to $\text{y}^2-2\text{ay}+\text{x}^2=\text{a}^2$ by eliminating a.
AnswerThe equation of the family of curves is
$y^2-2 a y+2+=a^2 \ldots(1)$
where a is a parameter.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}-2\text{a}\frac{\text{dy}}{\text{dx}}+2\text{x}=0$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{x}=2\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}=\text{a}$
Substituting the value of a in equation (2), we get
$\text{y}^2-2\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)\text{y}+\text{x}^2=\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)^2$
$\Rightarrow\frac{\text{y}^2\frac{\text{dy}}{\text{dx}}-2\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)\text{y}+\text{x}^2\frac{\text{dy}}{\text{dx}}}{\frac{\text{dy}}{\text{dx}}}=\frac{\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)}{\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$
$\Rightarrow\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\\=\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}^2$
$\Rightarrow(\text{x}^2-2\text{y}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-4\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{x}^2=0$
It is the required differential equation.
View full question & answer→Question 735 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0,\text{y}(0)=1,\text{y}(0)=3$
Function $\text{y}=\text{e}^\text{x}+\text{e}^{2\text{x}}$
Answer$\text{y}=\text{e}^{\text{x}}+\text{e}^{2\text{x}} ...(\text{i})$
Differentiating it with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}+2\text{e}^{2\text{x}} ...\text{(ii)}$
Again, differentiating it with respect to $x,$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{e}^{\text{x}}+4\text{e}^{2\text{x}}$
$=(3-2)\text{e}^{\text{x}}+(6-2)\text{e}^{2\text{x}}$
$=3\text{e}^{\text{x}}+6\text{e}^{2\text{x}}-2\text{e}^{\text{x}}-2\text{e}^{2\text{x}}$
$=3(\text{e}^\text{x}+2\text{e}^{2\text{x}})-2 (\text{e}^{\text{x}}+\text{e}^{2\text{x}})$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=3\frac{\text{dy}}{\text{dx}}-2\text{y}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
It is the given equation, so $y - e^x + 2e^{2x}$ is the solution of the given equation.
put $x = 0$ in equation $(i),y = e^0+ e^0$
$y = 1 + 1$
$y = 2$
so,
$y(0) = 2$
put $x - 0$ in equation $(ii),$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{0}+2\text{e}^{0}y' = 1 + 2$
$y' = 3$
so,
$y'(0) = 3$
View full question & answer→Question 745 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3\ \dots(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}}$
$\text{Q}=\text{x}^3$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{\log|\text{x}|}=\text{x}$
Multiplying both sides of (1) by x, we get
$\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}\Big)=\text{x x}^3$
$\Rightarrow\ \text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}^4$
Integrating both sides with respect to x, we get
$\text{xy}=\int\text{x}^4\text{dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^5}{5}+\text{C}$
$\Rightarrow\ 5\text{xy}=\text{x}^5+5\text{C}$
$\Rightarrow\ 5\text{xy}=\text{x}^5+\text{K}$ (where, K = 5C)
Hence, $5\text{xy}=\text{x}^5+\text{K}$ is the required solution.
View full question & answer→Question 755 Marks
Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.
AnswerThe equation of the family of parabolas with latus rectum 4a and axis parallel to the x-axis is given by
$(\text{y}-\beta)^2=4\text{a}(\text{x}-\text{a})\ ...(1)$
where a and Bare two arbitrary constants.
As this equation has two arbitrary constants, we shall get second order differential equation.
Differentiating equation (1) with respect to x, we get
$2(\text{y}-\beta)\frac{\text{dy}}{\text{dx}}=4\text{a}\ ...(2)$
Differentiating equation (2) with respect to x, we get
$(\text{y}-\beta)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\frac{\text{dy}}{\text{dx}}\Big)=0$
Now, from equation (2) we get,
$(\text{y}-\beta)=\frac{4\text{a}}{\frac{\text{dy}}{\text{dx}}}$
From (3) and (4), we get
$\frac{2\text{a}}{\frac{\text{dy}}{\text{dx}}}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
$\Rightarrow2\text{a}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=0$
It is the required differential equation.
View full question & answer→Question 765 Marks
Solve the following differential equation:
$\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
AnswerThe given differential equation is $\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
This differential equation can be written as
$\frac{\text{dx}}{\text{dy}} = \frac{\cot^{-1}\text{y}+\text{x}}{1+\text{y}^2}$
$\Rightarrow\frac{\text{dx}}{\text{dy}} + \Big(-\frac{1}{1+\text{y}^2}\Big)\text{x}= \frac{\cot^{-1}\text{y}}{1+\text{y}^2}$
This is a linear differential equation with $\text{P}=-\frac{1}{1+\text{y}^2}$ and $\text{Q}= \frac{\cot^{-1}\text{y}}{1+\text{y}^2}$
$\text{I}.\text{F}.=\text{e}^{-\int\frac{1}{1+\text{y}2}\text{dy}}=\text{e}^{\cot{^{-1}{\text{y}}}}$
Multiply the differential equation by integration factor (I.F.), we get
$\frac{\text{dx}}{\text{dy}}\text{e}^{\cot{^{-1}\text{y}}}-\frac{\text{x}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}=\frac{\cot^{-1}\text{y}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}$
$\Rightarrow\frac{\text{d}}{\text{dy}}\Big(\text{xe}^{\cot{^{-1}\text{y}}}\Big)=\frac{\cot^{-1}\text{y}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}$
Integrating both sides with respect y, we get
$\text{xe}^{\cot{^{-1}\text{y}}}=\int\frac{\cot^{-1}\text{y}}{\big(1+{\text{y}^2\big)}}\text{e}^{\cot{^{-1}\text{y}}} \text{dy}+\text{C}$
Putting $\text{t}=\cot^{-1}\text{y}$ and $\text{dt}=-\frac{1}{1+\text{y}^2}\text{dy},$ we get
$\text{xe}^{\cot{^{-1}\text{y}}}=-\int \text{te}^{\text{t}}\text{dt}+\text{C}$
$\Rightarrow\text{xe}^{\cot{^{-1}\text{y}}}=-\text{e}\big(\text{t}-1\big)+\text{C}$
$\Rightarrow\text{xe}^{\cot{^{-1}\text{y}}}=\text{e}^{\cot{^{-1}\text{y}}}\big(1-\cot^{-1}\text{y}\big) + \text{C}$
View full question & answer→Question 775 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$ |
$\text{y}=\frac{1}{4}(\text{x}\pm\text{a})^2$ |
AnswerWe have
$\text{y}=\frac{1}{4}(\text{x}\pm\text{a})^2\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{4}\times2(\text{x}\pm\text{a})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}(\text{x}\pm\text{a})$
Squaring both sides we get
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big) ^2=\Big[\frac{1}{2}(\text{x}\pm\text{a})\Big]^2$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)=\frac{1}{4}(\text{x}\pm\text{a})^2$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2=\text{y}$
$\therefore\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 785 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}},\text{x}\ne0$
Answer$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}}$
$\Rightarrow\text{dy}=\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
View full question & answer→Question 795 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\frac{\text{dy}}{\text{dx}}-\text{y}=\cos2\text{x}$
AnswerHere,
$\frac{\text{dy}}{\text{dx}}-\text{y}=\cos2\text{x}$
It is a linear differential equation. Comparing it with,
$\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$
$\text{P}=-1,\text{Q}=\cos2\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\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\cos2\text{x}\times\text{e}^{-\text{x}}\text{dx + C}\ \dots(\text{i})$
$\text{I}=\int\cos2\text{x}\text{e}^{-\text{x}}\text{dx}=\cos2\text{x}\times(-\text{e}^{-\text{x}})-\int\Big(\frac{\sin2\text{x}}{2}\Big)\text{e}^{-\text{x}}\text{dx}$ [Using integration by parts]
$\text{I}=-\text{e}^{-\text{x}}\cos2\text{x}-\frac{1}2\Big[\big(-\sin2\text{x}\text{e}^{-\text{x}}\big)+\int\frac{\cos2\text{x}}{2}\text{e}^{-\text{x}}\text{dx}\Big]$
$\text{I}=-\text{e}^{-\text{x}}\cos2\text{x}+\frac{1}2\sin2\text{x}\text{e}^{-\text{x}}-\frac{1}4\text{I}$
$\frac{5}4\text{I}=\frac{\text{e}^{-\text{x}}}{2}(\sin2\text{x}-2\cos2\text{x})$
$\text{I}=\frac{2}5\text{e}^{-\text{x}}(\sin2\text{x}-2\cos2\text{x})$
So, solution of the equation is given by
$\text{y}=\frac{2}5(\sin2\text{x}-2\cos2\text{x})+\text{C}\text{e}^{-\text{x}}$
View full question & answer→Question 805 Marks
Form the differential equation of all the circle which pass through the origin and whose centres lies in $y-$axis.
AnswerWe know that, equation of a dide with centre at $(h, k)$ and radius $r$ is given by,
$(x - h^2) + (y - k)^2 = r^2 ...(1)$
Here, entre lies, on y-axis, so $h = 0$
$x^2 + (y - k)^2 = r^2...(2)$
Also given that circle is passing through origin, so
$0 + k^2 = r^2$
$k^2 = r^2$^
So, equation $(2)$ becomes
$x^2 + (y - k)^2 = k^2$
$x^2 + y^2 - 2yk = 0$
$2yk = x^2+ y^2$
$\text{k}=\frac{\text{x}^2+\text{y}^2}{2\text{y}}$
Differentiating with respect to $x,$
$0=\frac{\Big(2\text{y}2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)-(\text{x}^2+\text{y}^2)2\frac{\text{dy}}{\text{dx}}}{(2\text{y})^2}$
$0=4\text{xy}+4\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}-2\text{y}^2\frac{\text{dy}}{\text{dx}}$
$0=2\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}+4\text{xy}$
$\text{x}^2\frac{\text{dy}}{\text{dx}}-\text{y}^2\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
View full question & answer→Question 815 Marks
Solve the following differential equation:
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
AnswerWe have,
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
$\Rightarrow\text{x dx + y dx}-\text{x dy}-\text{y dy}=\text{dx + dy}$
$\Rightarrow(\text{x + y}-1)\text{dx}=(\text{x + y}+1)\text{dy}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{x}+\text{y}-1}{\text{x}+\text{y}+1}$
Let $\text{ x} + \text{y} = \text{v}$
$\therefore 1+ \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{\text{v}-1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1}{\text{v}+1}+1$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1+\text{v}+1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{2\text{v}}{\text{v}+1}$
$\Rightarrow \frac{\text{v}+1}{2\text{v}}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}+1}{2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\int\text{dv}+\frac{1}{2}\int\frac{1}{\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\text{v}+\frac{1}{2}\log|\text{v}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{x}+\text{y})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{y}-\text{x})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{C}$
View full question & answer→Question 825 Marks
Solve the following differential equations:$\cos\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}},\text{y}(0)=\frac{\pi}{2}$
Answer$\cos\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}},\text{y}(0)=\frac{\pi}{2}$
$\Rightarrow\cos\text{y dy = e}^{\text{x}}\text{ dx}$
Integrating both sides, we get
$\int\cos\text{y dy}=\int\text{e}^{\text{x}}\text{ dx}$
$\Rightarrow\sin\text{y}=\text{e}^{\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=\frac{\pi}{2}.$
Substituting the valuse 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
$\sin\text{y}=\text{e}^{\text{x}}$
$\Rightarrow\text{y}=\sin^{-1}(\text{e}^{\text{x}})$
Hence, $\text{y}=\sin^{-1}(\text{e}^{\text{x}})$ is the required solution.
View full question & answer→Question 835 Marks
Solve the following equation
$\text{x}\cos^2\text{y dx}=\text{y}\cos^2\text{x dx}$
AnswerWe have,$\text{x}\cos^2\text{y dx}=\text{y}\cos^2\text{x dx}$
$\Rightarrow\frac{\text{x}}{\cos^2\text{x}}\ \text{dx}=\frac{\text{y}}{\cos^2\text{y}}\ \text{dy}$
$\Rightarrow\text{x}\sec^2\text{x dx}=\text{y}\sec^2\text{y dy}$
Integrating both sides, we get
$\int\text{x}\sec^2\text{x dx}=\int\text{y}\sec^2\text{y dy}$
$\Rightarrow\text{x}\int\sec\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sec^2\text{x dx}\Big\}\text{dx}\\=\text{y}\int\sec^2\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}(\text{y})\int\sec^2\text{y dy}\Big\}\text{dy}$
$\Rightarrow\text{x}\tan\text{x}-\int\int\tan\text{x dx}=\text{y}\tan\text{y}-\int\tan\text{y dy}$
$\Rightarrow\text{x}\tan\text{x}-\log|\sec\text{x}|=\text{y}\tan\text{y}-\log|\sec\text{y}|+\text{C}$
$\Rightarrow\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$
Hence, $\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$ is the required solution.
View full question & answer→Question 845 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
Answer$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
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}}=\text{v}+\sin\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}-\text{v}$
$\Rightarrow\ \frac{1}{\sin\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\sin\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \tan\frac{\text{v}}{2}=\text{Cx}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$
Hence, $\tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$ is the required solution.
View full question & answer→Question 855 Marks
Solve the following differential equation
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2\text{y}$
AnswerWe have $(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2\text{y}$$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{2\text{x}^3}{\text{x}-1}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\frac{2\text{x}^3}{\text{x}-1}\text{dx}$
$\Rightarrow\log|\text{y}|=2\int\frac{\text{x}^3-1+1}{\text{x}-1}\ \text{dx}$ $\Rightarrow\log|\text{y}|=2\int\frac{(\text{x}-1)(\text{x}^2+\text{x}+1)+1}{\text{x}-1}\ \text{dx}$ $\Rightarrow\log|\text{y}|=2\int(\text{x}^2+\text{x}+1)\text{dx}+2\int\frac{1}{\text{x}-1}\ \text{dx}$ $\Rightarrow\log|\text{y}|=\frac{2}{3}\text{x}^3+\text{x}^2+2\text{x}+\log|\text{x}-1|+\text{C}$ Hence, $\log|\text{y}|=\frac{2}{3}\text{x}^3+\text{x}^2+2\text{x}+\log|\text{x}-1|+\text{C}$ is the required solution.
View full question & answer→Question 865 Marks
Find the particular solution of the differential equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y},$ given that when x = 1, y = 0.
AnswerConsider the given equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y}$ This is a homogeneous equation. Substituting y = vx and $\frac{\text{dy}}{\text{dx}}=\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)$In the above equation, we have,
$(\text{x}-\text{vx})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=\text{x +2vx}$ $\Rightarrow\ (1-\text{v})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=1+2\text{v}$ $\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}(1-\text{v})}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\frac{\text{dx}}{\text{x}}$ Integrating both sides, we have, $\Rightarrow\ \int\frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{(1+\text{v}+\text{v}^2)}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\text{v}^2+\frac{1}4+\text{v}+\frac{3}4}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\big(\text{v}+\frac{1}2\big)^2+\big(\frac{\sqrt3}2\big)^2}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\times\frac{1}{\frac{\sqrt3}{2}}\tan^{-1}\frac{\text{v}+\frac{1}{2}}{\frac{\sqrt3}{2}}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{v}+1}{\sqrt3}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}{2}\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\text{C}\ \dots(1)$ Given that when x = 1, y = 0 Substituting the values, in the above equation, we get, $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\times0+1}{\sqrt3}-\frac{1}2\log(1+0+0^2)=\log1+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{1}{\sqrt3}-\frac{1}{2}\times0=0+\text{C}$ $\Rightarrow\ \text{C}=\sqrt3\times\frac{\pi}{6}$ $\Rightarrow\ \text{C}=\frac{\pi}{2\sqrt3}$ Thus equation (1) becomes, $\sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\frac{\pi}{2\sqrt3}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{2\sqrt3}=\log\text{x}+\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log\text{x}^2+\log\Big(\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log(\text{x}^2+\text{xy}+\text{y}^2)$
View full question & answer→Question 875 Marks
Solve the following differential equations:$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
AnswerWe have,
$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
$\Rightarrow(1+\text{x})(1+\text{y}^2)\text{dx}=-(1+\text{y})(1+\text{x}^2)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
Integrating both sides, we get
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\int\frac{1}{1+\text{x}^2}\text{dx}+\int\frac{\text{x}}{1+\text{x}^2}\text{dx}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\int\frac{\text{y}}{1+\text{y}^2}\text{dy}$
Substituting $1+\text{x}^2=\text{t}$ in the second integral of LHS and $1+\text{y}^2=\text{u}$ in the second integral of RHS, we get
$2\text{x dx = dt}$ and $2\text{y dy = du}$
$\therefore\int \frac{1}{1+\text{x}^2}\text{dx}+\frac{1}{2}\int\text{dt}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\frac{1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|\text{t}|=-\tan^{-1}\text{y}-\frac{1}{2}\log|\text{u}|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|1+\text{x}^2|=-\tan^{-1}\text{y}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log|1+\text{x}^2|+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$
Hence, $\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$ is the required solution.
View full question & answer→Question 885 Marks
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its intial mass.
AnswerLet A be the quantity of mass at any time t, So
$\frac{\text{dA}}{\text{dt}}\propto\text{A}$
$\frac{\text{dA}}{\text{dt}}=-\lambda\text{A}$
$\frac{\text{dA}}{\text{A}}=-\lambda\text{dt}$
$\int \frac{\text{dA}}{\text{A}}=-\lambda\int\text{dt}$
$\log\text{A}=-\lambda\text{t}+\text{C}\ ...(\text{i})$
Let intial of mass be A, So
$\log\text{A}_{0}=-\lambda(0)+\text{C}$
$\log(\text{A}_{0})=\text{C}$
Now, eq. (i),
$\log\text{A}=-\lambda\text{t}+\log\text{A}_{0}$
$\log\frac{\text{A}}{\text{A}_{0}}=-\lambda\text{t}$
Let be the time to half the mass $\text{A}=\frac{1}{2}\text{A}_{0}$
$\log\frac{\text{A}}{\text{A}_{0}}=-\lambda\text{t}$
$\log\frac{\text{A}}{\text{2A}}=-\lambda\text{t}$
$-\log2=-\lambda\text{t}$
$\frac{1}{\lambda}\log2=\text{t}$
Required time is $\frac{1}{\lambda}\log2$ units of proportion.
View full question & answer→Question 895 Marks
Solve the following differential equation
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
AnswerWe have,
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
$\Rightarrow\text{dy}=\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Integrating both sides, we get
$\Rightarrow\int\text{dy}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
$\Rightarrow\text{y}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{y}=\int\frac{1}{\text{t}^4}\ \text{dt}$
$=\frac{\text{t}^{-3}}{-3}+\text{C}$
$=\frac{-\sin^{-3}}{3}+\text{C}$
$=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$
hence, $\text{y}=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$ is the solution to the given differential equation.
View full question & answer→Question 905 Marks
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}=-\frac{1}{(\text{x}^2+1)^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{4\text{x}}{\text{x}^2+1}$
$\text{Q}=-\frac{1}{(\text{x}^2+1)^2}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{2\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}}$
$=\text{e}^{2\log|\text{x}^2+1|}$
$=(\text{x}^2+1)^2$
Multiplying both sides of (1) by $(x^2+ 1)^2$, we get
$(\text{x}^2+1)^2\Big(\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}\Big)=(\text{x}^2+1)^2\Big[-\frac{1}{(\text{x}^2+1)^2}\Big]$
$\Rightarrow\ (\text{x}^2+1)^2\frac{\text{dy}}{\text{dx}}+4\text{x}(\text{x}^2+1)\text{y}=-1$
Integrating both sides with respect to x, we get
$(\text{x}^2+1)^2\text{y}=-\int\text{dx + C}$
$\Rightarrow\ (\text{x}^2+1)^2\text{y}=-\text{x + C}$
Hence, $(\text{x}^2+1)^2\text{y}=-\text{x + C}$ is the required solution.
View full question & answer→Question 915 Marks
Solve the following differential equations:
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0,$ given that $\text{y}=\frac{\pi}{4},$ when $\text{x}=\sqrt{2}.$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\cot\text{y}$
$\Rightarrow\tan\text{y dy}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\tan\text{y dy}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\log|\sec\text{y}|=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\log(|\text{x}||\sec\text{y}|)=\log\text{C}$
$\Rightarrow\text{x}\sec\text{y = C}...(1)$
Given: $\text{x}=\sqrt{2},\text{y}=\frac{\pi}{4}.$
Substituting the values of x and y in (1), we get
$\sqrt{2}\sec\frac{\pi}{4}=\text{C}$
$\Rightarrow\text{C}=2$
Substituting the value of C in (1), we get
$\text{x}\sec\text{y}=2$
$\Rightarrow\text{x}=2\cos\text{y}$
Hence, $\text{x}=2\cos\text{y}$ is the required solution.
View full question & answer→Question 925 Marks
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Answer
Portion of the x-axis cut off between the origin and tangent at a point $=\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}=\text{OT}$
It is given, $\text{OT}=2\text{x}$
$\therefore \ \text{x}-\text{y}\frac{\text{dx}}{\text{dy}}=2\text{x}$
$-\text{x}=\text{y}\frac{\text{dx}}{\text{dy}}$
$-\int\frac{\text{dx}}{\text{dy}}=\int\frac{\text{dy}}{\text{y}}$
$\therefore\ \text{xy}=\text{k}$
Since the curve passes through the point (1, 2)
at $\text{x}=1, \text{y}=2$
$\therefore \text{k}=2$
$\therefore \text{xy}=2$ View full question & answer→Question 935 Marks
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
AnswerEquation of normal on point (x, y) on the curve
$\text{y}-\text{y}=\frac{-\text{dx}}{\text{dy}}(\text{x}-\text{x})$
Its passing through (3, 0)
$\Rightarrow\text{0}-\text{y}=\frac{-\text{dx}}{\text{dy}}(3-\text{x})$
$\Rightarrow \text{y}=\frac{\text{dx}}{\text{dy}}(3-\text{x})$
$\Rightarrow \text{y}\ \text{dy}=(3-\text{x})\text{dx}$
$\Rightarrow \int\text{y}\ \text{dy}=\int(3-\text{x})\text{dx}$
$\Rightarrow \frac{\text{y}^{2}}{2}=3\text{x}-\frac{\text{x}^{2}}{2}+\text{C}\ ...(\text{i})$
It passing through (3, 4),
$\frac{16}{2}=9-\frac{9}{2}+\text{C}$
$\frac{16}{2}=\frac{9}{2}+\text{C}$
$\text{C}=7$
Put $\text{C}=7$ is equation (i)
$\frac{\text{y}^{2}}{2}=3\text{x}-\frac{\text{x}^{2}}{2}+\frac{7}{2}$
$\text{y}^{2}=6\text{x}-\text{x}^{2}+{7}$
View full question & answer→Question 945 Marks
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x},\text{ y}(1)=0$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\frac{\log\text{x}}{\text{x}}\ ...(\text{1})$
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 $\text{Q}=\frac{\log\text{x}}{\text{x}}$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}{-\int\frac{1}{\text{x}}}\text{ dx}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\times\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\frac{\log\text{x}}{\text{x}^2}$
Integrsting both sides with respect to x, we get
$\text{y}\frac{1}{\text{x}}=\int\frac{1}{\text{x}^2}\times\log\text{x dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\log\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\frac{1}{\text{x}^2}\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}+\int\frac{1}{\text{x}^2}\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}-\frac{1}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\log\text{x}-1+\text{Cx}\ ...(\text{ii})$
Now,
$\text{y}(1)=0$
$\therefore\ 0=-0-1+\text{C}(1)$
$\Rightarrow\text{C}=1$
Putting the value of C in (2) we get
$\text{y}=-\log\text{x}-1+\text{x}$
$\Rightarrow\text{y}=\text{x}-1-\log\text{x}$
Hence, $\text{y}=\text{x}-1-\log\text{x}$ is the required solution.
View full question & answer→Question 955 Marks
Solve the following differential equation:
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
AnswerWe have,
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{\text{x}^2-\text{xy}+\text{y}^2}$
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2\text{x}^2}{\text{x}^2-\text{vx}^2+\text{v}^2\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}-\text{v}^3}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \frac{1-\text{v}+\text{v}^2}{\text{v}+\text{v}^3}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1+\text{v}^2}{\text{v}(1+\text{v}^2)}\text{dv}-\int\frac{1}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{1+\text{v}^2}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\text{v}|-\tan^{-1}|\text{v}|=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\big|\frac{\text{vx}}{\text{C}}\big|=\tan^{-1}\text{v}$
$\Rightarrow\ \big|\frac{\text{vx}}{\text{C}}\big|=\text{e}^{\tan^{-1}\text{v}}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ |\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$
Hence, $|\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$ is the required solution.
View full question & answer→Question 965 Marks
Show that aii curve for which the slope at any point (x, y) on its is $\frac{\text{x}^{2}+\text{y}^{2}}{\text{2xy}}$ are rectangular hyperbola.
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{y}^{2}}{\text{2xy}}$
Let y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
$\therefore \text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^{2}+\text{v}^{2}\text{x}^{2}}{2\text{vx}^{2}}$
$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{1}+\text{v}^{2}}{2\text{v}}-\text{v}$
$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{1}+\text{v}^{2}-2\text{v}^{2}}{2\text{v}}$
$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{1}+\text{v}^{2}}{2\text{v}}$
$\Rightarrow\frac{2\text{v}}{1-\text{v}^{2}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{2\text{v}}{1-\text{v}^{2}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow-\log|1-\text{v}^{2}|=\log|\text{x}|-\log|\text{C}|$
$\Rightarrow-\log\Big|\frac{1-\text{v}^{2}}{\text{C}}\Big|=-\log|\text{x}|$
$\Rightarrow 1-\text{v}^{2}=\frac{\text{C}}{\text{x}}$
$\Rightarrow \frac{\text{x}^{2}-\text{y}^{2}}{\text{x}^{2}}=\frac{\text{C}}{\text{x}}$
$\Rightarrow \text{x}^{2}-\text{y}^{2}=\text{Cx}$
View full question & answer→Question 975 Marks
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+1=0;\text{y}(-1)=0$
Answer$\text{x}\frac{\text{dy}}{\text{dx}}+1=0,\text{y}(-1)=0$
$\text{x}\frac{\text{dy}}{\text{dx}}=-1$
$\text{dy}=-\frac{\text{dx}}{\text{x}}$
$\int\text{dy}=\int-\frac{\text{dx}}{\text{x}}$
$\text{y}=-\log|\text{x}|+\text{C}$
Put x = -1 and y = 0
0 = 0 + c
c = 0
put c = 0 in equation (1),
$\text{y}=-\log|\text{x}|,\text{x}<0$
View full question & answer→Question 985 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=1$
$\text{Q}=\sin\text{x}$
$\therefore\ \text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\text{dx}}=\text{e}^{\text{x}}$
Multiplying both sides of (1) by $e^x$, we get
$\text{e}^\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\Big)=\text{e}^\text{x}\sin \text{x}$
$\Rightarrow\ \text{e}^\text{x}\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{x}}\text{y}=\text{e}^\text{x}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{\text{x}}=\int\text{e}^{\text{x}}\sin\text{x dx + C}$
$\Rightarrow\ \text{y}\text{e}^{\text{x}}=\frac{\text{e}^{\text{x}}}{2}(\sin\text{x}-\cos{\text{x}})+\text{C}$
$\Rightarrow\ \text{y}=\text{Ce}^{-\text{x}}+\frac{1}{2}(\sin\text{x}-\cos{\text{x}})$
Hence, $\text{y}=\text{Ce}^{-\text{x}}+\frac{1}{2}(\sin\text{x}-\cos{\text{x}})$ is the required solution.
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verify that $\text{y}^2=4\text{a}(\text{x}+\text{a})$ is a solution of the differential equation $\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=2\text{x}\frac{\text{dy}}{\text{dx}}.$
Answer$\text{y}^2=4\text{a}(\text{x}+\text{a})\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}\ ...(2)$
Now,
$\text{y}\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}$
$=\Big[\text{y}^2\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}\Big]\frac{1}{\text{y}}$
$=\Big[4\text{a}(\text{x}+\text{a})-4\text{a}(\text{x}+\text{a})\Big(\frac{2\text{a}}{\text{y}}\Big)^2\Big]\frac{1}{\text{y}}$
Using equation (1) and (2)
$=\Big[4\text{ax}+4\text{a}^2-\frac{16\text{a}3\text{x}}{\text{y}^2}-\frac{16\text{a}^4}{\text{y}^2}\Big]\frac{1}{\text{y}}$
$=\frac{4\text{a}}{\text{y}^3}[\text{xy}^2+\text{ay}^2-4\text{a}^2\text{x}-4\text{a}^3\Big]$
$=\frac{4\text{a}}{\text{y}^3}[\text{y}^2(\text{a}+\text{x})-4\text{a}^2(\text{x}+\text{a})]$
$\frac{4\text{a}}{\text{y}^3}(\text{a}+\text{x})(\text{y}^2-4\text{a}^2)$
View full question & answer→Question 1005 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2,\text{y}(0)=1$
Answer$\frac{\text{dy}}{\text{dx}}=1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2,\text{y}(0)=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{(1+\text{y}^2)}=(1+\text{x}^2)\text{dx}$
Integrating both sides, we get
$\int \frac{\text{dy}}{(1+\text{y}^2)}=\int(1+\text{x}^2)\text{dx}$
$\Rightarrow\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\frac{\pi}{4}=0+0+\text{C}$
$\Rightarrow\text{C}=\frac{\pi}{4}$
Substituting the value of C in (1), we get
$\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\frac{\pi}{4}$
Hence, $\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\frac{\pi}{4}$ is the required solution.
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