Question 513 Marks
For each of the differential equations in find a particular solution satisfying the given condition:
$\text{cos}\bigg(\frac{\text{dy}}{\text{dx}}\bigg) = \text{a}(\text{a} \in \text{R}); \ \text{y}=1\text{when x = }0$
AnswerGiven: Differention equation $\text{cos}\bigg(\frac{\text{dy}}{\text{dx}}\bigg) = \text{a}(\text{a} \in \text{R}); \ \text{y}=1\text{when x = }0$
$\ \Rightarrow \ \frac{\text{dy}}{\text{dx}}=\text{cos}^{-1}\text{a}$
$\Rightarrow \ \text{dy}=\Big(\text{cos}^{-1}\text{a}\Big)\text{dx}$
Integrating both sides, $\int 1\ \text{dy}=\int\Big(\text{cos}^{-1}\text{a}\Big)\text{dx}$
$\Rightarrow \ \text{y}=\Big(\text{cos}^{-1}\text{a}\Big)\int1\ \text{dx}$
$\ \Rightarrow \ \text{y}=\Big(\text{cos}^{-1}\text{a}\Big)\ \text{x+c}\ .....\text{(i)}$
Now putting y = 1 when x = 0 in eq. (i), we get c = 1
Putting c = 1 in eq. (i), $y = (\cos ^{-1} a) x + 1$
$\Rightarrow\frac{\text{y-1}}{\text{x}}=\cos^{-1} \text{a}$
$\Rightarrow cos\frac{\text{y-1}}{\text{x}}=\text{a}$
View full question & answer→Question 523 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $\text{y} – \text{cos}\ \text{y} = \text{x} \ :\ (\text{y} \ \text{sin} \ \text{y} + \text{cos} \ \text{y} + \text{x}) \text{y}' = \text{y}$
AnswerGiven: y-cos y = x ....(i)
To prove: y given by eq. (i) is a solution of differential equation
(y sin y + cos y + x) y' = y ....(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x, we have
$\text{y}'+(\text{sin}\ \text{y})\text{y}'=1 \ \Rightarrow\ \ \text{y}'(1+\text{sin}\ \text{y})=1$
$\Rightarrow\ \ \text{y}'=\frac{1}{1+\text{sin}\ \text{y}} \ ....(\text{iii})$
Putting the value of x from eq. (i) and value of y' from eq. (iii) in L.H.S. of eq. (ii),
$\text{(y sin y + cos y + x)}{\text{y}}' $ $\ \Rightarrow \ \text{(y sin y + cos y + y} -\text{cos} \ \text{y}) \frac{1}{1+\text{sin}\ \text{y}}$
$\Rightarrow \ \text{(y sin y + y)}\frac{1}{1+ \text{sin y}} $ $\ \Rightarrow \ \text{y}(\text{sin y} + 1)\frac{1}{1+\text{sin y}}= \text{y = R.H.S. of (ii)}$
Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x) y' = y.
View full question & answer→Question 533 Marks
verify that $\text{y}=\text{cx}+2\text{c}^2$ is a solution of the differential equation $2\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2-\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
AnswerWe have,
$\text{y}=\text{cx}+2\text{c}^2\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{c}\ ...(2)$
Now,
$2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$=2\text{c}^2+\text{cx}-\text{cx}-2\text{c}^2=0$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=0$
Hence, the given is the solution to the given differential equation.
View full question & answer→Question 543 Marks
Solve the following differential equations:
$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
Answer$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
$=(1+\text{x})+\text{y}(1+\text{x})$
$\text{xy}\frac{\text{dy}}{\text{dx}}=(1+\text{x})(1+\text{y})$
$\int\frac{\text{ydy}}{\text{y}+1}=\int\frac{1+\text{x}}{\text{x}}\text{dx}$
$\int\Big(1-\frac{1}{\text{y}+1}\Big)\text{dy}=\int\Big(\frac{1}{\text{x}}+1\Big)\text{dx}$
$\text{y}-\log|\text{y}+1|=\log|\text{x}|+\text{x}+\log|\text{c}|$
$\text{y}=\log|\text{cx(y+1})|+\text{x}$
View full question & answer→Question 553 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=0,\text{y}'(0)=1$Function $\text{y}=\sin\text{x}$
AnswerWe have,$\text{y}=\sin{\text{x}} ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x} ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
it is the given differential equation.
Here, $\text{y}=\sin\text{x}$ satisfies the given differential equation; hence, it is a solution.
Also, when $\text{x}=0,\text{y}=\sin0=0,\text{i.e.},\text{y}(0)=0.$
And, when $\text{x}=0,\text{y}'=\cos 0=1,\text{i.e.,}\text{y}'(0)=1.$
Hence, $\text{y}=\sin\text{x}$ is the solution to the given initial value problem.
View full question & answer→Question 563 Marks
Solve the following differential equations:$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{ dx}$
Answer$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{dx}$
$(1-\text{x}^2)\text{dy = dx}(\text{xy}^2-\text{xy})$
$(1-\text{x}^2)\text{dy = xy(y}-1)\text{dx}$
$\int\frac{\text{dy}}{\text{y(y}-1)}=\int\frac{\text{xdx}}{1-\text{x}^2}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=\frac{1}{2}\int\frac{2\text{x}}{1-\text{x}^2}\text{dx}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=-\frac{1}{2}\int\frac{-2\text{x}}{1-\text{x}^2}\text{dx}$
$\log|\text{y}-1|-\log|\text{y}|=-\frac{1}{2}\log|1-\text{x}^2|+\text{C}$
View full question & answer→Question 573 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y}=\sqrt{\text{a}^2-\text{x}^2}\text{x} \in(-\text{a, a}) :\ \text{x+y}\frac{\text{dy}}{\text{dx}}=\ 0 (\text{y}\neq0)$
AnswerGiven: $\text{y}=\sqrt{\text{a}^2-\text{x}^2},\ \text{x}\in(-\text{a, a}) \ ....(\text{i})$
To prove: y given by eq. (i) is a solution of differential equation $\text{x+y}\frac{\text{dy}}{\text{dx}} = 0 \ ....(\text{ii})$
Proof: From eq. (i), $\frac{\text{dy}}{\text{dx}}=\frac{1}{2}(\text{a}^2-\text{x}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{a}^2-\text{x}^2)=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$
$= \frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\ .....(\text{iii})$
Putting the values of y and $\frac{\text{dy}}{\text{dx}}$ from eq. (i) and (iii) in L.H.S. of eq. (ii),
$\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}+\sqrt{\text{a}^2-\text{x}^2}\bigg(\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\bigg)$ $=\text{x}-\text{x}=0=\text{R.H.S. of eq.(ii)}$
Hence, Function given by eq. (i) is a solution of $\text{x+y}\frac{\text{dy}}{\text{dx}}=0.$
View full question & answer→Question 583 Marks
Form the differential equation having $\text{y} = (\sin^{-1}\text{x})^{2} + \text{A}\cos^{-1} + \text{ B},$ where A and B are arbitrary constants, as its general solution.
AnswerWe have, $\text{y}=(\sin^{-1}\text{x})^2+\text{A}\cos^{-1}\text{x}+\text{B}$
On differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{(-\text{A})}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
Again, differentiating w.r.t.x, we get
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\frac{\text{dy}}{\text{dx}}.\frac{-2\text{x}}{2\sqrt{1+\text{x}^2}}=\frac{2}{{\sqrt{1-\text{x}}^2}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}.\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
which is the required differential equation.
View full question & answer→Question 593 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=4\text{ax}$
AnswerThe equation of the family of curves is
$\text{y}=4\text{ax}\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dt}}{\text{dx}}=4\text{a}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}-2\text{x}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 603 Marks
Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}=0.$
AnswerThe given differential equation is:
$\frac{\text{dy}}{\text{dx}}+\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}=0$
$\text{or}\ \ \frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}},\ \ \text{or}\ \ \frac{1}{\sqrt{1-\text{y}^2}}\text{dy}=-\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}$
Integrating both sides, w.r.t.x,
$\int\frac{1}{\sqrt{1-\text{y}^2}}\text{dy}=-\int\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}$
$\therefore\ \sin^{-1}\text{y}=-\sin^{-1}\text{x}+\sin^{-1}\text{c, where }\sin^{-1}\text{c}$ $\text{is an arbitrary constant}.$
$\therefore\ \sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\text{c}$
$\Rightarrow\ \ \sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]=\sin^{-1}\text{c}$
$\text{or}\ \ \text{x}\Big[\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]=\text{c,}$ $\text{where c is an arbitrary constant}.$
View full question & answer→Question 613 Marks
show that $\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}$ is a solution of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}\ ...(1)$
Differentiating both sides with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{be}^\text{x}+2\text{ce}^{2\text{x}}\ ...(2)$
Differentiating both sides with respect to x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{be}^\text{x}+4\text{ce}^{2\text{x}}\ ...(3)$
now,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3(\text{be}^\text{x}+2\text{ce}^{2\text{x}})+2(\text{be}^\text{x}+\text{ce}^{2\text{x}})$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}+2\text{be}^\text{x}+2\text{ce}^{2\text{x}}$
$=3\text{be}^\text{x}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}-6\text{ce}^{2\text{x}}$
$=0$
View full question & answer→Question 623 Marks
For each of the differential equations in find the general solution:
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5$
AnswerThe given differential equation is
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5 \ \ \text{or} \ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^5}{\text{x}^5}$
Separating the variables, we get,
$\frac{1}{\text{y}^5}\text{dy}=-\frac{1}{\text{x}^5}\text{dx}$
Integrating, $\int \frac{\text{1}}{\text{y}^5}\text{dy}=-\int \frac{1}{\text{x}^5}\ \text{dx} \ \text{or}$ $ \int \text{y}^{-5}\text{dy}=-\int\text{x}^{-5}\text{dx}$
$\therefore\ \ \frac{\text{y}^{-4}}{-4}=-\frac{\text{x}^{-4}}{-4}+\text{c}'\ \text{or}$ $\frac{1}{-4\text{y}^4}=\frac{1}{4\text{x}^4}+\text{c}'$
$\text{or} \ \frac{1}{\text{x}^4}+\frac{1}{\text{y}^4}=-4\text{c}'$
$\text{or}\ \text{x}^{-4}+\text{y}^{-4}=\text{c}, \ \text{where c}=-4\text{c}'.$
View full question & answer→Question 633 Marks
For each of the differential equations in find the general solution:
y log y dx – x dy = 0
AnswerGiven: Differential equation
$\text{y log y dx}- \text{x dy=0}\ \Rightarrow \ -\text{x dy =}-\text{y log y dx}$
$\Rightarrow \ \frac{\text{dy}}{\text{y log y}}= \frac{\text{dx}}{\text{x}} \ \ [\text{Separating variables}]$
Integrating both sides, $\int \frac{\text{dy}}{\text{y log y}}=\int\frac{\text{dx}}{\text{x}}$
Putting log y=t on L.H.S., we get $\frac{1}{\text{y}}=\frac{\text{dt}}{\text{dy}}\ \Rightarrow\ \frac{\text{dy}}{\text{y}}=\text{dt}$
Now eq. (i) becomes $\ \int \frac{\text{dt}}{\text{t}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow \ \ \text{log}|\text{t}|=\text{log}|\text{x}|+\text{log}|\text{c}|=\text{log}|\text{xc}|$
[If all the terms in the solution of a differential equation involve log, it is better to use log c or log |c| instead of c in the solution.]
$\Rightarrow \ |\text{t}|=|\text{xc}|\ \Rightarrow \ \text{t}=\pm\text{xc}$ $\Rightarrow \ \text{log} \ \text{y}=\pm \text{xc}=\text{ax}\ \text{where a =}\ \pm\text{c}$
$\Rightarrow \ \text{y = e}^{\text{ax}}$
View full question & answer→Question 643 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{-2\text{x}}$
AnswerThe given differential equation is $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}$ (where p = 3 and $Q = e ^{-2x}$)$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int3\text{dx}}=\text{e}^{3\text{x}}.$
The solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int(\text{e}^{-2\text{x}}\times\text{e}^{3\text{x}})+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int\text{e}^\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\text{e}^\text{x}+\text{C}$
$\Rightarrow\ \text{y}=\text{e}^{\text{-2}\text{x}}+\text{C e}^{\text{-3}\text{x}}$
This is the required general solution of the given differential equation.
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Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}\ (\text{x}\neq0),$ $\text{given that y}=0\ \text{when x}=\frac{\pi}{2}.$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}$
Comparing this equation with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},\ \ \text{P}=\cot\text{x}\ \text{and}\ \text{Q}=4\text{x}\ \text{cosec}\ \text{x}$
$\int\text{P}\ \text{dx}=\int\cot\text{x}\ \text{dx}=\log\sin\text{x}$ $\text{I.F}=\text{e}^{\int\text{p dx}}=\text{e}^{\log\sin\text{x}}=\sin\text{x}$
The general solution is $\text{y}(\text{I.F})=\int\text{Q}(\text{I.F}.)\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=\int4\text{x}\ \text{cosec}\ \text{x}\ \sin\text{x}\ \text{dx}+\text{c}$ $\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}.\frac{1}{\sin\text{x}}\sin\text{x}\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}\ \text{dx}+\text{c}=4\frac{\text{x}^2}{2}+\text{c}$ $\Rightarrow\ \ \text{y}\sin\text{x}=2\text{x}^2+\text{c}\ \ ...{(\text{i})}$
$\text{Now Putting y}=0,\text{x}=\frac{\pi}{2}\ \text{in eq. (i)},$ $0=2.\frac{\pi^{2}}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{-\pi^{2}}{2}$
$\text{Putting c}=\frac{-\pi^2}{2}\ \text{in eq. (i)},\ \ \text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
View full question & answer→Question 663 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{y}}(\text{e}^{\text{x}}+\text{e}^{-\text{x}})$
$\Rightarrow\text{e}^{-\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
$\Rightarrow-\text{e}^{-\text{y}}=\text{e}^{\text{x}}-\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$
Hence, $\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$ is the required solution.
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In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{xy} = \text{log}\ \text{y} + \text{C} \ \ : \ \text{y}'=\frac{\text{y}^2}{1-\text{xy}}\ (\text{xy} \neq1)$
AnswerGiven: xy = log y + C ....(i) To prove: y given by eq. (i) is a solution of differential equation $\text{y}'=\frac{\text{y}^2}{1-\text{xy}} \ ...(\text{ii})$ Proof: Differentiating both sides of eq. (i) w.r.t x, we have $\text{xy}'+\text{y}(1)=\frac{1}{\text{y}}\text{y}'+0 \ $ $\Rightarrow \ \text{xy}'-\frac{\text{y}'}{\text{y}}=-\text{y} $ $ \Rightarrow \ \text{y}'\bigg(\text{x}-\frac{\text{1}}{\text{y}}\bigg)=-\text{y}$ $\Rightarrow \ \text{y}'\bigg(\frac{\text{xy}-1}{\text{y}}\bigg)=-\text{y}\ $ $\Rightarrow \ \text{y}'(\text{xy}-1)=-\text{y}^2$ $ \Rightarrow \ \text{y}'=\frac{-\text{y}^2}{\text{xy}-1}$$\Rightarrow \ \ \text{y}'=\frac{-\text{y}^2}{-(1-\text{xy})}=\frac{\text{y}^2}{1-\text{xy}}$
Hence, Function (implicit) given by eq. (i) is a solution of ${\text{y}}'=\frac{\text{y}^2}{1-\text{xy}}.$
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For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2}(-2<\text{y}<2)$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2} \ \Rightarrow \ \ \text{dy}=\sqrt{4-\text{y}^2}\ \text{dx}$
$\Rightarrow \ \frac{\text{dy}}{\sqrt{4-\text{y}^2}}=\text{dx}$
Integrating both sides, $\int \frac{\text{dy}}{\sqrt{4-\text{y}^2}}\ \text{dy}=\int1\ \text{dx}$
$\Rightarrow \ \ \text{sin}^{-1}\frac{\text{y}}{2}=\text{x}+\text{c}\ \ \Bigg[\therefore\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\ \text{dx}=\text{sin}^{-1}\frac{\text{x}}{\text{a}}\Bigg]$
$\frac{\text{y}}{2}=\text{sin}(\text{x+c)} \ \Rightarrow \ \text{y}=2\text{sin}(\text{x+c)}$
View full question & answer→Question 693 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$
AnswerThe given differential equation is
$\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
Separating the variables and integrating,
$\int \text{dy}=\int\text{sin}^{-1}\text{x dx}$
$\therefore\ \int \text{1 dy}= \int \sin^{-1}\text{x}\cdot 1\ \text{dx}$
$\therefore \ \text{y}=(\text{sin}^{-1}\text{x}).\text{x}-\int\frac{1}{\sqrt{1-\text{x}^2}}.\text{x dx}$
$\text{or}\ \text{y = x sin}^{-1}\text{x}+\frac{1}{2}\int(1-\text{x}^2)^{-\frac{1}{2}}(-2\ \text{x}) \ \text{dx}$
$\therefore\ \text{y}=\text{x sin}^{-1}\text{x}+\frac{1}{2}\frac{(1-\text{x}^2)\frac{1}{2}}{\frac{1}{2}}+\text{c}$
$\therefore\ \text{y}= \text{x sin}^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$
which is the required solution.
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Represent the followinf families of curves by forming the corresponding differential equation:
$\text{y}=\text{e}^{\text{ax}}$
AnswerThe equation of the family of curves is
$\text{y}=\text{e}^{\text{ax}}\ ...(1)$
$\Rightarrow\log\text{y}=\text{ax}$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{y}}{\text{x}}$
$\Rightarrow\text{x}\frac{\text{dx}}{\text{dx}}=\text{y}\log\text{y}$
It is the required differential equation.
View full question & answer→Question 713 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
$\Rightarrow\text{dy}=\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\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 723 Marks
Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$Function $\text{y}=\text{e}^\text{-x}+2$
AnswerHere, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to x,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation. so,
$y = e^x+ 1$ is a solution of the equation
put x - 0 in equation (1),
$\Rightarrow y = e^0+ 1 = 2$
$y(0) = 2$
put x = 0 in equation (2),
$y' = e^0 = 1$
$y(0) = 1$
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Differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=1,\text{y}(1)=0$Function $\text{y}=\log\text{x}$
AnswerHere, y = logxDifferentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=1$
so, y=logx is a solution of the equation
If $\text{x}=1,\text{y}=\log1=0$
so,
y(1) = 0
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Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$
AnswerWe have,
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
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Solve the following differential equations:
$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
Answer$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
$\sqrt{1+\text{x}^2}\ \text{dy}=-\sqrt{1+\text{y}^2}\ \text{dx}$
$\int\frac{\text{dy}}{\sqrt{1+\text{y}^2}}=-\int\frac{\text{dx}}{\sqrt{1+\text{x}^2}}$
$\log|\text{y}+\sqrt{1+\text{y}^2}|=-\log|\text{x}+\sqrt{1+\text{x}^2}|=\log|\text{c}|$
$(\text{y}+\sqrt{1+\text{y}^2})(\text{x}+\sqrt{1+\text{x}^2})=\text{c}$
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Solve the following equation
$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
Answer$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
$\frac{\text{y}}{\text{y}-1}\text{dy}=\frac{\text{x}+1}{\text{x}}\ \text{dx}$
$\int\Big(1+\frac{1}{\text{y}-1}\Big)\text{dy}=\int\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$\text{y}+\log|\text{y}-1|=\text{x}+\log|\text{x}|+\text{C}$
$\text{y}-\text{x}=\log|\text{x}|-\log|\text{y}-1|+\text{C}$
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Prove that $\text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2$ is the general solution of differential equation $(\text{x}^3-3\text{x y}^2) \text{dx}=(\text{y}^3-3\text{x}^2\text{y})\text{dy, where c}$ is a parameter.
Answer$\text{Here},\ \ \text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2\ \ ...{(1)}$
Differentiating w.r.t.x, we get,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]$
$\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big]\ \ ...(2)$
Dividing (2) by (1), we get.
$\frac{\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}}{\text{x}^2-\text{y}^2}=\frac{2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)}{\text{x}^2+\text{y}^2}$
$\therefore\ \ \text{x}(\text{x}^2+\text{y}^2)-\text{y}(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}$ $=2\text{x}(\text{x}^2-\text{y}^2)+2\text{y}(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}$
$\therefore\ \ \big[2\text{y}(\text{x}^2-\text{y}^2)+\text{y}(\text{x}^2+\text{y}^2)\big]\frac{\text{dy}}{\text{dx}}$ $=\text{x}(\text{x}^2+\text{y}^2)-2\text{x}(\text{x}^2-\text{y}^2)$
$\therefore\ \ (2\text{x}^2\text{y}-2\text{y}^3+\text{x}^2\text{y}+\text{y}^3)\frac{\text{dy}}{\text{dx}}$ $=\text{x}^3+\text{xy}^2-2\text{x}^3+2\text{xy}^2$
$\therefore\ \ (3\text{x}^2\text{y}-\text{y}^3)\frac{\text{dy}}{\text{dx}}=3\text{xy}^2-\text{x}^3$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^3-3\text{xy}^2}{\text{y}^3-3\text{x}^2\text{y}}$ which is the required equation.
Hence the result.
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Solve the differential equation $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}.$
AnswerWe have $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}\ ....(\text{i})$
Take $\text{x}+\text{y}=\text{t}$
$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=\frac{\text{dt}}{\text{dx}}$
Substituting $\text{x}+\text{y}=\text{t}$ in equation (i) we get,
$\frac{\text{dt}}{\text{dx}}=\text{e}^\text{t}$
$\Rightarrow\text{e}^{-\text{t}\text{dt}}=\text{dx}$
$\Rightarrow-\text{e}^{-\text{t}}=\text{x}+\text{C}$
$\Rightarrow\frac{-1}{\text{e}^\text{x+y}}=\text{x}+\text{C}$
$\Rightarrow-1=(\text{x}+\text{C})\text{e}^\text{x+y}$
$\Rightarrow(\text{x}+\text{C})\text{ e}^\text{x+y}1=0$
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Form the differential equation having $\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$ where A and B are aribitrary constants, as its general solution.
Answer$\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$
$\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}\times\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\text{Ax}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)=0$
$\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)(-2\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)-0$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
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Solve the following differential equations:
$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
Answer$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
$\text{dy}=-(\text{x}+1)(\text{y}+1)\text{dx}$
$\int\frac{\text{dy}}{\text{y}+1}=-\int(\text{x}+1)\text{dx}$
$\log|\text{y}+1|=-\frac{\text{x}^2}{2}-\text{x}+\text{C}$
$\log|\text{y}+1|+\frac{\text{x}^2}{2}+\text{x = C}$
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In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
$y = e^{2x} (a + bx)$
Answer$y = e^{2x} (a + bx)$ ....(1) Differentiating both sides with respect to x, we get: $\text{y}'=2\text{e}^{2\text{x}}(\text{a+bx}) + \text{e}^{2\text{x}}\cdot\text{b}$ $\Rightarrow \text{y}'=\text{e}^{2\text{x}} (2\text{a}+2\text{bx}+\text{b}) \ ....(2)$ MUltiplying equation (1) by (2) and than subtracting it equation (2), we get:$\text{y}'-2\text{y = e}^{2\text{x}} (2\text{a + 2bx + b)}-\text{e}^{2\text{x}}(2\text{a + 2bx})$
$\Rightarrow \text{y}'-2\text{y}=\text{be}^{2\text{x}} \ ...(3)$
Differentiating both sides with respect to x, we get: $\text{y}''-2\text{y}'=2\text{be}^{2\text{x}} \ ...(4)$ Dividing equation (4) by equation (3), we get: $\frac{\text{y}''-2\text{y}'}{\text{y}'-2\text{y}}=2$ $\Rightarrow \text{y}''-2\text{y}'=2\text{y}'-4\text{y}$ $\Rightarrow \text{y}''-4\text{y}'+4\text{y}=0$ This is the required differential equation of the given curve.
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Represent the following families of curves by forming the corresponding differential equation:
$\text{y}^2=4\text{a}(\text{x}-\text{b})$
AnswerThe equation of the family of curves is
$\text{y}^2=4\text{a}(\text{x}-\text{b})\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$
Differentiating (2) with respect to x, we get
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
It is the required differential equation.
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Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$
AnswerWe have $\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$ $\Rightarrow\text{dx}=\frac{1}{\sin^2\text{y}}$ $\Rightarrow\text{dx}=\text{cosec}^2\text{y dy}$ Integrating both sides, we get $\int\text{dx}=\text{cosec}^2\text{y dy}$ $\Rightarrow\text{x}=-\cot\text{y}+\text{C}$$\Rightarrow\text{x}+\cot\text{y}=\text{C}$
Hence, $\Rightarrow\text{x}+\cot\text{y}=\text{C}$ is the required solution.
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For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$ |
: |
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$ |
AnswerThe given differential equation is
$\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$
$\Rightarrow\ \text{e}^{-\text{x}}\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}\ \ ...(1)$
Differentiating (1) twice w.r.t. .x. we get
$\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}(-1)=-\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{and}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\text{e}^{-\text{x}}(-1)-\Big\{-\text{y e}^{-\text{x}}+\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}\Big\}$ $=-\text{a}\cos\text{x}-\text{b}\sin\text{x}$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}=-\text{y x}^{-\text{x}}\ \ [\because \text{of}\ (1)]$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+2\text{y e}^{-\text{x}}=0$ $\ \text{or}\ \text{e}^{-\text{x}}\Big\{\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}\Big\}=0$
$\ \text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence the result.
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Show that $\text{y}=\text{Ae}^{\text{bx}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{\text{bx}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ABe}^{\text{Bx}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{AB}^2\text{e}^{\text{Bx}}$
$=\frac{(\text{ABe}^{\text{Bx}^2})}{(\text{Ae}^{\text{Bx}})}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Hence, the given function is the solution to the given differential equation.
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Solve the following equation:
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
Answer$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
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