Examples (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations
Example 1
Find the order and degree, if defined, of each of the following differential equations:
(i) $\frac{d y}{d x}-\cos x=0$
(ii) $x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$
(iii) $y^{\prime \prime \prime}+y^2+e^{y^{\prime}}=0$
Solution
(i) The highest order derivative present in the differential equation is $\frac{d y}{d x}$, so its order is one. It is a polynomial equation in $y^{\prime}$ and the highest power raised to $\frac{d y}{d x}$ is one, so its degree is one.
(ii) The highest order derivative present in the given differential equation is $\frac{d^2 y}{d x^2}$, so its order is two. It is a polynomial equation in $\frac{d^2 y}{d x^2}$ and $\frac{d y}{d x}$ and the highest power raised to $\frac{d^2 y}{d x^2}$ is one, so its degree is one.
(iii) The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.
Example 2
Verify that the function $y=e^{-3 x}$ is a solution of the differential equation
$
\frac{d^2 y}{d x^2}+\frac{d y}{d x}-6 y=0
$
Solution
Given function is $y=e^{-3 x}$. Differentiating both sides of equation with respect to $x$, we get
$
\frac{d y}{d x}=-3 e^{-3 x}
$
Now, differentiating (1) with respect to $x$, we have
$
\frac{d^2 y}{d x^2}=9 e^{-3 x}
$
Substituting the values of $\frac{d^2 y}{d x^2}, \frac{d y}{d x}$ and $y$ in the given differential equation, we get L.H.S. $=9 e^{-3 x}+\left(-3 e^{-3 x}\right)-6 . e^{-3 x}=9 e^{-3 x}-9 e^{-3 x}=0=$ R.H.S..
Therefore, the given function is a solution of the given differential equation.
Example 3
Verify that the function $y=a \cos x+b \sin x$, where, $a, b \in \mathbf{R}$ is a solution of the differential equation $\frac{d^2 y}{d x^2}+y=0$
Solution
The given function is
$
y=a \cos x+b \sin x
$
Differentiating both sides of equation (1) with respect to $x$, successively, we get
$
\begin{aligned}
\frac{d y}{d x} & =-a \sin x+b \cos x \\
\frac{d^2 y}{d x^2} & =-a \cos x-b \sin x
\end{aligned}
$
Substituting the values of $\frac{d^2 y}{d x^2}$ and $y$ in the given differential equation, we get L.H.S. $=(-a \cos x-b \sin x)+(a \cos x+b \sin x)=0=$ R.H.S.
Therefore, the given function is a solution of the given differential equation.
Example 4
Find the general solution of the differential equation $\frac{d y}{d x}=\frac{x+1}{2-y},(y \neq 2)$
Solution
We have
$
\frac{d y}{d x}=\frac{x+1}{2-y}
$
Separating the variables in equation (1), we get
$
(2-y) d y=(x+1) d x
$
Integrating both sides of equation (2), we get
$
\begin{aligned}
\int(2-y) d y & =\int(x+1) d x \\
2 y-\frac{y^2}{2} & =\frac{x^2}{2}+x+\mathrm{C}_1
\end{aligned}
$
or
$
x^2+y^2+2 x-4 y+2 C_1=0
$
or
$
x^2+y^2+2 x-4 y+\mathrm{C}=0 \text {, where } \mathrm{C}=2 \mathrm{C}_1
$
which is the general solution of equation (1).
Example 5
Find the general solution of the differential equation $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$.
Solution
Since $1+y^2 \neq 0$, therefore separating the variables, the given differential equation can be written as
$
\frac{d y}{1+y^2}=\frac{d x}{1+x^2}
$
Integrating both sides of equation (1), we get
$
\begin{aligned}
\int \frac{d y}{1+y^2} & =\int \frac{d x}{1+x^2} \\
\tan ^{-1} y & =\tan ^{-1} x+\mathrm{C}
\end{aligned}
$
or
which is the general solution of equation (1).
Example 6
Find the particular solution of the differential equation $\frac{d y}{d x}=-4 x y^2$ given that $y=1$, when $x=0$.
Solution
If $y \neq 0$, the given differential equation can be written as
$
\frac{d y}{y^2}=-4 x d x
$
Integrating both sides of equation (1), we get
$
\int \frac{d y}{y^2}=-4 \int x d x
$
or
$
\begin{aligned}
-\frac{1}{y} & =-2 x^2+\mathrm{C} \\
y & =\frac{1}{2 x^2-\mathrm{C}}
\end{aligned}
$
or
Substituting $y=1$ and $x=0$ in equation (2), we get, $\mathrm{C}=-1$.
Now substituting the value of $\mathrm{C}$ in equation (2), we get the particular solution of the given differential equation as $y=\frac{1}{2 x^2+1}$.
Example 7
Find the equation of the curve passing through the point $(1,1)$ whose differential equation is $x d y=\left(2 x^2+1\right) d x(x \neq 0)$.
Solution
The given differential equation can be expressed as
$
\begin{aligned}
d y^* & =\left(\frac{2 x^2+1}{x}\right) d x * \\
d y & =\left(2 x+\frac{1}{x}\right) d x
\end{aligned}
$
or Integrating both sides of equation (1), we get
$
\begin{aligned}
\int d y & =\int\left(2 x+\frac{1}{x}\right) d x \\
y & =x^2+\log |x|+\mathrm{C}
\end{aligned}
$
or
Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point $(1,1)$. Therefore substituting $x=1, y=1$ in equation (2), we get $\mathrm{C}=0$.
Now substituting the value of $\mathrm{C}$ in equation (2) we get the equation of the required curve as $y=x^2+\log |x|$.
Example 8
Find the equation of a curve passing through the point $(-2,3)$, given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2 x}{y^2}$.
Solution
We know that the slope of the tangent to a curve is given by $\frac{d y}{d x}$.
so,
$
\frac{d y}{d x}=\frac{2 x}{y^2}
$
Separating the variables, equation (1) can be written as
$
y^2 d y=2 x d x
$
Integrating both sides of equation (2), we get
$
\int y^2 d y=\int 2 x d x
$
or
$
\frac{y^3}{3}=x^2+\mathrm{C}
$
Substituting $x=-2, y=3$ in equation (3), we get $\mathrm{C}=5$.
Substituting the value of $\mathrm{C}$ in equation (3), we get the equation of the required curve as
$
\frac{y^3}{3}=x^2+5 \quad \text { or } \quad y=\left(3 x^2+15\right)^{\frac{1}{3}}
$
Example 9
In a bank, principal increases continuously at the rate of $5 \%$ per year. In how many years Rs 1000 double itself?
Solution
Let $\mathrm{P}$ be the principal at any time $t$. According to the given problem,
$
\begin{aligned}
& \frac{d p}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P} \\
& \frac{d p}{d t}=\frac{\mathrm{P}}{20}
\end{aligned}
$
or separating the variables in equation (1), we get
$
\frac{d p}{\mathrm{P}}=\frac{d t}{20}
$
Integrating both sides of equation (2), we get
$
\log \mathrm{P}=\frac{t}{20}+\mathrm{C}_1
$
or
$
\mathrm{P}=e^{\frac{t}{20}} \cdot e^{\mathrm{C}_1}
$
or
$
\begin{aligned}
& \mathrm{P}=\mathrm{C} e^{\frac{t}{20}}\left(\text { where } e^{\mathrm{C}_1}=\mathrm{C}\right) \\
& \mathrm{P}=1000, \text { when } t=0
\end{aligned}
$
Now
Substituting the values of $\mathrm{P}$ and $t$ in (3), we get $\mathrm{C}=1000$. Therefore, equation (3), gives
$
\mathrm{P}=1000 e^{\frac{t}{20}}
$
Let $t$ years be the time required to double the principal. Then
$
2000=1000 e^{\frac{t}{20}} \Rightarrow t=20 \log _e 2
$
Example 10
Show that the differential equation $(x-y) \frac{d y}{d x}=x+2 y$ is homogeneous and solve it.
Solution
The given differential equation can be expressed as
$
\frac{d y}{d x}=\frac{x+2 y}{x-y}
$
Let
$
\mathrm{F}(x, y)=\frac{x+2 y}{x-y}
$
Now
$
\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda(x+2 y)}{\lambda(x-y)}=\lambda^0 \cdot f(x, y)
$
Therefore, $\mathrm{F}(x, y)$ is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.
Alternatively,
$
\frac{d y}{d x}=\left(\frac{1+\frac{2 y}{x}}{1-\frac{y}{x}}\right)=g\left(\frac{y}{x}\right)
$
R.H.S. of differential equation (2) is of the form $g\left(\frac{y}{x}\right)$ and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution
$
y=v x
$
Differentiating equation (3) with respect to, $x$ we get
$
\frac{d y}{d x}=v+x \frac{d v}{d x}
$
Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1) we get
$
v+x \frac{d v}{d x}=\frac{1+2 v}{1-v}
$
or
$
x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v
$
or
$
x \frac{d v}{d x}=\frac{v^2+v+1}{1-v}
$
or
$
\frac{v-1}{v^2+v+1} d v=\frac{-d x}{x}
$
Integrating both sides of equation (5), we get
$
\int \frac{v-1}{v^2+v+1} d v=-\int \frac{d x}{x}
$
or
$
\frac{1}{2} \int \frac{2 v+1-3}{v^2+v+1} d v=-\log |x|+\mathrm{C}_1
$
$
\frac{1}{2} \int \frac{2 v+1}{v^2+v+1} d v-\frac{3}{2} \int \frac{1}{v^2+v+1} d v=-\log |x|+\mathrm{C}_1
$
or
$
\frac{1}{2} \log \left|v^2+v+1\right|-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d v=-\log |x|+\mathrm{C}_1
$
or
$
\begin{aligned}
& \frac{1}{2} \log \left|v^2+v+1\right|-\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+\mathrm{C}_1 \\
& \frac{1}{2} \log \left|v^2+v+1\right|+\frac{1}{2} \log x^2=\sqrt{3} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)+\mathrm{C}_1
\end{aligned}
$
(Why?)
Replacing $v$ by $\frac{y}{x}$, we get
or
$
\frac{1}{2} \log \left|\frac{y^2}{x^2}+\frac{y}{x}+1\right|+\frac{1}{2} \log x^2=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+\mathrm{C}_1
$
or
$
\frac{1}{2} \log \left|\left(\frac{y^2}{x^2}+\frac{y}{x}+1\right) x^2\right|=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C_1
$
or
$
\log \left|\left(y^2+x y+x^2\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 \mathrm{C}_1
$
or
$
\log \left|\left(x^2+x y+y^2\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+\mathrm{C}
$
which is the general solution of the differential equation (1)
Example 11
Show that the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ is homogeneous and solve it.
Solution
The given differential equation can be written as
$
\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}
$
It is a differential equation of the form $\frac{d y}{d x}=\mathrm{F}(x, y)$.
Here
$
\mathrm{F}(x, y)=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}
$
Replacing $x$ by $\lambda x$ and $y$ by $\lambda y$, we get
$
\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left[y \cos \left(\frac{y}{x}\right)+x\right]}{\lambda\left(x \cos \frac{y}{x}\right)}=\lambda^0[\mathrm{~F}(x, y)]
$
Thus, $\mathrm{F}(x, y)$ is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation.
To solve it we make the substitution
$
y=v x
$
Differentiating equation (2) with respect to $x$, we get
$$
\frac{d y}{d x}=v+x \frac{d v}{d x}
$$
Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1), we get
$
v+x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v}
$
or
$
\begin{aligned}
& x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v}-v \\
& x \frac{d v}{d x}=\frac{1}{\cos v}
\end{aligned}
$
or
$
\cos v d v=\frac{d x}{x}
$
Therefore
$
\int \cos v d v=\int \frac{1}{x} d x
$
or
$
\sin v=\log |x|+\log |C|
$
or
$
\sin v=\log |\mathrm{C} x|
$
Replacing $v$ by $\frac{y}{x}$, we get
$
\sin \left(\frac{y}{x}\right)=\log |\mathrm{C} x|
$
which is the general solution of the differential equation (1).
Example 12
Show that the differential equation $2 y e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0$ is homogeneous and find its particular solution, given that, $x=0$ when $y=1$.
Solution
The given differential equation can be written as
$
\frac{d x}{d y}=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}}
$
Let
Then
$
\begin{gathered}
\mathrm{F}(x, y)=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}} \\
\mathrm{~F}(\lambda x, \lambda y)=\frac{\lambda\left(2 x e^{\frac{x}{y}}-y\right)}{\lambda\left(2 y e^{\frac{x}{y}}\right)}=\lambda^0[\mathrm{~F}(x, y)]
\end{gathered}
$
Thus, $\mathrm{F}(x, y)$ is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve it, we make the substitution
$
x=v y
$
Differentiating equation (2) with respect to $y$, we get
$
\frac{d x}{d y}=v+y \frac{d v}{d y}
$
Substituting the value of $x$ and $\frac{d x}{d y}$ in equation (1), we get
$
v+y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}
$
or
$
y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}-v
$
or
$
y \frac{d v}{d y}=-\frac{1}{2 e^v}
$
or
$
2 e^v d v=\frac{-d y}{y}
$
or
$
\int 2 e^v \cdot d v=-\int \frac{d y}{y}
$
or
$
2 e^v=-\log |y|+\mathrm{C}
$
and replacing $v$ by $\frac{x}{y}$, we get
$
2 e^{\frac{x}{y}}+\log |y|=\mathrm{C}
$
Substituting $x=0$ and $y=1$ in equation (3), we get
$
2 e^0+\log |1|=\mathrm{C} \Rightarrow \mathrm{C}=2
$
Substituting the value of $\mathrm{C}$ in equation (3), we get
$
2 e^{\frac{x}{y}}+\log |y|=2
$
which is the particular solution of the given differential equation.
Example 13
Show that the family of curves for which the slope of the tangent at any point $(x, y)$ on it is $\frac{x^2+y^2}{2 x y}$, is given by $x^2-y^2=c x$.
Solution
We know that the slope of the tangent at any point on a curve is $\frac{d y}{d x}$.
$\text { Therefore, } \quad \frac{d y}{d x}=\frac{x^2+y^2}{2 x y}$
or
$
\frac{d y}{d x}=\frac{1+\frac{y^2}{x^2}}{\frac{2 y}{x}}
$
Clearly, (1) is a homogenous differential equation. To solve it we make substitution
$
y=v x
$
Differentiating $y=v x$ with respect to $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =v+x \frac{d v}{d x} \\
v+x \frac{d v}{d x} & =\frac{1+v^2}{2 v}
\end{aligned}
$
or
$
\begin{aligned}
x \frac{d v}{d x} & =\frac{1-v^2}{2 v} \\
\frac{2 v}{1-v^2} d v & =\frac{d x}{x}
\end{aligned}
$
or
$
\frac{2 v}{v^2-1} d v=-\frac{d x}{x}
$
Therefore
$
\int \frac{2 v}{v^2-1} d v=-\int \frac{1}{x} d x
$
or
$
\log \left|v^2-1\right|=-\log |x|+\log \left|C_1\right|
$
or
$
\log \left|\left(v^2-1\right)(x)\right|=\log \left|C_1\right|
$
or
$
\left(v^2-1\right) x= \pm \mathrm{C}_1
$
Replacing $v$ by $\frac{y}{x}$, we get
$
\left(\frac{y^2}{x^2}-1\right) x= \pm \mathrm{C}_1
$
or
$
\left(y^2-x^2\right)= \pm \mathrm{C}_1 x \text { or } x^2-y^2=\mathrm{C} x
$
Example 14
Find the general solution of the differential equation $\frac{d y}{d x}-y=\cos x$.
Solution
Given differential equation is of the form
$
\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \text {, where } \mathrm{P}=-1 \text { and } \mathrm{Q}=\cos x
$
Therefore
$
\text { I. } \mathrm{F}=e^{\int-1 d x}=e^{-x}
$
Multiplying both sides of equation by I.F, we get
$
e^{-x} \frac{d y}{d x}-e^{-x} y=e^{-x} \cos x
$
or
$
\frac{d y}{d x}\left(y e^{-x}\right)=e^{-x} \cos x
$
On integrating both sides with respect to $x$, we get
$
y e^{-x}=\int e^{-x} \cos x d x+\mathrm{C}
$
Let
$
\begin{aligned}
\mathrm{I} & =\int e^{-x} \cos x d x \\
& =\cos x\left(\frac{e^{-x}}{-1}\right)-\int(-\sin x)\left(-e^{-x}\right) d x
\end{aligned}
$
$
\begin{aligned}
& =-\cos x e^{-x}-\int \sin x e^{-x} d x \\
& =-\cos x e^{-x}-\left[\sin x\left(-e^{-x}\right)-\int \cos x\left(-e^{-x}\right) d x\right] \\
& =-\cos x e^{-x}+\sin x e^{-x}-\int \cos x e^{-x} d x \\
\mathrm{I} & =-e^{-x} \cos x+\sin x e^{-x}-\mathrm{I} \\
2 \mathrm{I} & =(\sin x-\cos x) e^{-x} \\
\mathrm{I} & =\frac{(\sin x-\cos x) e^{-x}}{2}
\end{aligned}
$
Substituting the value of I in equation (1), we get
$
\begin{aligned}
y e^{-x} & =\left(\frac{\sin x-\cos x}{2}\right) e^{-x}+\mathrm{C} \\
y & =\left(\frac{\sin x-\cos x}{2}\right)+\mathrm{C} e^x
\end{aligned}
$
which is the general solution of the given differential equation.
Example 15
Find the general solution of the differential equation $x \frac{d y}{d x}+2 y=x^2(x \neq 0)$. Solution The given differential equation is
$
x \frac{d y}{d x}+2 y=x^2
$
Dividing both sides of equation (1) by $x$, we get
$
\frac{d y}{d x}+\frac{2}{x} y=x
$
which is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, where $\mathrm{P}=\frac{2}{x}$ and $\mathrm{Q}=x$.
So
$
\text { I.F }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2\left[\text { as } e^{\log f(x)}=f(x)\right]
$
Therefore, solution of the given equation is given by
$
y \cdot x^2=\int(x)\left(x^2\right) d x+\mathrm{C}=\int x^3 d x+\mathrm{C}
$
or
$
y=\frac{x^2}{4}+\mathrm{C} x^{-2}
$
which is the general solution of the given differential equation.
Example 16
Find the general solution of the differential equation $y d x-\left(x+2 y^2\right) d y=0$.
Solution
The given differential equation can be written as
$
\frac{d x}{d y}-\frac{x}{y}=2 y
$
This is a linear differential equation of the type $\frac{d x}{d y}+\mathrm{P}_1 x=\mathrm{Q}_1$, where $\mathrm{P}_1=-\frac{1}{y}$ and $\mathrm{Q}_1=2 y$. Therefore I.F $=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log (y)^{-1}}=\frac{1}{y}$
Hence, the solution of the given differential equation is
$
x \frac{1}{y}=\int(2 y)\left(\frac{1}{y}\right) d y+\mathrm{C}
$
or
$
\frac{x}{y}=\int(2 d y)+\mathrm{C}
$
or
$
\frac{x}{y}=2 y+C
$
or
$
x=2 y^2+\mathrm{C} y
$
which is a general solution of the given differential equation.
Example 17
Find the particular solution of the differential equation
$
\frac{d y}{d x}+y \cot x=2 x+x^2 \cot x(x \neq 0)
$
given that $y=0$ when $x=\frac{\pi}{2}$.
Solution
The given equation is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, where $\mathrm{P}=\cot x$ and $\mathrm{Q}=2 x+x^2 \cot x$. Therefore
$
\text { I.F }=e^{\int \cot x d x}=e^{\log \sin x}=\sin x
$
Hence, the solution of the differential equation is given by
$
y \cdot \sin x=\int\left(2 x+x^2 \cot x\right) \sin x d x+\mathrm{C}
$
or
$
\begin{aligned}
& y \sin x=\int 2 x \sin x d x+\int x^2 \cos x d x+\mathrm{C} \\
& y \sin x=\sin x\left(\frac{2 x^2}{2}\right)-\int \cos x\left(\frac{2 x^2}{2}\right) d x+\int x^2 \cos x d x+\mathrm{C}
\end{aligned}
$
Or
$
y \sin x=x^2 \sin x-\int x^2 \cos x d x+\int x^2 \cos x d x+\mathrm{C}
$
or
$
y \sin x=x^2 \sin x+C
$
Substituting $y=0$ and $x=\frac{\pi}{2}$ in equation (1), we get
$
\begin{aligned}
& 0=\left(\frac{\pi}{2}\right)^2 \sin \left(\frac{\pi}{2}\right)+C \\
& C=\frac{-\pi^2}{4}
\end{aligned}
$
Substituting the value of $\mathrm{C}$ in equation (1), we get
$
y \sin x=x^2 \sin x-\frac{\pi^2}{4}
$
or
$
y=x^2-\frac{\pi^2}{4 \sin x}(\sin x \neq 0)
$
which is the particular solution of the given differential equation.
Example 18
Find the equation of a curve passing through the point $(0,1)$. If the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$ coordinate (abscissa) and the product of the $x$ coordinate and $y$ coordinate (ordinate) of that point.
Solution
We know that the slope of the tangent to the curve is $\frac{d y}{d x}$.
Therefore,
$
\frac{d y}{d x}=x+x y
$
or
$
\frac{d y}{d x}-x y=x
$
This is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, where $\mathrm{P}=-x$ and $\mathrm{Q}=x$.
Therefore,
$
\text { I.F }=e^{\int-x d x}=e^{\frac{-x^2}{2}}
$
Hence, the solution of equation is given by
$
\begin{aligned}
y \cdot e^{\frac{-x^2}{2}} & =\int(x)\left(e^{\frac{-x^2}{2}}\right) d x+\mathrm{C} \\
\mathrm{I} & =\int(x) e^{\frac{-x^2}{2}} d x
\end{aligned}
$
Let
Let $\frac{-x^2}{2}=t$, then $-x d x=d t$ or $x d x=-d t$.
Therefore, $\quad \mathrm{I}=-\int e^t d t=-e^t=-e^{\frac{-x^2}{2}}$
Substituting the value of I in equation (2), we get
$
\begin{aligned}
y e^{\frac{-x^2}{2}} & =-e^{\frac{-x^2}{2}}+\mathrm{C} \\
y & =-1+\mathrm{C} e^{\frac{x^2}{2}}
\end{aligned}
$
Now (3) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through $(0,1)$. Substituting $x=0$ and $y=1$ in equation $(3)$ we get
$
1=-1+\mathrm{C} \cdot e^0 \text { or } \mathrm{C}=2
$
Substituting the value of $\mathrm{C}$ in equation (3), we get
$
y=-1+2 e^{\frac{x^2}{2}}
$
which is the equation of the required curve.
Example 19
Verify that the function $y=c_1 e^{a x} \cos b x+c_2 e^{a x} \sin b x$, where $c_1, c_2$ are arbitrary constants is a solution of the differential equation
$
\frac{d^2 y}{d x^2}-2 a \frac{d y}{d x}+\left(a^2+b^2\right) y=0
$
Solution
The given function is
$
y=e^{a x}\left[c_1 \cos b x+c_2 \sin b x\right]
$
Differentiating both sides of equation (1) with respect to $x$, we get
$
\frac{d y}{d x}=e^{a x}\left[-b c_1 \sin b x+b c_2 \cos b x\right]+\left[c_1 \cos b x+c_2 \sin b x\right] e^{a x} \cdot a
$
or $\quad \frac{d y}{d x}=e^{a x}\left[\left(b c_2+a c_1\right) \cos b x+\left(a c_2-b c_1\right) \sin b x\right]$
Differentiating both sides of equation (2) with respect to $x$, we get
$
\begin{aligned}
\frac{d^2 y}{d x^2}= & e^{a x}\left[\left(b c_2+a c_1\right)(-b \sin b x)+\left(a c_2-b c_1\right)(b \cos b x)\right] \\
& +\left[\left(b c_2+a c_1\right) \cos b x+\left(a c_2-b c_1\right) \sin b x\right] e^{a x} \cdot a \\
= & e^{a x}\left[\left(a^2 c_2-2 a b c_1-b^2 c_2\right) \sin b x+\left(a^2 c_1+2 a b c_2-b^2 c_1\right) \cos b x\right]
\end{aligned}
$
Substituting the values of $\frac{d^2 y}{d x^2}, \frac{d y}{d x}$ and $y$ in the given differential equation, we get
$
\begin{aligned}
\text { L.H.S. }= & \left.e^{a x}\left[a^2 c_2-2 a b c_1-b^2 c_2\right) \sin b x+\left(a^2 c_1+2 a b c_2-b^2 c_1\right) \cos b x\right] \\
& -2 a e^{a x}\left[\left(b c_2+a c_1\right) \cos b x+\left(a c_2-b c_1\right) \sin b x\right] \\
& +\left(a^2+b^2\right) e^{a x}\left[c_1 \cos b x+c_2 \sin b x\right] \\
= & e^{a x}\left[\begin{array}{l}
\left(a^2 c_2-2 a b c_1-b^2 c_2-2 a^2 c_2+2 a b c_1+a^2 c_2+b^2 c_2\right) \sin b x \\
+\left(a^2 c_1+2 a b c_2-b^2 c_1-2 a b c_2-2 a^2 c_1+a^2 c_1+b^2 c_1\right) \cos b x
\end{array}\right] \\
= & e^{a x}[0 \times \sin b x+0 \cos b x]=e^{a x} \times 0=0=\text { R.H.S. }
\end{aligned}
$
Hence, the given function is a solution of the given differential equation.
Example 20
Find the particular solution of the differential equation $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ given that $y=0$ when $x=0$.
Solution
The given differential equation can be written as
$
\frac{d y}{d x}=e^{(3 x+4 y)}
$
or
$
\frac{d y}{d x}=e^{3 x} \cdot e^{4 y}
$
Separating the variables, we get
$
\frac{d y}{e^{4 y}}=e^{3 x} d x
$
Therefore
$
\int e^{-4 y} d y=\int e^{3 x} d x
$
or
$
\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+\mathrm{C}
$
or
$
4 e^{3 x}+3 e^{-4 y}+12 \mathrm{C}=0
$
Substituting $x=0$ and $y=0$ in (2), we get
$
4+3+12 \mathrm{C}=0 \text { or } \mathrm{C}=\frac{-7}{12}
$
Substituting the value of $\mathrm{C}$ in equation (2), we get
$
4 e^{3 x}+3 e^{-4 y}-7=0
$
which is a particular solution of the given differential equation.
Example 21
Solve the differential equation
$
(x d y-y d x) y \sin \left(\frac{y}{x}\right)=(y d x+x d y) x \cos \left(\frac{y}{x}\right) .
$
Solution
The given differential equation can be written as
$
\begin{aligned}
& {\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x} \\
& \frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}
\end{aligned}
$
Dividing numerator and denominator on RHS by $x^2$, we get
$
\frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y^2}{x^2}\right) \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}
$
Clearly, equation (1) is a homogeneous differential equation of the form $\frac{d y}{d x}=g\left(\frac{y}{x}\right)$. To solve it, we make the substitution
$
y=v x
$
or
$
\begin{gathered}
\frac{d y}{d x}=v+x \frac{d v}{d x} \\
v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}
\end{gathered}
$
(using (1) and (2))
or
$
x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v}
$
or
$
\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=\frac{2 d x}{x}
$
Therefore $\int\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=2 \int \frac{1}{x} d x$
or
$
\int \tan v d v-\int \frac{1}{v} d v=2 \int \frac{1}{x} d x
$
or $\log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|$
or
$
\log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|
$
or
$
\frac{\sec v}{v x^2}= \pm \mathrm{C}_1
$
Replacing $v$ by $\frac{y}{x}$ in equation (3), we get
$
\frac{\sec \left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)\left(x^2\right)}=\mathrm{C} \text { where, } \mathrm{C}= \pm \mathrm{C}_1
$
or
$
\sec \left(\frac{y}{x}\right)=\mathrm{C} x y
$
which is the general solution of the given differential equation.
Example 22
Solve the differential equation
$
\left(\tan ^{-1} y-x\right) d y=\left(1+y^2\right) d x .
$
Solution
The given differential equation can be written as
$
\frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}
$
Now (1) is a linear differential equation of the form $\frac{d x}{d y}+\mathrm{P}_1 x=\mathrm{Q}_1$,
where, $\quad \mathrm{P}_1=\frac{1}{1+y^2}$ and $\mathrm{Q}_1=\frac{\tan ^{-1} y}{1+y^2}$.
Therefore, $\quad$ I.F $=e^{\int \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$
Thus, the solution of the given differential equation is
$
x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^2}\right) e^{\tan ^{-1} y} d y+\mathrm{C}
$
Let
$
\mathrm{I}=\int\left(\frac{\tan ^{-1} y}{1+y^2}\right) e^{\tan ^{-1} y} d y
$
Substituting $\tan ^{-1} y=t$ so that $\left(\frac{1}{1+y^2}\right) d y=d t$, we get
$
\mathrm{I}=\int t e^t d t=t e^t-\int 1 \cdot e^t d t=t e^t-e^t=e^t(t-1)
$
or
$
\mathrm{I}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)
$
Substituting the value of I in equation (2), we get
$
x . e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+\mathrm{C}
$
or
$
x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y}
$
which is the general solution of the given differential equation.
