Exercise 10.7 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.7$
Solve the following Linear differential equations:
Question 1.
$
\cos x \frac{d y}{d x}+y \sin x=1
$
Solution:
$
\cos x \frac{d y}{d x}+y \sin x=1
$
Dividing throughout by ' $\cos \mathrm{x}$ '
$
\frac{d y}{d x}+y \tan x=\sec x
$
This is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ Where $P=\tan x$, $Q=\sec x$
$
\begin{aligned}
\int \mathrm{P} d x & =\int \tan x d x=\log \sec x \\
\text { I.F. }=e^{\int \mathrm{P} d x} & =e^{\log \sec x}=\sec x \\
\text { Solution is } y e^{\int \mathrm{P} d x} & =\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c \\
y \sec x & =\int \sec x \cdot \sec x \cdot d x+c \\
y \sec x & =\int \sec ^2 x d x+c \\
y \sec x & =\tan x+c(\div \sec x) \\
y & =\sin x+c \cos x
\end{aligned}
$
Question 2.
$
\left(1-x^2\right) \frac{d y}{d x}-x y=1
$
Solution:

$
\left(1-x^2\right) \frac{d y}{d x}-x y=1
$
Dividing throughout by $\left(1-x^2\right)$
$
\frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{1}{1-x^2}
$
This is a linear differential equation of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Where $\mathrm{P}=\frac{-x}{1-x^2}$ and $\mathrm{Q}=\frac{1}{1-x^2}$
$
\begin{aligned}
& \int \mathrm{P} d x=\frac{1}{2} \int \frac{-2 x}{1-x^2} d x=\frac{1}{2} \log \left(1-x^2\right)=\log \left(1-x^2\right)^{1 / 2} \\
& \text { I.F }=e^{\int P d x}=e^{\log \left(1-x^2\right)^{1 / 2}}=\left(1-x^2\right)^{1 / 2}=\sqrt{1-x^2} \\
& \text { Solution is } y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c \\
& y \sqrt{1-x^2}=\int \frac{1}{1-x^2} \sqrt{1-x^2} d x+c \\
& y \sqrt{1-x^2}=\int \frac{1}{\sqrt{1-x^2}} d x+c \\
& y \sqrt{1-x^2}=\sin ^{-1} x+c \quad\left(\div \sqrt{1-x^2}\right) \\
& y=\frac{\sin ^{-1} x}{\sqrt{1-x^2}}+c\left(1-x^2\right)^{\frac{-1}{2}} \\
&
\end{aligned}
$
Question 3.
$
\frac{d y}{d x}+\frac{y}{x}=\sin x
$
Solution:

$
\frac{d y}{d x}+\frac{y}{x}=\sin x
$
This is a linear differential equation of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$
$
\begin{aligned}
& \text { Where } \mathrm{P}=\frac{1}{x} \text { and } \mathrm{Q}=\sin x \\
& \int \mathrm{P} d x=\int \frac{1}{x} d x=\log x \\
& \text { I.F. }=e^{\int P d x}=e^{\log x}=x \\
& \text { Solution is } y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c \\
& y x=\int x \sin x d x+c \\
& \text { Consider } \int x \sin x d x \\
& x y=-x \cos x+\sin x+c \\
& x y+x \cos x=\sin x+c \\
& x(y+\cos x)=\sin x+c \\
&
\end{aligned}
$
Question 4.
$
\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}
$

Solution:
$
\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}
$
Dividing throughout by $\left(x^2+1\right)$
$
\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{\sqrt{x^2+4}}{x^2+1}
$
This is a Linear differential equation of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ where $\mathbf{P}=\frac{2 x}{1+x^2}$ and $\mathrm{Q}=\frac{\sqrt{x^2+4}}{x^2+1}$
$
\begin{aligned}
& \int \mathrm{P} d x=\int \frac{2 x}{1+x^2} d x=\log \left(1+x^2\right) \quad\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x)\right] \\
& \text { I.F }=e^{\int \mathrm{P} d x}=e^{\log \left(1+x^2\right)}=1+x^2 \\
& \text { Solution is } y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c \\
& y\left(1+x^2\right)=\int \frac{\sqrt{x^2+4}}{x^2+1}\left(1+x^2\right) d x+c \\
& y\left(1+x^2\right)=\int \sqrt{x^2+4} d x+c \\
& \int \sqrt{a^2+x^2} d x=\frac{x}{2} \sqrt{a^2+x^2}+\frac{a^2}{2} \log \left(x+\sqrt{a^2+x^2}\right)+c \\
& \therefore y\left(1+x^2\right)=\frac{x}{2} \sqrt{x^2+4}+2 \log \left(x+\sqrt{x^2+4}\right)+c \\
&
\end{aligned}
$
Question 5.
$
\left(2 x-10 y^3\right) d y+y d x=0
$
Solution:

$
\begin{aligned}
\left(2 x-10 y^3\right) d y+y d x & =0 \\
y d x & =\left(10 y^3-2 x\right) d y \\
y \frac{d x}{d y} & =10 y^3-2 x \\
y \frac{d x}{d y}+2 x & =10 y^3 \quad(\div y) \\
\frac{d x}{d y}+\frac{2 x}{y} & =10 y^2
\end{aligned}
$
This is a Linear differential equation of the form $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$ where $\mathrm{P}=\frac{2}{y}$ and $\mathrm{Q}=10 y^2$
$
\begin{gathered}
\int \mathrm{P} d y=2 \int \frac{1}{y} d y=2 \log y=\log y^2 \\
\text { I.F. }=e^{\int \mathrm{P} d y}=e^{\log y^2}=y^2
\end{gathered}
$
Solution is $\quad x e^{\int \mathrm{P} d y}=\int \mathrm{Q} e^{\int \mathrm{P} d y} d y+c$
$
\begin{aligned}
& x y^2=\int 10 y^2\left(y^2\right) d y+c \\
& x y^2=10 \int y^4 d y+c \\
& x y^2=10 \frac{y^5}{5}+c \Rightarrow x y^2=2 y^5+c
\end{aligned}
$
Question 6.
$
x \sin x \frac{d y}{d x}+(x \cos x+\sin x) y=\sin x
$
Solution:

$
x \sin x \frac{d y}{d x}+(x \cos x+\sin x) y=\sin x
$
Dividing throughout by ' $x \sin x$ '
$
\frac{d y}{d x}+\left(\cot x+\frac{1}{x}\right) y=\frac{1}{x}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+P y=Q$
$
\begin{aligned}
& \text { where } \mathrm{P}=\cot x+\frac{1}{x} \text { and } \mathrm{Q}=\frac{1}{x} \\
& \int \mathrm{P} d x=\int\left(\cot x+\frac{1}{x}\right) d x=\log \sin x+\log x=\log (x \sin x) \\
& \text { I.F }=e^{\int \mathrm{P} d x}=e^{\log (x \sin x)}=x \sin x \\
&
\end{aligned}
$
Solution is $\quad y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c$
$
\begin{aligned}
y(x \sin x) & =\int \frac{1}{x} \cdot x \sin x d x+c \\
y x \sin x & =\int \sin x d x+c \\
y x \sin x & =-\cos x+c \\
x y \sin x+\cos x & =c
\end{aligned}
$
Question 7.
$
\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^2}=0
$
Solution:

$
\begin{aligned}
\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^2} & =0 \\
\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y} & =-\sqrt{1-x^2} \\
\frac{d y}{d x} & =-\frac{\left(y-e^{\sin ^{-1} x}\right)}{\sqrt{1-x^2}} \\
\frac{d y}{d x}+\frac{y}{\sqrt{1-x^2}} & =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}
\end{aligned}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
$
\text { where } \begin{array}{r}
\mathrm{P}=\frac{1}{\sqrt{1-x^2}} \text { and } \mathrm{Q}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} \\
\int P d x=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x \\
\text { I.F } e^{\int P d x}=e^{\sin ^{-1} x}
\end{array}
$
Solution is $y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{Pdx}} d x+c$
$
\begin{aligned}
& y e^{\sin ^{-1} x}=\int \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} e^{\sin ^{-1} x} d x+c \\
& y e^{\sin ^{-1} x}=\int \frac{\left(e^{\sin ^{-1} x}\right)^2}{\sqrt{1-x^2}} d x+c \\
& \text { Consider } \int \frac{\left(e^{\sin ^{-1} x}\right)^2}{\sqrt{1-x^2}} d x
\end{aligned}
$
Put $t=\sin ^{-1} x$

$
\begin{aligned}
d t & =\frac{1}{\sqrt{1-x^2}} d x \\
\therefore \int \frac{\left(e^{\sin ^{-1} x}\right)^2}{\sqrt{1-x^2}} d x & =\int\left(e^t\right)^2 d t \\
& =\int e^{2 t} d t \\
& =\frac{e^{2 t}}{2}=\frac{e^{2 \sin ^{-1} x}}{2}
\end{aligned}
$
Substituting in (1) $\Rightarrow y e^{\sin ^{-1} x}=\frac{e^{2 \sin ^{-1} x}}{2}+c$
Question 8.
$
\frac{d y}{d x}+\frac{y}{(1-x) \sqrt{x}}=1-\sqrt{x}
$
Solution:

$
\frac{d y}{d x}+\frac{y}{(1-x) \sqrt{x}}=1-\sqrt{x}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ where $\mathrm{P}=\frac{1}{(1-x) \sqrt{x}}$ and $\mathrm{Q}=1-\sqrt{x}$
Let $u=\sqrt{x} \Rightarrow d u=\frac{1}{2 \sqrt{x}} d x$
$
\begin{aligned}
2 d u & =\frac{d x}{\sqrt{x}} \\
\therefore \int \frac{1}{(1-x) \sqrt{x}} d x & =2 \int \frac{d u}{1-u^2} \\
& =2 \frac{1}{2}[\log |(u+1)|-\log |(u-1)|]=\log \left|\frac{(u+1)}{(u-1)}\right|=\log \left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right| \\
\text { I.F. } e^{\int P d x} & =e^{\log \left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right|}=\left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right|
\end{aligned}
$
Solution is $y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c$
$
\begin{aligned}
y\left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right| & =\int\left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right|(1-\sqrt{x}) d x+c \\
y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right) & =\int(\sqrt{x}+1) d x+c=\frac{2}{3} x^{3 / 2}+x+c \\
\therefore y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right) & =\frac{2}{3} x \sqrt{x}+x+c
\end{aligned}
$

Question 9.
$
\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0
$
Solution:
$
\begin{aligned}
\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right) & =0 \\
\left(1+x+x y^2\right) \frac{d y}{d x} & =-\left(y+y^3\right) \\
\frac{d x}{d y} & =-\frac{\left(1+x+x y^2\right)}{y+y^3} \\
\frac{d x}{d y} & =-\frac{1}{y\left(1+y^2\right)}-\frac{x\left(1+y^2\right)}{y\left(1+y^2\right)} \\
\frac{d x}{d y}+\frac{x}{y} & =-1 / y\left(1+y^2\right)
\end{aligned}
$
This is a Linear differential equation of the type $\frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}$
$
\begin{aligned}
& \text { where } \mathrm{P}=\frac{1}{y} \text { and } \mathrm{Q}=-1 / y\left(1+y^2\right) \\
& \int \mathrm{P} d y=\int \frac{1}{y} d y=\log y \\
& \text { I.F }=e^{\int \mathrm{P} d y}=e^{\log y}=y
\end{aligned}
$
Solution is $x e^{\int \mathrm{P} d y}=\int \mathrm{Q} e^{\int \mathrm{P} d y} d y+c$
$
x y=-\int \frac{1}{y\left(1+y^2\right)} y d y+c
$

$
\begin{aligned}
x y & =-\int \frac{1}{1+y^2} d y+c \\
x y & =-\tan ^{-1} y+c \\
x y+\tan ^{-1} y & =c
\end{aligned}
$
Question 10.
$
\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x}
$
Solution:
$
\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$
where $\mathrm{P}=\frac{1}{x \log x}$ and $\mathrm{Q}=\frac{\sin 2 x}{\log x}$
$
\begin{aligned}
e^{\int P d x} & =\int \frac{1}{x \log x} d x=\log (\log x) \quad\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x)\right] \\
\text { I.F }=e^{\int P d x} & =e^{\log (\log x)=\log x}
\end{aligned}
$
Solution is $y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c$
$
\begin{aligned}
& y \log x=\int \frac{\sin 2 x}{\log x} \log x d x+c \\
& y \log x=\int \sin 2 x d x+c
\end{aligned}
$

$
\begin{aligned}
y \log x= & =-\frac{\cos 2 x}{2}+c \\
y \log x+\frac{\cos 2 x}{2} & =c
\end{aligned}
$
Question 11.
$
(x+a) \frac{d y}{d x}-2 y=(x+a)^4
$
Solution:

$
(x+a) \frac{d y}{d x}-2 y=(x+a)^4
$
Dividing throughout by $(x+a)$
$
\frac{d y}{d x}-\frac{2 y}{(x+a)}=(x+a)^3
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ where $\mathrm{P}=\frac{-2}{(x+a)}$ and $\mathrm{Q}=(x+a)^3$
$
\begin{aligned}
& \int \mathrm{P} d x=-2 \int \frac{d x}{x+a}=-2 \log (x+a)=\log \frac{1}{(x+a)^2} \\
& \text { I.F }=e^{\int \mathrm{P} d x}=e^{\log \frac{1}{(x+a)^2}}=\frac{1}{(x+a)^2}
\end{aligned}
$
Solution is $y e^{\int \mathrm{Pdx}}=\int \mathrm{Q} e^{\int \mathrm{Pdx}} d x+c$
$
\begin{aligned}
& y \frac{1}{(x+a)^2}=\int(x+a)^3 \frac{1}{(x+a)^2} d x+c \\
& y \frac{1}{(x+a)^2}=\int(x+a) d x+c \\
& \frac{y}{(x+a)^2}=\frac{(x+a)^2}{2}+c \\
& 2 y=(x+a)^4+2 c(x+a)^2
\end{aligned}
$
Question 12.
$
\frac{d y}{d x}=\frac{\sin ^2 x}{1+x^3}-\frac{3 x^2}{1+x^3} y
$
Solution:

$
\begin{aligned}
\frac{d y}{d x} & =\frac{\sin ^2 x}{1+x^3}-\frac{3 x^2}{1+x^3} y \\
\frac{d y}{d x}+\frac{3 x^2 y}{1+x^3} & =\frac{\sin ^2 x}{1+x^3}
\end{aligned}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$
where $\mathrm{P}=\frac{3 x^2}{1+x^3}$ and $\mathrm{Q}=\frac{\sin ^2 x}{1+x^3}$
$
\begin{aligned}
\int \mathrm{P} d x & =\int \frac{3 x^2}{1+x^3} d x=\log \left(1+x^3\right) \quad\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x)\right] \\
\text { I.F }= & e^{\int \mathrm{P} d x}=e^{\log \left(1+x^3\right)}=1+x^3
\end{aligned}
$
Solution is $y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c$
$
\begin{aligned}
& y\left(1+x^3\right)=\int \frac{\sin ^2 x}{1+x^3}\left(1+x^3\right) d x+c \\
& y\left(1+x^3\right)=\int \sin ^2 x d x+c \\
& y\left(1+x^3\right)=\int\left(\frac{1-\cos 2 x}{2}\right) d x+c \\
& y\left(1+x^3\right)=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]+c \\
& y\left(1+x^3\right)=\frac{x}{2}-\frac{\sin 2 x}{4}+c
\end{aligned}
$
Question 13.
$
x \frac{d y}{d x}+y=x \log x
$
Solution:

$
x \frac{d y}{d x}+y=x \log x
$
Dividing throughout by ' $x$ '
$
\frac{d y}{d x}+\frac{y}{x}=\log x
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
where $\mathrm{P}=\frac{1}{x}$ and $\mathrm{Q}=\log x$
$
\begin{aligned}
\int \mathrm{P} d x & =\int \frac{1}{x} d x=\log x \\
\text { I.F } & =e^{\int \mathrm{P} d x}=e^{\log x}=x
\end{aligned}
$
Solution is
$
\begin{aligned}
y e^{\int P d x} & =\int Q e^{\int P d x} d x+c \\
y x & =\int x \log x d x+c
\end{aligned}
$
Now, Using integration by parts method to integrate $\int x \log x d x$
Let $u=\log x \quad \int d v=\int x d x$
$
\begin{aligned}
& d u=\frac{1}{x} d x \quad v=\frac{x^2}{2} \\
& \int u d v=u v-\int v d u \\
& \int x \log x d x=\frac{x^2}{2} \log x-\int \frac{x^2}{2} \frac{1}{x} d x \\
& =\frac{x^2}{2} \log x-\frac{x^2}{4}+c=\frac{x^2}{2}\left[\log x-\frac{1}{2}\right]+c \\
&
\end{aligned}
$
$\therefore$ The solution is $x y=\frac{x^2}{2}\left[\log x-\frac{1}{2}\right]+c$ (or) $4 x y=2 x^2 \log x-x^2+4 c$

Question 14.
$
x \frac{d y}{d x}+2 y-x^2 \log x=0
$
Solution:
$
\begin{aligned}
x \frac{d y}{d x}+2 y-x^2 \log x & =0 \\
x \frac{d y}{d x}+2 y & =x^2 \log x \\
\frac{d y}{d x}+\frac{2 y}{x} & =x \log x
\end{aligned}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
$
\begin{aligned}
& \text { where } \mathrm{P}=\frac{2}{x} \text { and } \mathrm{Q}=x \log x \\
& \int \mathrm{P} d x=2 \int \frac{1}{x} d x=2 \log x=\log x^2 \\
& \text { I.F. }=e^{\int \mathrm{P} d x}=e^{\log x^2}=x^2
\end{aligned}
$
Solution is $y e^{\int P d x}=\int Q e^{\int P d x} d x+c$
$
\begin{aligned}
& y\left(x^2\right)=\int(x \log x) x^2 d x+c \\
& y\left(x^2\right)=\int x^3 \log x d x+c
\end{aligned}
$

Using integration by parts method
Let $u=\log x \quad \int d v=\int x^3 d x$
$
\begin{aligned}
& d u=\frac{1}{x} d x \quad v=\frac{x^4}{4} \\
& \int u d v=u v-\int v d u \\
& \int x^3 \log x d x=\frac{x^4}{4} \log x-\int \frac{x^4}{4} \frac{1}{x} d x=\frac{x^4}{4} \log x-\frac{x^4}{16}+c \\
&
\end{aligned}
$
$\therefore$ The solution is $y x^2=\frac{x^4}{4} \log x-\frac{x^4}{16}+c$
Question 15.
$\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^2}$, given that $y=2$ when $x=1$
Solution:

$
\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^2}
$
This is a Linear differential equation of the type $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ where $\mathrm{P}=\frac{3}{x}$ and $\mathrm{Q}=\frac{1}{x^2}$
$
\begin{aligned}
& \int \mathrm{P} d x=3 \int \frac{1}{x} d x=3 \log x=\log x^3 \\
& \text { I.F }=e^{\int \mathrm{P} d x}=e^{\log x^3}=x^3
\end{aligned}
$
Solution is $y e^{\int P d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c$
$
\begin{aligned}
& y x^3=\int \frac{1}{x^2} x^3 d x+c \\
& y x^3=\int x d x+c \\
& y x^3=\frac{x^2}{2}+c
\end{aligned}
$
Given $y=2$ when $x=1$
(2) $(1)^3=\frac{1}{2}+c \Rightarrow c=2-\frac{1}{2}=\frac{3}{2}$
$\therefore$ The solution is $y x^3=\frac{x^2}{2}+\frac{3}{2}$
$
2 x^3 y=x^2+3
$