Exercise 4.4 - Chapter 4 Differential Equations 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $4.4$
Solve the following:
Question 1.

$\frac{d y}{d x}-\frac{y}{x}=x$
Solution:
The given equation can be written as $\frac{d y}{d x}-\left(\frac{1}{x}\right) y=x$ It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ Here $\mathrm{P}=-\frac{1}{x}$, and $\mathrm{Q}=x$
$
\begin{aligned}
\int \mathrm{P} d x & =\int-\frac{1}{x} d x=-\log x=\log \left(\frac{1}{x}\right) \\
\text { The required solution is } y & =e^{\int p d x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x} \\
& =\int \mathrm{Q}(\text { I.F }) d x+c \\
\frac{y}{x} & =\int x\left(\frac{1}{x}\right) d x+c \\
\frac{y}{x} & =x+c
\end{aligned}
$
Question 2.
$
\frac{d y}{d x}+\mathrm{y} \cos \mathrm{x}=\sin \mathrm{x} \cos \mathrm{x}
$
Solution:
The given equation can be written as $\frac{d y}{d x}+(\cos \mathrm{x}) \mathrm{y}=\sin \mathrm{x} \cos \mathrm{x}$
It is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q} d \mathrm{x}$
Here $P=\cos x, Q=\sin x \cos x$

$\int P d x=\int \cos x d x=\sin x$
$
\text { I.F }=e^{\int p d x}=e^{\sin x}
$
The required solution is $y$ (I.F) $=\int Q($ I.F $) d x+c$
$
y e^{\sin x}=\int \sin x \cos x e^{\sin x} d x+c
$
Put $\sin x=t$. Then $\cos x d x=d t$
(Using integration by parts)
$\begin{aligned} \Rightarrow & y e^{\sin x} & =\sin x e^{\sin x}-e^{\sin x}+c \\ \text { (or) } & y e^{\sin x} & =e^{\sin x}(\sin x-1)+c\end{aligned}$
Question 3.
$
\mathrm{x} \frac{d y}{d x}+2 \mathrm{y}=\mathrm{x}^4
$
Solution:
The given equation can be written as $\frac{d y}{d x}+\frac{2}{x} y=x^3$
It is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Here $\mathrm{P}=\frac{2}{x}$ and $\mathrm{Q}=x^3$
$
\begin{aligned}
\int \mathrm{P} d x & =\int \frac{2}{x} d x=2 \log x=\log x^2 \\
\text { I.F } & =e^{\int p d x}=e^{\log x^2}=x^2
\end{aligned}
$
The required solution is $y$ (I.F) $=\int Q$ (I.F) $d x+c$
$
\begin{aligned}
y x^2 & =\int\left(x^3\right)\left(x^2\right) d x+c \\
x^2 y & =\int x^5 d x+c
\end{aligned}
$
(or) $x^2 y=\frac{x^6}{6}+c$ is the solution

Question 4.
$
\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}
$
Solution:
The given equation is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Where $\mathrm{P}=\frac{3 x^2}{1+x^3}$ and $\mathrm{Q}=\frac{1+x^2}{1+x^3}$
$
\begin{array}{r}
\int \mathrm{P} d x=\int \frac{3 x^2}{1+x^3} d x=\log \left(1+x^3\right) \\
\text { I.F }=e^{\int p d x}=e^{\log \left(1+x^3\right)}=1+x^3
\end{array}
$
The required solution is $y$ (I.F) $=\int Q($ I.F $) d x+c$
$
\begin{aligned}
\left(1+x^3\right) y & =\int \frac{1+x^2}{1+x^3}\left(1+x^3\right) d x+c \\
& =\int\left(1+x^2\right) d x+c \\
y\left(1+x^3\right) & =x+\frac{x^3}{3}+c \text { is the solution }
\end{aligned}
$

Question 5.
$\frac{d y}{d x}+\frac{y}{x}=x e^x$
Solution:
The given equation is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
where $\mathrm{P}=\frac{1}{x}, \mathrm{Q}=x e^x$
Now $\quad \int \mathrm{P} d x=\int \frac{1}{x} d x=\log x$
The solution is
Consider,
$
\begin{aligned}
\int x^2 e^x d x & =x^2 e^x-\int 2 x e^x=x^2 e^x-2\left[x e^x-\int e^x\right] \\
& =x^2 e^x-2 x e^x+2 e^x \quad \text { (Using integration by parts) }
\end{aligned}
$
Thus the solution is
$
\begin{aligned}
x y & =x^2 e^x-2 x e^x+2 e^x+c \\
\text { (or) } x y & =e^x\left(x^2-2 x+2\right)+c
\end{aligned}
$
Question 6.
$
\frac{d y}{d x}+\mathrm{y} \tan \mathrm{x}=\cos ^3 \mathrm{x}
 

Solution:
The given equation is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
where $\mathrm{P}=\tan \mathrm{x}, \mathrm{Q}=\cos ^3 \mathrm{x}$
Now $\int \mathrm{P} \mathrm{dx}=\int \tan \mathrm{x} d \mathrm{x}=\log \sec \mathrm{x}$
$
\begin{aligned}
\text { I.F } & =e^{\int p d x}=e^{\log \sec x}=\sec x \\
y(\text { I.F) } & =\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d x+c \\
y \sec x & =\int \cos ^3 x(\sec x) d x+c \\
& =\int \cos ^2 x d x+c=\int\left(\frac{1+\cos 2 x}{2}\right) d x+c \\
y \sec x & =\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)+c
\end{aligned}
$
Question 7.
If $\frac{d y}{d x}+2 \mathrm{y} \tan \mathrm{x}=\sin \mathrm{x}$ and if $\mathrm{y}=0$ when $\mathrm{x}=\frac{\pi}{3}$ express $\mathrm{y}$ in terms of $\mathrm{x}$

Solution:
$
\frac{d y}{d x}+2 \mathrm{y} \tan \mathrm{x}=\sin \mathrm{x}
$
Here $P=2 \tan x$ and $Q=\sin x$
$\int P d x=\int 2 \tan x=2 \log \sec x=\log \sec ^2 x$
$
\text { I.F }=e^{\int p d x}=e^{\log \sec ^2 x}=\sec ^2 x
$
The solution is $\operatorname{y~sec}^2 \mathrm{x}=\sec \mathrm{x}-2$

Question 8.
$\frac{d y}{d x}+\frac{y}{x}=x e^x$
Solution:
$
\begin{aligned}
& \frac{d y}{d x}+\frac{y}{x}=x e^x \\
& \mathrm{P}=\frac{1}{x}, \mathrm{Q}=x e^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 { The solution is }=\int x^2 e^x d x+c \\
& x y=e^x\left(x^2-2 x+2\right)+c
\end{aligned}
$
Question 9.
A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. A man invests $₹ 1,00,000$ in the bank deposit which accrues interest, $8 \%$ per year compounded continuously. How much will he get after 10 years? $\left(\mathrm{e}^{0.8}=2.2255\right)$

Solution:
Let $\mathrm{P}$ be the principal at time ' $\mathrm{t}$ '
According to the given condition,
$
\begin{aligned}
\frac{d P}{d t} & =\frac{8}{100} \mathrm{P}=0.08 \mathrm{P} \\
\int \frac{d p}{p} & =\int 0.08 d t+c \\
\log \mathrm{P} & =0.08 t+c \\
\mathrm{P} & =e^{0.08 t} e^c=c_1 e^{0.08 t}
\end{aligned}
$
Given $\quad \mathrm{P}=1,00,000$ when $t=0$
$
\begin{array}{ll}
\Rightarrow & c_1=1,00,000 \\
\Rightarrow & \mathrm{P}=1,00,000 e^{0.08 t}
\end{array}
$
To find $\mathrm{P}$ when $t=10$
$
\begin{aligned}
& P=100000 e^{0.8}=1,00,000(2.2255) \\
& P=222,550

$\text { The man will get } ₹ 2,22,550 \text { after } 10 \text { years. }$
\end{aligned}
$