Exercise 4.2 - Chapter 4 Differential Equations 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $4.2$
Question 1.
Solve
(i) $\frac{d y}{d x}=a e^y$
(ii) $\frac{1+x^2}{1+y}=x y \frac{d y}{d x}$
Solution:
(i) $\frac{d y}{d x}=a e^y$
Separating the variables, we get $\frac{d y}{e^y}=a d x$
Integrating we get
$
\begin{aligned}
\int \frac{d y}{e^y} & =\int a d x \\
\int e^{-y} d y & =\int a d x \\
-e^{-y} & =a x+c
\end{aligned}
$
(or) $e^{-y}+a x+c=0$ is the general solution
(ii) $\frac{1+x^2}{1+y}=x y \frac{d y}{d x}$
Separating the variables,
$\begin{array}{ll}\frac{\left(1+x^2\right) d x}{x} d x=y(1+y) d y \\ \Rightarrow & \left(\frac{1}{x}+x\right) d x=\left(y+y^2\right) d x\end{array}$
$
\int\left(\frac{1}{x}+x\right) d x=\int\left(y+y^2\right) d x
$
$\Rightarrow \quad \log x+\frac{x^2}{2}=\frac{y^2}{2}+\frac{y^3}{3}+c$ is the general solution

Question 2.
Solve: $\mathrm{y}(1-\mathrm{x})-\mathrm{x} \frac{d y}{d x}=0$
Solution:
$
\mathrm{y}(1-\mathrm{x})-\mathrm{x} \frac{d y}{d x}=0
$
Separating the variables,
$
\begin{aligned}
y(1-x) & =\frac{x d y}{d x} \\
\frac{(1-x)}{x} d x & =\frac{d y}{y}
\end{aligned}
$
Integrating both the sides,
$
\int\left(\frac{1}{x}-1\right) d x=\int \frac{d y}{y}
$
$\log \mathrm{x}-\mathrm{x}=\log \mathrm{y}+\mathrm{c}$ is the general solution
Question 3.
Solve:
(i) $y d x-x d y=0$
(ii) $\frac{d y}{d x}+\mathrm{e}^{\mathrm{x}}+\mathrm{y} \mathrm{e}^{\mathrm{x}}=0$
Solution:
(i) $y d x-x d y=0$
Separating the variables,
$y d x=x d y$
Integrating,
$
\int \frac{1}{y+1} d y=\int-e^x d x
$
$\log (\mathrm{y}+1)=-\mathrm{e}^{\mathrm{x}}+\mathrm{c}$ is the general solution
Question 4.
Solve: $\cos x(1+\cos y) d x+\sin y(1+\sin x) d y=0$
Solution:
$\cos x(1+\cos y) d x+\sin y(1+\sin x) d y=0$
Separating the variables,
$\cos x(1+\cos y) d x=-\sin y(1+\sin x) d y$

$
\frac{\cos x}{1+\sin x} d x=\frac{-\sin y}{1+\cos y} d y
$
Integrating, we get
$
\begin{aligned}
\int \frac{\cos x}{1+\sin x} d x & =\int \frac{-\sin y}{1+\cos y} d y \\
\log (1+\sin x) & =\log (1+\cos y)+\log c \\
(\because d(1+\sin x)=\cos x ; & d(1+\cos y)=-\sin y) \\
\Rightarrow \quad 1+\sin x & =c(1+\cos y) \text { is the required solution }
\end{aligned}
$
Question 5.
Solve: $(1-\mathrm{x}) \mathrm{dy}-(1+\mathrm{y}) \mathrm{dx}=0$
Solution:
$
(1-x) d y-(1+y) d x=0
$
Separating the variables,
$
\begin{aligned}
& (1-\mathrm{x}) \mathrm{dy}=(1+\mathrm{y}) \mathrm{dx} \\
& \frac{d y}{1+y}=\frac{d x}{1-x}
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{1}{1+y} d y=\int \frac{1}{1-x} d x \\
& \log (\mathrm{y}+1)=-\log (1-\mathrm{x})+\log \mathrm{c} \\
& \log (\mathrm{y}+1)+\log (1-\mathrm{x})=\log \mathrm{c} \\
& (\mathrm{y}+1)(1-\mathrm{x})=\mathrm{c}
\end{aligned}
$
Question 6.
Solve:
(i) $\frac{d y}{d x}=\mathrm{y} \sin 2 \mathrm{x}$
(ii) $\log \left(\frac{d y}{d x}\right)=\mathrm{ax}+\mathrm{by}$

Solution:
(i) $\frac{d y}{d x}=\mathrm{y} \sin 2 \mathrm{x}$
Separating the variables, $\frac{d y}{y}=\sin 2 \mathrm{xdx}$
Integrating, we get
$\int \frac{1}{y} d y=\int \sin 2 x d x$
$\log y=-\frac{\cos 2 x}{2}+c$ is the required solution
(ii) $\log \left(\frac{d y}{d x}\right)=a x+b y$
$
\begin{aligned}
& \Rightarrow \quad \frac{d y}{d x}=e^{a x+b y} \\
& \Rightarrow \quad \frac{d y}{d x}=e^{a x} e^{b y} \\
& \frac{d y}{e^{b y}}=e^{a x} d x \\
&
\end{aligned}
$
Integrating, we get
(or)
$
\begin{aligned}
\int \frac{1}{e^{b y}} d y & =\int e^{a x} d x \\
\int e^{-b y} d y & =\int e^{a x} d x \\
\frac{e^{-b y}}{-b} & =\frac{e^{a x}}{a}+c \\
\frac{e^{a x}}{a} & =-\frac{e^{-b y}}{b}+c
\end{aligned}
$
Question 7.
Find the curve whose gradient at any point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ on it is $\frac{x-a}{y-b}$ and which passes through the origin.
Solution:
The gradient at any point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ on the curve is given by $\frac{d y}{d x}$

According to the problem
$
\frac{d y}{d x}=\frac{x-a}{y-b}
$

Separating the variables,
$(y-b) d y=(x-a) d x$
Integrating,
$\int(y-b) d y=\int(x-a) d x$
$
\frac{(y-b)^2}{2}=\frac{(x-a)^2}{2}+c
$
The curve passes through the origin, $y=0, x=0$
So
$
\begin{aligned}
\frac{b^2}{2} & =\frac{a^2}{2}+c \\
c & =\frac{b^2-a^2}{2}
\end{aligned}
$
$
\Rightarrow \quad c=\frac{b^2-a^2}{2}
$
Thus the curve becomes
$
\frac{(y-b)^2}{2}=\frac{(x-a)^2}{2}+\frac{b^2-a^2}{2}
$
(or) $(y-b)^2=(x-a)^2+b^2-a^2$ is the required equation of the curve.