Exercise 4.3 - Chapter 4 Differential Equations 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $4.3$
Solve the following homogeneous differential equations
Question 1

$\mathrm{x} \frac{d y}{d x}=\mathrm{x}+\mathrm{y}$
Solution:
$
x \frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}=\frac{x+y}{x}
$
It is a homogeneous differential equation,
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ the equation becomes,
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{x+v x}{x}=\frac{(1+v) x}{x} \\
x & =1+v \\
x \frac{d v}{d x} & =1+v-v=1 \\
d v & =\frac{d x}{x}
\end{aligned}
$
On integration
$
\begin{aligned}
\int d v & =\int \frac{d x}{x} \\
v & =\log x+\log c
\end{aligned}
$
Now, replace
$
\begin{aligned}
v & =\frac{y}{x} \\
\Rightarrow \frac{y}{x} & =\log x+\log c \\
\frac{y}{x} & =\log c x \quad \text { (or) } \quad c x=e^{\frac{y}{x}} \\
x & =\frac{1}{c} e^{\frac{y}{x}} \\
x & =k e^{\frac{y}{x}}, k=\frac{1}{c}
\end{aligned}
$

Question 2.
$(\mathrm{x}-\mathrm{y}) \frac{d y}{d x}=\mathrm{x}+3 \mathrm{y}$

Solution:
$
\begin{aligned}
(x-y) \frac{d y}{d x} & =x+3 y \\
\frac{d y}{d x} & =\frac{x+3 y}{x-y}
\end{aligned}
$
This is a homogeneous equation
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
We have,
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{x+3 v x}{x-v x}=\frac{x(1+3 v)}{x(1-v)}=\frac{1+3 v}{1-v} \\
x \frac{d v}{d x} & =\frac{1+3 v}{1-v}-v \\
& =\frac{1+3 v-v+v^2}{1-v}=\frac{1+2 v+v^2}{1-v}
\end{aligned}
$
Separating the variables,
$
\frac{1-v}{1+2 v+v^2} d v=\frac{d x}{x}
$
On integration,
$
\int \frac{1-v}{1+2 v+v^2} d v=\int \frac{1}{x} d x
$
Writing $1-v=2-1-v=2-(v+1)$
We get, $\int \frac{2-(v+1)}{(v+1)^2} d v=\int \frac{1}{x} d x$

$\begin{aligned}
& \int \frac{2}{(v+1)^2} d v-\int \frac{1}{v+1} d v=\int \frac{1}{x} d x \\
& \frac{-2}{v+1}-\log (v+1)=\log x+\log c \\
& \frac{-2}{v+1}=\log [c x(v+1) \\
& \text { Replace } v=\frac{y}{x} \\
& \frac{-2}{\frac{y}{x}+1}=\log \left[c x\left(\frac{y}{x}+1\right)\right] \\
& \frac{-2 x}{x+y}=\log \left[c x \frac{(x+y)}{x}\right] \\
& \Rightarrow \quad c(x+y)=e^{\frac{-2 x}{x+y}} \\
& x+y=k e^{\frac{-2 x}{x+y}}, k=\frac{1}{c} \\
&
\end{aligned}$

Question 3.
$x \frac{d y}{d x}-y=\sqrt{x^2+y^2}$
Solution:
$
\begin{aligned}
x \frac{d y}{d x}-y & =\sqrt{x^2+y^2} \\
\frac{d y}{d x} & =\frac{\sqrt{x^2+y^2}+y}{x}
\end{aligned}
$
It is a homogeneous equation
Now put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
We get,
$
\begin{aligned}
& \text { We get, } \begin{aligned}
v+x \frac{d v}{d x} & =\frac{\sqrt{x^2+v^2 x^2}+v x}{x} \\
& =\frac{x\left[\sqrt{1+v^2}+v\right]}{x}=\sqrt{1+v^2}+v \\
\Rightarrow \quad x \frac{d v}{d x} & =\sqrt{1+v^2} \text { (or) } \frac{d v}{\sqrt{1+v^2}}=\frac{d x}{x}
\end{aligned}
\end{aligned}
$

On integration,
$
\begin{aligned}
& \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x} \\
& \log \left(v+\sqrt{1+v^2}\right)=\log x+\log c \\
& v+\sqrt{1+v^2}=c x \\
& \text { Replace } v=\frac{y}{x} \\
& \frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=c x \\
& \frac{y}{x}+\sqrt{\frac{x^2+y^2}{x^2}}=c x \\
& \Rightarrow \quad y+\sqrt{x^2+y^2}=c x^2 \text { is the required solution }
\end{aligned}
$
Question 4.
$
\frac{d y}{d x}=\frac{3 x-2 y}{2 x-3 y}
$
Solution:
$
\frac{d y}{d x}=\frac{3 x-2 y}{2 x-3 y}
$
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
We get, $v+x \frac{d v}{d x} \quad=\frac{3 x-2 v x}{2 x-3 v x}=\frac{x(3-2 v)}{x(2-3 v)}$
$
\begin{aligned}
x \frac{d v}{d x} & =\frac{3-2 v}{2-3 v}-v \\
& =\frac{3-2 v-2 v+3 v^2}{2-3 v}=\frac{3-4 v+3 v^2}{2-3 v}
\end{aligned}
$
Separating the variables,
$
\frac{2-3 v}{3-4 v+3 v^2} d v=\frac{d x}{x}
$

Integrating,
$
\begin{aligned}
\int \frac{23}{3-4 v+3 v} d v & =\int \frac{d x}{x} \\
-\frac{1}{2} \int \frac{6 v-4}{3 v^2-4 v+3} d v & =\int \frac{d x}{x} \\
-\frac{1}{2} \log \left(3 v^2-4 v+3\right) & =\log x+\log c
\end{aligned}
$
$
\frac{1}{\sqrt{3 v^2-4 v+3}}=c x
$
Replace $v=\frac{y}{x}$ we get,
$
\frac{1}{\sqrt{3 \frac{y^2}{x^2}-4 \frac{y}{x}+3}}=c x
$
Squaring both the sides,
$
\begin{gathered}
\frac{1}{3 \frac{y^2}{x^2}-4 \frac{y}{x}+3}=c^2 x^2 \\
\frac{x^2}{3 y^2-4 x y+3 x^2}=c^2 x^2
\end{gathered}
$
(or) $3 y^2-4 x y+3 x^2=k$ is the required solution

Question 5.
$
\left(y^2-2 x y\right) d x=\left(x^2-2 x y\right) d y
$
Solution:
$
\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-2 x y}
$
This is a homogeneous differential equation
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$ in (1)
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{v^2 x^2-2 x^2 v}{x^2-2 x^2 v}=\frac{v^2-2 v}{1-2 v} \\
x \frac{d v}{d x} & =\frac{v^2-2 v}{1-2 v}-v=\frac{v^2-2 v-v+2 v^2}{1-2 v} \\
& =\frac{3 v^2-3 v}{1-2 v}=\frac{3\left(v^2-v\right)}{1-2 v}
\end{aligned}
$

On separating the variables,
$
\begin{aligned}
& \frac{1-2 v}{3\left(v^2-v\right)} d v=\frac{d x}{x} \\
& -\frac{1}{3} \int \frac{2 v-1}{v^2-v} d v=\int \frac{d x}{x} \\
& -\frac{1}{3} \log \left(v^2-v\right)=\log x+\log c \\
& \log \left(v^2-v\right)^{-\frac{1}{3}}=\log c x \\
& \frac{1}{\left(v^2-v\right)^{\frac{1}{3}}}=c x \\
& \text { Replace } v=\frac{y}{x} \\
& \frac{1}{\left(\frac{y^2}{x^2}-\frac{y}{x}\right)^{\frac{1}{3}}}=c x \\
& {\left[\frac{\left(x y^2-x^2 y\right)}{x^3}\right]^{\frac{1}{3}} c x=1} \\
& \left(x y^2-x^2 y\right)^{\frac{1}{3}} c=1 \\
& x y^2-x^2 y=\frac{1}{c^3} \text { (or) }\left(x y-y^2\right) x=-\frac{1}{c^3} \\
& \left(x y-y^2\right) x=k \text {, where } k=-\frac{1}{c^3} \\
&
\end{aligned}
$
Question 6.
The slope of the tangent to a curve at any point $(x, y)$ on it is given by $\left(y^3-2 y x^2\right) d x+\left(2 x y^2-x^3\right)$ $d y=0$ and the curve passes through $(1,2)$. Find the equation of the curve.
Solution:

Given the slope is $\left(y^3-2 y x^2\right) d x+\left(2 x y^2-x^3\right) d y=0$
$
\begin{aligned}
\left(y^3-2 y x^2\right) d x & =-\left(2 x y^2-x^3\right) d y \\
\frac{d y}{d x} & =\frac{y^3-2 y x^2}{x^3-2 x y^2}
\end{aligned}
$
This is a homogeneous differential equation
$
\text { Put } \begin{aligned}
y=v x \text { and } \frac{d y}{d x} & =v+x \frac{d v}{d x} \\
v+x \frac{d v}{d x} & =\frac{v^3 x^3-2 v x^3}{x^3-2 x^3 v^2}=\frac{v^3-2 v}{1-2 v^2} \\
x \frac{d v}{d x} & =\frac{v^3-2 v}{1-2 v^2}-v \\
& =\frac{v^3-2 v-v+2 v^3}{1-2 v^2}=\frac{3 v^3-3 v}{1-2 v^2}
\end{aligned}
$
Separating the variables,
$
\begin{gathered}
\frac{1-2 v^2}{3\left(v^3-v\right)} d v=\frac{d x}{x} \\
\frac{1}{3} \int \frac{1-2 v^2}{v\left(v^2-1\right)} d v=\int \frac{d x}{x}
\end{gathered}
$
Using partial fraction method,
$
\begin{aligned}
& \frac{1-2 v^2}{v\left(v^2-1\right)}=\frac{1-2 v^2}{v(v+1)(v-1)}=\frac{\mathrm{A}}{v}+\frac{\mathrm{B}}{v+1}+\frac{\mathrm{C}}{v-1} \\
& 1-2 v^2=\mathrm{A}(v+1)(v-1)+\mathrm{B} v(v-1)+\mathrm{C} v(v+1) \\
&
\end{aligned}
$

$\begin{aligned}
& \text { Put, } v=0 \Rightarrow \mathrm{A}=-1 \\
& v=1 \Rightarrow \mathrm{C}=-\frac{1}{2} \\
& v=-1 \Rightarrow \mathrm{B}=-\frac{1}{2} \\
& \frac{1}{3}\left[\int \frac{-1}{v} d v-\int \frac{1}{2(v+1)} d v-\frac{1}{2} \int \frac{1}{v-1} d v\right]=\int \frac{d x}{x} \\
& \frac{1}{3}\left[-\log v-\frac{1}{2} \log (v+1)-\frac{1}{2} \log (v-1)\right]=\log x+\log c \\
& -\frac{1}{3}[\log v+\log \sqrt{v+1}+\log \sqrt{v-1}]=\log c x \\
& -\frac{1}{3} \log v \sqrt{v^2-1}=\log c x
\end{aligned}$

$
\begin{aligned}
& \Rightarrow \quad \frac{1}{\left(v \sqrt{v^2-1}\right)^{\frac{1}{3}}}=c x \\
& \frac{1}{\left(v \sqrt{v^2-1}\right)}=c^3 x^3 \\
& \frac{1}{\frac{y}{x} \sqrt{\frac{y^2-x^2}{x^2}}}=c^3 x^3 \Rightarrow \frac{x^2}{y \sqrt{y^2-x^2}}=c^3 x^3 \\
& y \sqrt{y^2 x^2}=\frac{1}{c^3 x}=\frac{1}{k x}, k=c^3 \\
&
\end{aligned}
$
The curve passes through $(1,2)(i e) x=1, y=2$
$
2 \sqrt{3}=\frac{1}{k}
$
Thus, $\quad y \sqrt{y^2-x^2}=\frac{2 \sqrt{3}}{x}$ (or) $\quad x y \sqrt{y^2-x^2}=2 \sqrt{3}$ is the required solution
Question 7.
An electric manufacturing company makes small household switches. The company estimates the marginal revenue function for these switches to be $\left(x^2+y^2\right) d y=x y d x$ where $x$ represents the number of units (in thousands). What is the total revenue function?
Solution:

Given that MR is $\left(x^2+y^2\right) d y+x y d x$
(ie) $\mathrm{MR}=\frac{d y}{d x}=\frac{x y}{x^2+y^2}$
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{v x^2}{x^2+x^2 v^2}=\frac{v}{1+v^2} \\
x \frac{d v}{d x} & =\frac{v}{1+v^2}-v \\
& =\frac{v-v-v^3}{1+v^2}=\frac{-v^3}{1+v^2}
\end{aligned}
$
Separating the variables,
$
\begin{aligned}
\frac{-\left(1+v^2\right)}{v^3} d v & =\frac{d x}{x} \\
-\int \frac{1+v^2}{v^3} d v & =\int \frac{d x}{x} \\
-\int \frac{1}{v^3} d v-\int \frac{1}{v} d v & =\int \frac{d x}{x} \\
\frac{1}{2 v^2}-\log v & =\log x+\log c \\
\log v+\log x+\log c & =\frac{1}{2 v^2} \\
v x c & =e^{\frac{1}{2 v^2}} \\
\text { Replace } v=\frac{y}{x}\left(\frac{y}{x}\right) x c & =e^{\frac{x^2}{2 y^2}} \text { (or) } y=\frac{1}{c} e^{\frac{x^2}{2 y^2}} \\
y & =k e^{\frac{x^2}{2 y^2}}, \frac{1}{c}=k \text { is the required total revenue function }
\end{aligned}
$