Exercise 10.6-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions

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Additional problems
Question 1.
Solve: $(2 \sqrt{x y}) \mathrm{dy}+\mathrm{y} \mathrm{dx}=0$
Solution:
The given equation is $\frac{d y}{d x}=\frac{-y}{2 \sqrt{x y}-x}$
$
\begin{aligned}
\text { Put } y & =v x \\
\text { L.H.S. } & =v+x \frac{d v}{d x} ; \text { R.H.S. }=\frac{-v}{2 \sqrt{v}-1}=\frac{v}{1-2 \sqrt{v}} \\
\therefore v+x \frac{d v}{d x} & =\frac{v}{1-2 \sqrt{v}} \\
\Rightarrow x \frac{d v}{d x} & =\frac{2 v \sqrt{v}}{1-2 \sqrt{v}} \Rightarrow\left(\frac{1-2 \sqrt{v}}{v \sqrt{v}}\right) d v=2 \frac{d x}{x} \\
\Rightarrow-2 v^{-1 / 2}-2 \log v & =2 \log x+2 \log c \\
-v^{-1 / 2} & =\log (v x c) \\
-\sqrt{\frac{x}{y}} & =\log (c y) \Rightarrow c y=e^{-\sqrt{x / y}} \text { (or) } y e^{\sqrt{x / y}}=c
\end{aligned}
$

Question 2 .
Solve: $\left(x^3+3 x y^2\right) d x+\left(y^3+3 x^2 y\right) d y=0$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =-\frac{x^3+3 x y^2}{y^3+3 x^2 y} \\
\text { Put } y & =v x \\
\text { L.H.S. } & =v+x \frac{d v}{d x} ; \text { R.H.S. }=\frac{x^3+3 x y^2}{y^3+3 x^2 y}=-\left(\frac{1+3 v^2}{v^3+3 v}\right)
\end{aligned}
$
$
\begin{aligned}
& \therefore v+x \frac{d v}{d x}=-\left(\frac{1+3 v^2}{v^3+3 v}\right) \\
& \Rightarrow x \frac{d v}{d x}=-\frac{v^4+6 v^2+1}{v^3+3 v} \\
& \Rightarrow \frac{4 d x}{x}=-\frac{4 v^3+12 v}{v^4+6 v^2+1} d v
\end{aligned}
$
Integrating, we have
$
4 \log x=-\log \left(v^4+6 v^2+1\right)+\log c
$
$\log \left[x^4\left(v^4+6 v^2+1\right)\right]=\log c$
i.e., $x^4\left(v^4+6 v^2+1\right)=c$ (or) $y^4+6 x^2 y^2+x^4=c$

Question 3.
Solve: $\frac{d y}{d x}=\frac{y(x-2 y)}{x(x-3 y)}$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =\frac{y(x-2 y)}{x(x-3 y)} \\
\text { Put } y & =v x \\
\text { R.H.S. } & =\frac{v x(x-2 v x)}{x(x-3 v x)}=\frac{x^2 v(1-2 v)}{x^2(1-3 v)}=\frac{v(1-2 v)}{1-3 v}=\frac{v-2 v^2}{1-3 v}
\end{aligned}
$
Which is in terms of $\mathrm{v}$ alone.
$\Rightarrow$ the given problem comes under homogeneous type.

$\begin{aligned}
& \text { Now } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}=\frac{v-2 v^2}{1-3 v} \\
& \therefore x \frac{d v}{d x}=\frac{v-2 v^2}{1-3 v}-v=\frac{v-2 v^2-v+3 v^2}{1-3 v}=\frac{v^2}{1-3 v} \\
& \text { So, } x d v=\frac{v^2}{1-3 v} d x \\
& \frac{1-3 v}{v^2} d v=\frac{d x}{x} ; \quad\left(\frac{1}{v^2}-\frac{3}{v}\right) d v=\frac{d x}{x} \\
& \int\left(\frac{1}{v^2}-\frac{3}{v}\right) d v=\int \frac{d x}{x} ; \quad \frac{-1}{v}-3 \log v=\log x+\log c \\
& \frac{-1}{v}=3 \log v+\log x+\log c=\log c v^3 x\left[\text { But } y=v x \Rightarrow v=\frac{y}{x}\right] \\
& \Rightarrow \frac{-1}{\frac{y}{x}}=\log c x\left(\frac{y^3}{x^3}\right) \\
& y^3=\frac{x^2}{c} e^{\frac{-x}{y}}=k x^2 e^{\frac{-x}{y}}
\end{aligned}$

Question 4.
Solve: $\left(x^2+y^2\right) d y=x y d x$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =\frac{x y}{x^2+y^2} \\
\text { Put } y & =v x \\
\operatorname{RHS} & =\frac{x v x}{x^2+v^2 x^2}=\frac{v x^2}{x^2\left(1+v^2\right)}=\frac{v}{1+v^2}
\end{aligned}
$
Which is a function in $\mathrm{v}$ alone.

$\Rightarrow$ the given problem comes under homogeneous type.
$
\begin{aligned}
y=v x \Rightarrow \frac{d y}{d x} & =v+x \frac{d v}{d x}=\frac{v}{1+v^2} \\
\therefore x \frac{d v}{d x}=\frac{v}{1+v^2}-v & =\frac{v-v-v^3}{1+v^2}(\text { i.e., }) x \frac{d v}{d x}=\frac{-v^3}{1+v^2} \\
\therefore x d v & =\frac{-v^3}{1+v^2} d x \\
\frac{1+v^2}{v^3} d v & =\frac{-d x}{x} \\
\int\left(\frac{1}{v^3}+\frac{1}{v}\right) d v & =\int \frac{-d x}{x} \\
\frac{-1}{2 v^2}+\log v & =-\log x+\log c ; \frac{-1}{2 v^2}+\log v x=\log c \\
\frac{y}{y}=v x \Rightarrow v & =\frac{y}{x} \\
\frac{-1}{2 y^2}+\log y & =\log c ; \log y=\log c+\frac{x^2}{2 y^2} \\
\therefore y & =e
\end{aligned}
$

Question 5.
Find the equation of the curve passing through $(1,0)$ and which has slope $1+\frac{y}{x}$ at $(x, y)$.

Solution:
$
\text { Slope of the curve is } \begin{aligned}
\frac{d y}{d x} & =1+\frac{y}{x} \\
\text { Put } y & =v x \\
\therefore v+x \frac{d v}{d x} & =1+\frac{v x}{x}=1+v
\end{aligned}
$
$
\text { So, } \begin{aligned}
x \frac{d v}{d x} & =1+v-v=1 \\
x d v & =d x \Rightarrow \int d v=\int \frac{d x}{x} \\
v & =\log x+c \text { (i.e.)) } \frac{y}{x}=\log x+c
\end{aligned}
$
Given the curve passes through $(1,0) \Rightarrow$ at $x=1, y=0$
$
\begin{aligned}
& 0=\log 1+c \Rightarrow c=0 \\
& \therefore \frac{y}{x}=\log x \Rightarrow y=x \log x
\end{aligned}
$