#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 26 Maths Textbook Solution.

Answer:$\mathrm{c}|2 x-y|^{\frac{5}{8}}\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}=e^{-\frac{3}{8} \tan ^{-1\left(\frac{y}{2 x}\right)}} \\$

Given: $\left(x^{2}+y^{2}\right) \frac{d y}{d x}=8 x^{2}-3 x y+2 y^{2}$

To solve:  We have to solve the given differential equation.

Hint: In homogeneous equation we put $x=vy$ and $\frac{dx}{dy}=v+y\frac{dv}{dy}$

Solution: Here,

$\left(x^{2}+y^{2}\right) \frac{d y}{d x}=8 x^{2}-3 x y+2 y^{2}$

$\Rightarrow \frac{dx}{dy}=\frac{8x^{2}-3xy+2y^{2}}{\left ( x^{2}+y^{2} \right )}$

It is homogeneous equation.

Put $y=vy$ and $\frac{dx}{dy}=v+x\frac{dv}{dx}$

So, $v+x \frac{d v}{d x}=\frac{8 x^{2}-3 x v x+2 x^{2} v^{2}}{x^{2}+x^{2} v^{2}}$

$\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}}{1+v^{2}}-v \\$

$\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}-v^{3}}{1+v^{2}}-v \\$

Separating the variables and integrating we get

$\Rightarrow \int \frac{1+v^{2}}{8-4 v+2 v^{2}-v^{3}} d v=\int \frac{1}{x} d x \\$

$\Rightarrow \int \frac{1+v^{2}}{(2-v)\left(v^{2}+4\right)} d v=\int \frac{1}{x} d x \rightarrow(A) \\$

$\Rightarrow \frac{1+v^{2}}{(2-v)\left(v^{2}+4\right)}=\frac{A x+B}{4+v^{2}}+\frac{c}{2-v}[\text { Using partial fraction }] \\$

$\Rightarrow 1+v^{2}=v^{2}(-A+C)+v(2 A+B)+2 B+4 C$

Comparing the coefficient of like power V.

\begin{aligned} &-A+C=1 \rightarrow(1) \\ &2 A-B=0 \\ &\Rightarrow B=2 A \rightarrow(2) \\ &\text { And } 2 B+4 C=1 \rightarrow(3) \end{aligned}

Solving equation $\left ( 1 \right ),\left ( 2 \right )$& $\left ( 3 \right )$we get

$A=\frac{3}{8},B=\frac{3}{4},C=\frac{5}{8}$

Using equation $\left ( A \right )$we get

\begin{aligned} &\Rightarrow \int \frac{-\frac{3}{8} x-\frac{3}{4}}{4+v^{2}} d v+\frac{5}{8} \int \frac{c}{2-v} d v=\int \frac{d x}{x} \\ &\Rightarrow-\frac{3}{8} \int \frac{v+2}{4+v^{2}} d v+\frac{5}{8} \int \frac{1}{2-v} d v=\int \frac{d x}{x} \end{aligned}

$\Rightarrow-\frac{3}{8} \int \frac{v}{4+v^{2}} d v+\frac{2 x 3}{8} \int \frac{1}{4+v^{2}} d v+\frac{5}{8} \int \frac{1}{2-v} d v=\int \frac{d x}{x} \\$

$\Rightarrow-\frac{3}{16} \log \left|4+v^{2}\right|-\frac{3}{8} \tan ^{-1} \frac{v}{2}+\frac{5}{8} \log |2-v|=\log x+\log c \\$

$\Rightarrow\left[\log \left[4+v^{2}\right]_{16}^{\frac{8}{16}}+\log e^{\frac{3}{8} \tan ^{-1}\left(\frac{v}{2}\right)}+\log |2-v|^{\frac{5}{8}}\right]=\log c x \\$

$\Rightarrow\left(4+v^{2}\right)^{\frac{3}{16}} \times e^{\frac{8}{\operatorname{s}} \tan ^{-1}\left(\frac{v}{2}\right)} \times(2-v)^{\frac{5}{8}}=\frac{c}{x}$

Put $y=vx$

$\Rightarrow\left(4+\frac{y^{2}}{x^{2}}\right)^{\frac{3}{16}} \times e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times(2-v)^{\frac{5}{8}}=\frac{c}{x} \\$

$\Rightarrow\left(\frac{4 x^{2}+y^{2}}{\left(x^{2}\right)^{\frac{8}{16}}}\right)^{\frac{8}{16}} \times \frac{(2-y)^{\frac{5}{8}}}{x^{\frac{5}{8}}}=e^{\frac{a}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times \frac{c}{x} \\$

$\Rightarrow \frac{\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}(2-y)^{\frac{5}{8}}}{x}=e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times \frac{c}{x}\left[\therefore x^{\left.\frac{6}{16}, x^{\frac{5}{8}}=x^{\frac{6}{16}}+\frac{5}{8}=x^{\frac{8}{8}}=x\right]}\right. \\$

$\Rightarrow\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}(2-y)^{\frac{5}{8}}=c e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \\$

$\Rightarrow \mathrm{c}|2 x-y|^{\frac{5}{8}}\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}=e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \text { where } c=\frac{1}{c}$

This is required solution.