#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 6 Maths Textbook Solution.

Answer: $\left ( x^{3}+y^{3} \right )^{2}=4x^{2}y^{2}$

Given:$\left(y^{4}-2 x^{3} y\right) d x+\left(x^{4}-2 x y^{3}\right) d x=0, y(1)=1$

To find: we have to find the solution of given differential equation.

Hint: we will put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Solution: we have,

$\left(y^{4}-2 x^{3} y\right) d x+\left(x^{4}-2 x y^{3}\right) d x=0, y(1)=1$

$\frac{dy}{dx}=\frac{2x^{3}y-y^{4}}{x^{4}-2xy^{3}}$                                  .....(i)

It is homogeneous equation.

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

So,

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

Separating the variables and integrating both side we get

$\int \frac{1-2 v^{3}}{v\left(v^{3}+1\right)}=\int \frac{d x}{x}\\$        ....(ii)

$\text { take, }\\$

$\frac{1-2 v^{3}}{v\left(v^{3}+1\right)}=\frac{1-2 v^{3}}{v(v+1)\left(v^{2}-v+1\right)}=\frac{A}{v}+\frac{B}{v-1}+\frac{v+1}{v^{2}-v+1}\\$

$1-2 v^{3}=v^{3}(A+B+C)+v^{2}(-B+C+1)+v(B+D)+A$

$Comparing \: the \: coefficient \: of \: like \: powers \: of\: \mathrm{v},$

$\mathrm{A}=1$

$B+D=0$

$-B+C+1=0$

$A+B+C=-2$

$On\: \: solving$

$\mathrm{A}=1, \mathrm{~B}=-1, \mathrm{C}=-2, \mathrm{x}=!$

Using (ii)

\begin{aligned} &\int\left(\frac{1}{v}-\frac{1}{v-1}-\frac{2 v+1}{v^{2}-v+1}\right) d v=\int \frac{d x}{x}\\ &\Rightarrow \log |\downarrow|-\log |v+1|-\log \left|v^{2}-v+1\right|=\log x+\log c\\ &\Rightarrow \log \left|\frac{v}{v^{3}+1}\right|=\log |x c|\\ &\text { Putting } \mathrm{v}=\frac{y}{x}\\ &\Rightarrow \frac{\frac{x}{y}}{y^{3}+x^{3}} \times x^{3}=\mathrm{x} \mathrm{c}\\ &\Rightarrow \frac{x y}{y^{3}+x^{3}}=c\\ \end{aligned}....(iii)

It is given that y=! when x=1

Putting y=1, x=1 in equation (iii) we get