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

Answers (1)

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}


\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},





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


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