Question

The solution of differential equation $x^2{y}^{2}dy=(1-x{y}^3)dx$ is

• A. $x^3{y}^3=x^2+C$
• B. $2x^3{y}^3=3x^2+C$
• C. $x^3{y}^3=x^2+x+C$
• D. $x^3{y}^3=3x^2+C$

Answer 1

Answer: (B)

$2x^3{y}^3=3x^2+C$

$x^2{y}^{2}dy=(1-x{y}^3)dx$

Divides the entire equation by $x^2$

$\Rightarrow y^{2}dy=(\frac{1}{x^2}-\frac{y^3}{x})dx$

$y^2\frac{dy}{dx}=(\frac{1}{x^2}-\frac{y^3}{x})$

$y^2\frac{dy}{dx}+\frac{1}{x}.y^3=\frac{1}{x^2}$

Let $y^3=x$

$\Rightarrow 3y^2\frac{dy}{dx}=\frac{dz}{dx}$

$\Rightarrow y^2\frac{dy}{dx}=\frac{1}{3}.\frac{dz}{dx}$

substituting, we get

$\Rightarrow \frac{1}{3}\frac{dz}{dx}+\frac{1}{x}.z=\frac{1}{x^2}$

Here we will use a standard result,

A linear equation of the form $\frac{dz}{dx}+P\left(x\right).z=Q\left(x\right)$ has a solution,

$z{e}^{\int p\left(x\right)dx}=\int Q\left(x\right)e^{\int p\left(x\right)dx}dx$

Here,

$\Rightarrow \frac{dz}{dx}+\frac{3}{x}.z=\frac{3}{x^2}$

$\Rightarrow P\left(x\right)=\frac{3}{x},Q\left(x\right)=\frac{3}{x^2}$

so the solution will be

$\Rightarrow z{e}^{3\int \frac{dx}{x}}=\int \frac{3}{x^2}{e}^{3\int \frac{dx}{x}}dx$

$\Rightarrow z{e}^{3lnx}=\int \frac{3}{x^2}{e}^{3lnx}dx$

$\Rightarrow z{e}^{ln{x}^3}=\int \frac{3}{x^2}{e}^{ln{x}^3}dx$

$\Rightarrow z.x^3=\int \frac{3}{x^2}{x}^{3}dx=3\int xdx$

$\Rightarrow z.x^3=\frac{3x^2}{2}+c$ ( c=integration constant )

Resubstituting, $z=y^3$ we get

$\Rightarrow 2x^3{y}^3=3x^2+c$ $( c'=2c )

Hence, solved.