Question

Solution of $y^{2}x+y-x\frac{dy}{dx}=0$ is

• A. $\frac{y^2}{5}+\frac{y^2}{4x^2}=c$
• B. $\frac{x^2}{2}+\frac{x}{y}=C^{\prime }$
• C. $\frac{y^2}{4}+\frac{y^5}{5x^4}=c$
• D. $\frac{x^2}{5}+\frac{x^4}{4y^4}=c$

Answer 1

The correct option is B $\frac{x^2}{2}+\frac{x}{y}=C^{\prime }$

We have,

$y^{2}x+y=x\frac{dy}{dx}\frac{x}{y^2}\frac{dy}{dx}=\frac{1}{y}+x\frac{1}{y^2}\frac{dy}{dx}=\frac{1}{xy}+1\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{xy}=1$

Let $-\frac{1}{y}=z$

$\frac{dz}{dz}=\frac{1}{y^2}\frac{dy}{dx}\frac{dz}{dz}+\frac{z}{x}=1z{e}^{\int \frac{dx}{x}}=\int e^{\int \frac{dx}{x}}\times 1dxzx=\frac{x^2}{2}+C\frac{-x}{y}=\frac{x^2}{2}+C\frac{x^2}{2}+\frac{x}{y}=C^{\prime }$

Hence, option $B$ is the correct answer.