Question

Solution of $y^{5}x+y-x\frac{dy}{dx}=0$ is

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

Answer 1

Answer: (D)

$\frac{x^5}{5}+\frac{x^4}{4y^4}=c$

We have,

$y^{5}x+y-x\frac{dy}{dx}=0x\frac{dy}{dx}=y^{5}x+y\frac{dy}{dx}=y^5+\frac{y}{x}\frac{1}{y^5}\frac{dy}{dx}=1+\frac{1}{x{y}^4}\frac{1}{y^5}\frac{dy}{dx}-\frac{1}{x{y}^4}=1$

Let

$z=\frac{1}{y^4}$

So,

$\frac{dz}{dx}=\frac{-4}{y^5}\frac{dy}{dx}-\frac{1}{4}\frac{dz}{dx}-\frac{z}{x}=1\frac{1}{4}\frac{dz}{dz}+\frac{z}{x}=-1\Rightarrow \frac{dz}{dz}+\frac{4z}{x}=-4$

Therefore,

$z{e}^{\int \frac{4}{x}dx}=\int e^{\int \frac{y}{x}dx}\left(-4\right)dxz{x}^4=-\frac{4}{5}{x}^5+C\frac{x^4}{y^4}+\frac{4}{5}{x}^5=C\frac{x^4}{4y^4}+\frac{x^5}{5}=C^{\prime }$

$Hence,optionDisthecorrectanswer.$