Question

Solution of $(x{y}^4+y)dx-xdy=0$ is

• A. $\frac{x^4}{4}+{\left(\frac{x}{y}\right)}^3=c$
• B. $\frac{x^4}{4}+\frac{1}{3}{\left(\frac{x}{y}\right)}^3=c$
• C. $\frac{x^4}{4}+3{\left(\frac{x}{y}\right)}^3=c$
• D. None of these

Answer 1

The correct option is B $\frac{x^4}{4}+\frac{1}{3}{\left(\frac{x}{y}\right)}^3=c$

$\begin{array}{c}\left(x{y}^4+y\right)dx=xdy\\ x{y}^4+y=x\frac{dy}{dx}\\ \frac{x}{y^4}\frac{dy}{dx}=x+\frac{1}{y^3}\\ \frac{x}{y}\frac{dy}{dx}-\frac{1}{y^3}=x\\ \\ \\ Now,\\ -\frac{1}{y^3}=z\\ \frac{dz}{dx}=\frac{3}{y^4}\frac{dy}{dx}\\ Then\\ \frac{x}{3}\frac{dz}{dx}+z=x\\ \frac{dz}{dx}=\frac{3}{y^4}\frac{dy}{dx}\\ \frac{1}{3}\frac{dz}{dx}+\frac{z}{x}=1\\ \frac{dz}{dx}+z\left(\frac{3}{x}\right)=3\\ z{e}^{\int \frac{3}{x}dx}=\int e^{\int \frac{3}{x}dx}3dx\\ z{x}^3=\frac{3}{4}{x}^4+C\\ \frac{x^3}{y^3}+\frac{3}{4}{x}^4=C^{\prime }\\ \frac{1}{3}{\left(\frac{x}{y}\right)}^3+\frac{x^4}{y}=C^{\prime }\\ Hence,theoptionBisthecorrectanswer.\end{array}$