Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x-1\right)\frac{dy}{dx}=2x^{3}y \\ \Rightarrow \frac{1}{y}dy=\frac{2x^3}{x-1}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y}dy=\int \frac{2x^3}{x-1}dx \\ \Rightarrow \log \left|y\right|=2\int \frac{x^3-1+1}{x-1}dx \\ \Rightarrow \log \left|y\right|=2\left[\int \frac{x^3-1}{x-1}dx+\int \frac{1}{x-1}dx\right] \\ \Rightarrow \log \left|y\right|=2\left[\int \frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}dx+\int \frac{1}{x-1}dx\right] \\ \Rightarrow \log \left|y\right|=2\left[\int \left(x^2+x+1\right)dx+\int \frac{1}{x-1}dx\right] \\ \Rightarrow \log \left|y\right|=2\left[\frac{x^3}{3}+\frac{x^2}{2}+x+\log \left|x-1\right|\right]+C \\ \Rightarrow \log \left|y\right|=\frac{2}{3}{x}^3+x^2+2x+\log {\left|x-1\right|}^2+C \\ \Rightarrow y=e^{\frac{2}{3}{x}^3+x^2+2x+\log {\left|x-1\right|}^2+\mathrm{C}} \\ \Rightarrow y=e^C\times e^{\log {\left|x-1\right|}^2}\times e^{\frac{2}{3}{x}^3+x^2+2x} \\ \Rightarrow y={\mathrm{C}}_1{\left|x-1\right|}^2{e}^{\frac{2}{3}{x}^3+x^2+2x}\left[\because e^{\ln x}=x\mathrm{and}\mathrm{where},C_1=e^{\mathrm{C}}\right] \\ \therefore y={\mathrm{C}}_1{\left|x-1\right|}^2{e}^{\frac{2}{3}{x}^3+x^2+2x}\mathrm{is}\mathrm{required}\mathrm{solution}. \end{gathered}$$