Mathematics - Differential Equation Question with Solution | TestHub

Given, $\frac{dy}{dx}-\frac{y}{x+1}=e^{3x}(x+1)$ Above is linear differential equation of form $\begin{array}{rl}& \frac{dy}{dx}+Px=Q\\ & \therefore \mathrm{IF}=e^{\int Pdx}=e^{\int -\frac{1}{x+1}dx}=e^{-\ln (1+x)}=\frac{1}{1+x}\end{array}$ $\Rightarrow$ Solution will be $\begin{array}{rl}y\cdot \mathrm{IF}& =\int Q\cdot \mathrm{IF}dx\\ y\cdot \frac{1}{1+x}& =\int e^{3x}(x+1)\times \frac{1}{(1+x)}dx\\ \frac{y}{1+x}& =\frac{e^{3x}}{3}+C^{\prime }\\ \frac{3y}{1+x}& =e^{3x}+C^{\prime }\left[\because C=3C^{\prime }\right]\end{array}$