Question

$\frac{dy}{dx}+\frac{y}{x}=x^3$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\frac{y}{x}=x^3.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=\frac{1}{x} \\ Q=x^3 \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\log \left|x\right|} \\ =x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}x,\mathrm{we}\mathrm{get} \\ x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x{x}^3 \\ \Rightarrow x\frac{dy}{dx}+y=x^4 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ xy=\int x^{4}dx+C \\ \Rightarrow xy=\frac{x^5}{5}+C \\ \Rightarrow 5xy=x^5+5C \\ \Rightarrow 5xy=x^5+K\left(\mathrm{where},K=5C\right) \\ \mathrm{Hence},5xy=x^5+K\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$