Question

Find the equation to the curve satisfying x (x + 1) $\frac{dy}{dx}-y$ = x (x + 1) and passing through (1, 0).

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\left(x+1\right)\frac{dy}{dx}-y=x\left(x+1\right) \\ \Rightarrow \frac{dy}{dx}-\frac{y}{x\left(x+1\right)}=1 \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-\frac{1}{x\left(x+1\right)} \\ Q=1 \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{-\int \frac{1}{x\left(x+1\right)}dx} \\ =e^{-\int \frac{1}{x}-\frac{1}{x+1}dx} \\ =e^{-\log \left|\frac{x}{x+1}\right|} \\ =\frac{x+1}{x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow \left(\frac{x+1}{x}\right)y=\int \frac{x+1}{x}dx+C \\ \Rightarrow \left(\frac{x+1}{x}\right)y=\int dx+\int \frac{1}{x}dx+C \\ \Rightarrow \left(\frac{x+1}{x}\right)y=x+\log \left|x\right|+C \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{throught}\mathrm{the}\mathrm{point}\left(1,0\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ \Rightarrow \left(\frac{1+1}{1}\right)0=1+\log \left|1\right|+C \\ \Rightarrow C=-1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get} \\ \left(\frac{x+1}{x}\right)y=x+\log \left|x\right|-1 \\ \Rightarrow y=\frac{x}{x+1}\left(x+\log \left|x\right|-1\right) \end{gathered}$$