Question

The integrating factor of all differential equation (x² + 1) $\frac{dy}{dx}$ + 2xy = x² - 1 is _______________.

Answer 1

Given: (x² + 1) $\frac{dy}{dx}$ + 2xy = x² − 1


$$\begin{gathered} \left(x^2+1\right)\frac{dy}{dx}+2xy=x^2-1 \\ \Rightarrow \frac{dy}{dx}+\frac{2xy}{x^2+1}=\frac{x^2-1}{x^2+1} \\ \Rightarrow \frac{dy}{dx}+y\left(\frac{2x}{x^2+1}\right)=\frac{x^2-1}{x^2+1} \\ \mathrm{Here},P=\left(\frac{2x}{x^2+1}\right)\mathrm{and}Q=\left(\frac{x^2-1}{x^2+1}\right) \\ \mathrm{Integrating}\mathrm{factor}=e^{\int Pdx} \\ =e^{\int \left(\frac{2x}{x^2+1}\right)dx} \\ =e^{\int \frac{dt}{t}}\left\{\begin{array}{l}\mathrm{Put}{x}^2+1=t\\ \Rightarrow 2xdx=dt\end{array}\right. \\ =e^{\log t} \\ =t \\ =x^2+1 \end{gathered}$$


Hence, the integrating factor of all differential equation (x² + 1) $\frac{dy}{dx}$ + 2xy = x² − 1 is $\underline{x^2+1}.$