## Calculus (3rd Edition)

$$y= x^2+2x.$$
This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ -\int \frac{1}{x} dx}=e^{-\ln x}= \frac{1}{x} .$$ Now the general solution is \begin{align} y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ & =x\left( \int dx+C\right)\\ & =x\left( x+C\right)\\ & =x^2+Cx \end{align} Since, $y(1)=3$, then $C=2$ Thus, the general solution is: $$y= x^2+2x.$$