## Calculus (3rd Edition)

$$y=\frac{1}{2}x^2+\frac{3}{2}.$$ $$y= 2e^{x-1}.$$
For the equation $$\frac{dy}{dx}=x\Longrightarrow y=\frac{1}{2}x^2+c.$$ Using the condition $y(1)=2$, we obtain that $2=\frac{1}{2}+c$, then $c=3/2$ and hence the solution is $$y=\frac{1}{2}x^2+\frac{3}{2}.$$ For the equation $$\frac{dy}{dx}=y\Longrightarrow \frac{dy}{y}=dx \Longrightarrow \ln y =x+c.$$ Using the condition $y(1)=2$, we obtain that $\ln 2=1+c$, then $c=-1+\ln 2$ and hence the solution is $$y=e^{x-1+\ln 2}=2e^{x-1}.$$