## Calculus: Early Transcendentals 8th Edition

The solution of the given initial-value problem $$\frac{d r}{d t}+2 t r=r , \quad \quad r(0)=5$$ is $$r(t)=5 e^{t-t^{2}}$$
The initial-value problem is: $$\frac{d r}{d t}+2 t r=r$$ $$\frac{d r}{d t}+2 t r=r \Rightarrow \frac{d r}{d t}=r-2 t r=r(1-2 t)$$ separating and integrating both sides, we get: $$\int \frac{d r}{r}=\int(1-2 t) d t \Rightarrow \ln |r|=t-t^{2}+C$$ $\Rightarrow$ $$|r|=e^{t-t^{2}+C}=k e^{t-t^{2}}$$ where where $k$ is an arbitrary constant. Since $$r(0)=5,5=k e^{0}=k .$$ Thus the solution of the given initial-value problem is $$r(t)=5 e^{t-t^{2}}$$