Chapter 10 - Differential Equations - Chapter Review - Review Exercises - Page 561: 2

False

$\frac{dy}{dx}=2y$ $\frac{dy}{dx}-2y=0$ The integrating factor is $I(x)=e^{\int -2dx}=e^{-2x}$ Multiplying each term by I(x) gives $e^{-2x}\frac{dy}{dx}-2e^{-2x}y=0$ The left side can now be replaced by $D_{x}(e^{-2x}y)$ to get $D_{x}(e^{-2x}y)=0$ Integrating on both sides gives $e^{-2x}y=\int 0dx$ $e^{-2x}y=0x + C$ $y=\frac{C}{e^{-2x}}$ or $y=Ce^{2x}$ The function $y=e^{2x}+5$ is in the form $y=e^{2x}+C$ which is different than the function we found. Hence, the function $y=e^{2x}+5$ does not satisfy the differential equation $\frac{dy}{dx}=2y$