Answer
See the explanation.
Work Step by Step
Verify that $y(0)=5$:
$y(0)=5e^{2\cdot 0}+0$
$y(0)=5\cdot e^0+0$
$y(0)=5\cdot 1$
$y(0)=5$ (True)
Find the derivative of $y$:
$\frac{dy}{dx}=\frac{d}{dx}(5e^{2x}+x)$
$\frac{dy}{dx}=5\cdot 2e^{2x}+1$
$\frac{dy}{dx}=10e^{2x}+1$
Show that $y=5e^{2x}+x$ satisfies the differential equation:
$\frac{dy}{dx}-2y=1-2x$ (Substitute $\frac{dy}{dx}=10e^{2x}+1$)
$10e^{2x}+1-2y=1-2x$ (Substitute $y=5e^{2x}+x$)
$10e^{2x}+1-2(5e^{2x}+x)=1-2x$
$10e^{2x}+1-10e^{2x}-2x=1-2x$
$1-2x=1-2x$
$\therefore$ The left-hand side = The right-hand side
