Show that the equation f''(x) + 4f'(x) + 4f(x) = 0
is satisfied if f(x) = (3x-5)e^-2x (thats e to the power of -2x, not e to the power of -2 times x)
any help would be greatly appreciated. can't seem to get it to equal 0
Initial equation
$\displaystyle f(x)=(3x-5)e^{-2x}$
$\displaystyle f(x)=3xe^{-2x}-5e^{-2x}$
First derivative, remember to use the chain rule and the multiplication rule. The multiplication rule says that the derivative of f(x)g(x) is f'(x)g(x)+g'(x)f(x). And the chain rule in this case says the derivative of $\displaystyle e^{-2x}$ is $\displaystyle e^{-2x}\left(\frac{dy}{dx}-2x\right)$ which becomes $\displaystyle -2e^{-2x}$
$\displaystyle f\prime(x)=3\left(e^{-2x}+(-2)xe^{-2x}\right)-5(-2)e^{-2x}$
$\displaystyle f\prime(x)= 3e^{-2x}-6xe^{-2x}+10e^{-2x}$
$\displaystyle f\prime(x) = 13e^{-2x}-6xe^{-2x}$
Second derivative, remember the same rules.
$\displaystyle f\prime\prime(x)=12(-2)e^{-2x}-6\left(e^{-2x} + (-2)xe^{-2x}\right)$
$\displaystyle f\prime\prime(x)=-24e^{-2x}-6e^{-2x}+12xe^{-2x}$
$\displaystyle f\prime\prime(x)=12xe^{-2x}-32e^{-2x}$
Now just plug them into the equation
$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=(12xe^{-2x}-32e^{-2x})+4(13e^{-2x}-6xe^{-2x})+4(3xe^{-2x}-5e^{-2x})$
and simplify
$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=12xe^{-2x}-32e^{-2x}+52e^{-2x}-24xe^{-2x}+12xe^{-2x}-20e^{-2x}$
$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=(12+12-24)xe^{-2x}+(52-32-20)e^{-2x}$
$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=0$