1. ## urgent, calculus equation

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

2. Originally Posted by s0urgrapes
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
The best thing would be for you to post some working and someone will spot where you're going wrong.

However, here's a tip. A least it's what I would do...

Don't do all the differentiation.

$\displaystyle f(x)=(3x-5)e^{-2x}$

$\displaystyle f'(x)=3e^{-2x}-2f(x)$

etc..

3. Originally Posted by s0urgrapes
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
$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=12xe^{-2x}-32e^{-2x}+52e^{-2x}-24xe^{-2x}+12xe^{-2x}-20e^{-2x}$

$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=(12+12-24)xe^{-2x}+(52-32-20)e^{-2x}$

$\displaystyle f\prime\prime(x)+4f\prime(x)+4f(x)=0$