1. ## help plz

How would you solve this and find y?
(e^(2x)(ydoubleprime) = 4(e^4x + 1)
I did this:
ydoubleprime = 4(e^4x + 1)/(e^(2x)
..but that doesnt seem to help, taking the integration of that looks difficult. Is there a different approach I could try? Thanks a lot!

2. Originally Posted by davecs77
How would you solve this and find y?
(e^(2x)(ydoubleprime) = 4(e^4x + 1)
I did this:
ydoubleprime = 4(e^4x + 1)/(e^(2x)
..but that doesnt seem to help, taking the integration of that looks difficult. Is there a different approach I could try? Thanks a lot!
What's so hard?

$e^{2x}y^{\prime \prime} = 4(e^{4x} + 1)$

$y^{\prime \prime} = 4\frac{e^{4x} + 1}{e^{2x}}$

$y^{\prime \prime} = 4(e^{2x} + e^{-2x})$

You can either do this directly, or observe that the RHS can also be considered as:
$y^{\prime \prime} = 8 \cdot cosh(2x)$

-Dan