Thank you very much! So I am just not understanding what I'm doing... Unfortunately I am in Houston and all the schools are closed down because of Harvey or I would go get tutor help. can you provide me steps on how you solved these? I would be very appreciative. Thanks for the support!
it's pretty much just plug and chug
$y = e^{-4x}$
$y' = -4 e^{-4x}$
$y''=16 e^{-4x}$
and just plug these into the possible answers listed. The $e^{-4x}$ will divide out and the algebra will either return 0 or not.
In problem 2
$y = C_1 e^{4x} + C_2 x e^{4x}$
$y' = e^{4 x} \left(4 C_2 x+4 C_1+C_2\right)$
$y'' = e^{4 x} \left(16 C_2 x+16 C_1+8 C_2\right)$
and again plug all this in, divide out the $e^{4x}$ term, and the algebra will either be 0 or not.
In problem 3
$y' = 3x(1+y^2)$
$\dfrac{dy}{1+y^2} = 3x~dx$
$\tan^{-1}(y) = \dfrac 3 2 x^2 + C$
$y = \tan\left(\dfrac 3 2 x^2 + C\right)$