# Thread: The program says Im wrong but Im not so sure.... can someone verify?

1. ## The program says Im wrong but Im not so sure.... can someone verify?

I just took an online quiz but I think some of the answers may be correct. can someone verify my answers truly are incorrect. Thanks you very very much!!

2. ## Re: The program says Im wrong but Im not so sure.... can someone verify?

i) d is correct

ii) d is correct

iii) e is correct

$y(x) = \tan\left( \dfrac{3x^2}{2} + c \right)$

3. ## Re: The program says Im wrong but Im not so sure.... can someone verify?

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!

4. ## Re: The program says Im wrong but Im not so sure.... can someone verify?

Originally Posted by thatsmessedup
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

$y' = 3x(1+y^2)$

$\dfrac{dy}{1+y^2} = 3x~dx$

$\tan^{-1}(y) = \dfrac 3 2 x^2 + C$

$y = \tan\left(\dfrac 3 2 x^2 + C\right)$