Calculus: Solve the following differential equation

[laker_gurl3]

Solve the following differential equation:

8y'' - 4x = 0 Where y= 2 and y'=-3 and x= 0

first i did this:

y' = (1/4)x^2

y' = (1/4)x^2 - 3

then antideriv again to get the final answer:

y = (x^3)/12 - 3x + 2

is the final answer 2 or this one :y = (x^3)/12 - 3x + 2

THANK YOU!

 

[OlderDan]

Solve the following differential equation:

8y'' - 4x = 0 Where y= 2 and y'=-3 and x= 0

first i did this:

y' = (1/4)x^2

y' = (1/4)x^2 - 3

then antideriv again to get the final answer:

y = (x^3)/12 - 3x + 2

is the final answer 2 or this one :y = (x^3)/12 - 3x + 2

THANK YOU!

I think they are looking for the function y = f(x) rather than the specific value y = f(0). You would not need to solve the equation to find the latter, because it is given. The wording of the intitial conditions would be better stated as

Where y= 2 and y'=-3 at (or when) x= 0

 

[laker_gurl3]

so did i get it wrong this is all it said for the question:

Solve the following differential equation:

8y'' - 4x = 0 Where y= 2 and y'=-3 at x= 0

 

[OlderDan]

so did i get it wrong this is all it said for the question:

Solve the following differential equation:

8y'' - 4x = 0 Where y= 2 and y'=-3 at x= 0

You did it correctly. I thought you were just trying to decide between the function you found and the specific value y = 2.

y = (x^3)/12 - 3x + 2 looks good to me.