Solve the following differential equation: y" + y' = 8x^2

Step-by-step explanation:

Let's  find a particular solution. We need a function of the form such that

and

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a =  8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is  .

Now, let's find the solution of the homogeneus equation with the method of constants coefficients. Let

then

and .

So, and the solution is

.