Question

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

Answer 1

`y=y_p+y_h = (8)/(3)x^3-8x^2+16x+ C_1 + C_2e^(-x)`

Explanation:

Let's find a particular solution. We need a function of the form
`y= ax^3+bx^2+cx+d` such that

`y'= 3ax^2+2bx+c` and

`y''= 6ax+2b`

`y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2`

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
`y_p= (8)/(3)x^3-8x^2+16x`.

Now, let's find the solution
`y_p` of the homogeneus equation
`y''+y'=0` with the method of constants coefficients. Let
`y=e^(\lambda x)`

`y'=\lambda e^(\lambda x)`

`y''=\lambda^2e^(\lambda x)`

then
`\lambda e^(\lambda x)+\lambda^2 e^(\lambda x) = 0`

`e^(\lambda x)(\lambda +\lambda^2)= 0`

`(\lambda +\lambda^2)= 0`

`\lambda (1+\lambda)= 0`

`\lambda =0` and
`\lambda)= -1`.

So,
`y_h = C_1 + C_2e^(-x)` and the solution is

`y=y_p+y_h =(8)/(3)x^3-8x^2+16x+ C_1 + C_2e^(-x)`.