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

Answer:[tex]y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}[/tex]Step-by-step explanation:Let's  find a particular solution. We need a function of the form [tex]y= ax^3+bx^2+cx+d[/tex] such that [tex]y'= 3ax^2+2bx+c[/tex] and [tex]y''= 6ax+2b[/tex][tex]y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2[/tex]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  [tex]y_p= \frac{8}{3}x^3-8x^2+16x[/tex].Now, let's find the solution [tex]y_p[/tex] of the homogeneus equation [tex]y''+y'=0[/tex] with the method of constants coefficients. Let [tex]y=e^{\lambda x}[/tex][tex]y'=\lambda e^{\lambda x}[/tex][tex]y''=\lambda^2e^{\lambda x}[/tex]then [tex]\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0[/tex][tex]e^{\lambda x}(\lambda +\lambda^2)= 0[/tex][tex](\lambda +\lambda^2)= 0[/tex][tex]\lambda (1+\lambda)= 0[/tex][tex]\lambda =0[/tex] and [tex]\lambda)= -1[/tex].So, [tex]y_h = C_1 + C_2e^{-x}[/tex] and the solution is[tex]y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}[/tex].