We're going to assume that the solution is a quadratic polynomial, since we have the sum of the function and its derivative adds up to [tex]4x^2[/tex]. So let [tex]f(x)=ax^2+bx+c \implies f'(x)=2ax+b \text{ and } f''(x)=2a.[/tex]
Plug all these into the given DE to get
[tex]2ax+3ax^2+3bx+3c=4x^2[/tex] or
[tex]3ax^2+(2a+3b)x+3c=4x^2[/tex]
From the above equation the left hand side should have a coffienet of 4 for [tex]x^2[/tex], and all other coffienents must be 0. That is
[tex]3a=4,2a+3b=0[/tex] and [tex]3c=0[/tex].
You can easily solve the system above and find [tex]a=\frac{4}{3},b=-\frac{8}{9}[/tex] and [tex]c=0[/tex].
Thus [tex]f(x)=\frac{4}{3}x^2-\frac{8}{9}x[/tex]
