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Verified Answer :

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 \(4x^2\). So let \(f(x)=ax^2+bx+c \implies f'(x)=2ax+b \text{ and } f''(x)=2a.\)
Plug all these into the given DE to get

\(2ax+3ax^2+3bx+3c=4x^2\) or
\(3ax^2+(2a+3b)x+3c=4x^2\)

From the above equation the left hand side should have a coffienet of 4 for \(x^2\), and all other coffienents must be 0. That is
\(3a=4,2a+3b=0\) and \(3c=0\).

You can easily solve the system above and find \(a=\frac{4}{3},b=-\frac{8}{9}\) and \(c=0\).

Thus \(f(x)=\frac{4}{3}x^2-\frac{8}{9}x\)




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