# Undetermined Coefficients for higher order differential equations

I have the following fourth order differential equation and was asked to find the general solution for it by using the method of undetermined coefficients.

$y^{(4)} +2y'' +y = (t-1)^2$

So, solving for the characteristic equation,

$r^4+2r^2 +1 =0$

I got,

$r=±i$

and from there I obtained the particular solution

$y_p=c_1 \cos t + c_2 \sin t + c_3t \cos t +c_4t\sin t$

Now, my question is: how do I use the forcing function to come up with a "guessed" equation to solve for the general solution?

• The notation $y^4$ is unfortunate, because it suggests $y\cdot y\cdot y\cdot y$, particularly when you use $y''$ in the same expression. – almagest Apr 20 '16 at 7:52
• Oh, im sorry! I'll fix that! Thanks for pointing it out1 – Marco Neves Apr 20 '16 at 7:52

For a right-hand side of the form $(t-1)^2=t^2-2t+1$, you propose a particular solution of the form: $$y_p = At^2+Bt+C$$ Substitution into the differential equation will give you a system of 3 linear equations in the undetermined coefficients $A$, $B$ and $C$.
Remark: when your proposed solution is already a part of the homogeneous solutions (which is not the case for your differential equation; but which would be the case if the RHS was, for example, $\sin t$), you will need to multiply the proposed solution with a sufficiently large power of $t$ so that it is no longer contained in the homogeneous solution.
• Correct: for a polynomial, you propose a polynomial of the same degree (careful: this means that if the right-hand side is just $t^3$, you still take $At^3+Bt^2+Ct+D$). – StackTD Apr 20 '16 at 7:46