# Finding particular solutions of $y^{(4)} + 2y'' + y = x \sin x$

We have a non homogeneous ODE $$y^{(4)} + 2y'' + y = x \sin x$$

with characteristic equation I get $$(l^2+1)^2 = 0$$ so $$l = -i ,i$$ and so the answer of homogeneous ODE is a linear combination of $$\sin x , \cos x , x \sin x , x\cos x$$.

For finding the Particular solution first I assumed $$y_p = (Ax+B)(C\sin x + D \cos x)$$ and it didn't work. Then $$y_p = x(Ax+B)(C\sin x + D \cos x)$$ and it didn't work. At last, $$y_p = x^2(Ax+B)(C\sin x + D \cos x)$$ worked and the answer was $$-1/24 x^3 \sin x -1/8 x^2 \cos x$$ worked but it took a lot of time to find that the other two don't work.

I want to know is there any way to guess the leading $$x^n$$ term and not testing different situations? (In this case $$n=2$$) I know it can be solved with a way involving Wronskian and Cramer's rule (but that way needs a 4x4 determinant which takes time to calculate) but I want to solve with undetermined coefficients rule so I want to find a better way for guessing the answer format.

$$y_p=(k_0+k_1x)(A\cos x+B\sin x)$$ but this should be checked with the complementary solution to prevent the similarity

the complementary solution was $$y_c=c_1\sin x+c_2\cos x+c_3x\sin x+c_4x\sin x$$ so the particular solution will be $$y_p=(k_0x^2+k_1x^3)(A\cos x+B\sin x)$$

Generally speaking when solving linear ODE

• your homogeneous equation has root $$r$$ with multiplicity $$m$$ .
• the full equation has a RHS of the form $$P(x)e^{rx}$$ with $$P$$ polynomial.

Then you need to search for a particular solution in the form $$Q(x)e^{rx}$$ with $$Q$$ polynomial and $$\deg(Q)=\deg(P)+m$$

Although since the homogeneous solution will already have vanishing terms $$(C_0+C_1x+\cdots+C_{m-1}x^{m-1})e^{rx}$$, you can ignore them in the polynomial Q.

In case of a linear combination of such terms in the RHS, you can also search for a linear combination of particular solutions for each.

Note: in the special case of $$RHS = P(x)$$, consider the root $$r=0$$ since $$e^{rx}=1$$, and the same rule applies.

In your case $$\pm i$$ are roots with multiplicity $$m=2$$.

Your polynomial $$P(x)=x$$ is of degree $$1$$ and $$\sin(x)$$ is a linear combination of $$e^{ix},\ e^{-ix}$$.

So we need to search for $$Q_1(x)e^{ix}$$ and $$Q_2(x)e^{-ix}$$ with $$Q$$ of degree $$3$$ while ignoring already vanishing terms $$(ax+b)e^{\pm ix}$$, meaning that we search for a particular solution of the form $$(ax^3+bx^2)e^{ix}+(cx^3+dx^2)e^{-ix}$$

Which itself is equivalent in searching for $$(ax^3+bx^2)(A\cos x+B\sin x)$$

You can use operator calculus to solve this iteratively. We have

$$[d’’+2d’+1]y=[d’+1][d’+1]y=[d’+1]w=x\sin x$$

Solve for $$w$$ then solve $$[d’+1]y=w$$.

Short answer: The power of $$x$$ added to the particular solution is exactly equal to the multiplicity of the characteristic root. In this case, $$r=\pm i$$ has a mutiplicity of $$2$$. Putting it all together

$$y_p(x) = \underbrace{x^2}_{\text{multiplicity =2}}\ \underbrace{(Ax+b)}_{\text{ansatz for } "x"}\underbrace{(A\sin x + B\cos x)}_{\text{ansatz for } "\sin x"}$$