How to solve DE: $y'''+y''+ y'+ y = 1 + x + e^{x} + \sin{x}$

Question

How to find the particular solution for:

$y'''+y''+ y'+ y = 1 + x + e^{x} + \sin{x}$

It seems as though the only thing I'd need to solve this would be a good substitution, correct? It doesn't seem like Let $y=e^{x}$ works as a substitution.

Answer 1

(This is the same method @J.G. gave, I just add explanation of why we take this solution)

First you solve the homogeneous part by setting $y_h=e^{\gamma x}$

Then you have the "scary" equation: $y'''+y''+ y'+ y = 1 + x + e^{x} + \sin{x}$

We can break it to $4$ different equations: $$y_1'''+y_1''+ y_1'+ y_1 = 1 + x\\y_2'''+y_2''+ y_2'+ y_2 = e^x\\y_3'''+y_3''+ y_3'+ y_3 = \sin{x}\\y_p=y_1+y_2+y_3$$(Can you see why?)

Solving $y_i$ shouldn't be a problem using undetermined coefficients method:$$y_1=C_1+C_2x\\y_2=C_3e^x\\y_3=C_4\sin(x)+C_5\cos(x)\\y_p=C_1+C_2x+C_3e^x+C_4\sin(x)+C_5\cos(x)$$When you will work it out you will see that the homogeneous part contains $y_3$ in it, so instead of $y_3$ we will use $x\cdot y_3$ and you will get the the guess of:$$y_p=C_1+C_2x+C_3e^x+C_4x\sin(x)+C_5x\cos(x)$$

Answer 2

Hint: try the Ansatz $y=a+bx+ce^x+dx\sin x +fx\cos x$.

Answer 3

You have $$ e^{-x} , \sin x, \cos x$$ as your solutions for the homogeneous equation $$y'''+y''+ y'+ y = 0$$

For your particular solution You consider $$A+Bx +C e^{x} + x(D \sin x +E \cos x)$$