Method of undetermined coefficients

We’re motivated by the idea that there’s no systematic method to find the particular solution of non-homogeneous ODEs. We make an educated guess using the method of undetermined coefficients. If we have an ODE like:

$$a_n(t)x^n+a_{n-1}(t)x^{n-1}+\cdots +a_1(t)x^{\prime }+a_0(t)x=g(t)$$

then we guess the terms of a solution based on what $g(t)$ is composed of, and substitute the derivatives of the particular solution into the left-hand side of the ODE to determine the coefficients. Recall that:

$$y=y_c+y_p$$

Note that if our guess already appears in the complementary solution, we need to make it linearly independent from the complementary solution first. For example, if we had $e^x$ in the forcing function and in the complementary equation, our guess must look like:

$$y_p=Ax{e}^x$$

Example

Take the second-order ODE:

$$y^{\prime \prime }+4y^{\prime }-2y=2x^2-3x+6$$

If we assume that $y_p=A{x}^2+Bx+C$, then we can substitute in:

$$y_p^{\prime \prime }+4y_p^{\prime }-2y_p=2x^2-3x+6$$

The corresponding coefficients of the left side are equal to the coefficients of the right side.

Table of guesses

$$1\to A$$ $$5x^2+7x+2\to A{x}^2+Bx+C$$ $$5e^{2t}\to A{e}^{2t}$$ $$t{e}^{2t}\to (At+B)e^{2t}$$ $$t{e}^t\to (At+B)t{e}^t$$ $$t^2{e}^t\to (A{t}^2+Bt+C)e^t$$ $$\sin t\to A\sin t+B\cos t$$ $$e^t\sin t\to A{e}^t\sin t+B{e}^t\cos t$$ $$t^2{e}^t\cos t\to (A{t}^2+Bt+C)e^t\cos t+(D{t}^2+Et+F)e^t\sin t$$

Addendums

This method tends to be really easy and quick to apply. But the forcing function that we must guess for has to be a polynomial, exponential, $\sin$ or $\cos$, because they have a finite amount of derivatives. For functions like $\arctan x$ or $\ln x$, we have an infinite set of derivative functions. This motivates the idea of variation of parameters as a more systematic and reliable method.