Method of undetermined coefficients

Typical forms of the particular integral

In order to find the particular integral, we need to 'guess' its form, with some coefficients left as variables to be solved for. This takes the form of the first derivative of the complementary function. Below is a table of some typical functions and the solution to guess for them.

If a term in the above particular integral for y appears in the homogeneous solution, it is necessary to multiply by a sufficiently large power of x in order to make the solution independent. If the function of x is a sum of terms in the above table, the particular integral can be guessed using a sum of the corresponding terms for y.

Examples

Example 1

Find a particular integral of the equation

${\displaystyle y^{″}+y=t\cos t.}$

The right side t cos t has the form

${\displaystyle P_n{e}^{\alpha t}\cos \beta t}$

with n = 2, α = 0, and β = 1.

Since α + iβ = i is a simple root of the characteristic equation

${\displaystyle {\lambda }^2+1=0}$

we should try a particular integral of the form

${\displaystyle \begin{array}{cc}{y}_p& =t\left[F_1(t)e^{\alpha t}\cos \beta t+G_1(t)e^{\alpha t}\sin \beta t\right]\\ & =t\left[F_1(t)\cos t+G_1(t)\sin t\right]\\ & =t\left[\left(A_{0}t+A_1\right)\cos t+\left(B_{0}t+B_1\right)\sin t\right]\\ & =\left(A_0{t}^2+A_{1}t\right)\cos t+\left(B_0{t}^2+B_{1}t\right)\sin t.\end{array}}$

Substituting y_p into the differential equation, we have the identity

${\displaystyle \begin{array}{cc}t\cos t& =y_p^{″}+y_p\\ & ={\left[\left(A_0{t}^2+A_{1}t\right)\cos t+\left(B_0{t}^2+B_{1}t\right)\sin t\right]}^{″}+\left[\left(A_0{t}^2+A_{1}t\right)\cos t+\left(B_0{t}^2+B_{1}t\right)\sin t\right]\\ & =\left[2A_0\cos t+2\left(2A_{0}t+A_1\right)(-\sin t)+\left(A_0{t}^2+A_{1}t\right)(-\cos t)+2B_0\sin t+2\left(2B_{0}t+B_1\right)\cos t+\left(B_0{t}^2+B_{1}t\right)(-\sin t)\right]\\ & +\left[\left(A_0{t}^2+A_{1}t\right)\cos t+\left(B_0{t}^2+B_{1}t\right)\sin t\right]\\ & =[4B_{0}t+(2A_0+2B_1)]\cos t+[-4A_{0}t+(-2A_1+2B_0)]\sin t.\end{array}}$

Comparing both sides, we have

${\displaystyle \{\begin{array}{l}1=4B_0\\ 0=2A_0+2B_1\\ 0=-4A_0\\ 0=-2A_1+2B_0\end{array}}$

which has the solution

${\displaystyle A_0=0,A_1=B_0=\frac{1}{4},B_1=0.}$

We then have a particular integral

${\displaystyle y_p=\frac{1}{4}t\cos t+\frac{1}{4}{t}^2\sin t.}$

Example 2

Consider the following linear nonhomogeneous differential equation:

${\displaystyle \frac{dy}{dx}=y+e^x.}$

This is like the first example above, except that the nonhomogeneous part (${\displaystyle e^x}$) is not linearly independent to the general solution of the homogeneous part (${\displaystyle c_1{e}^x}$); as a result, we have to multiply our guess by a sufficiently large power of x to make it linearly independent.

Here our guess becomes:

${\displaystyle y_p=Ax{e}^x.}$

By substituting this function and its derivative into the differential equation, one can solve for A:

${\displaystyle \frac{d}{dx}\left(Ax{e}^x\right)=Ax{e}^x+e^x}$

${\displaystyle Ax{e}^x+A{e}^x=Ax{e}^x+e^x}$

${\displaystyle A=1.}$

So, the general solution to this differential equation is:

${\displaystyle y=c_1{e}^x+x{e}^x.}$

Example 3

Find the general solution of the equation:

${\displaystyle \frac{dy}{dt}=t^2-y}$

${\displaystyle t^2}$ is a polynomial of degree 2, so we look for a solution using the same form,

${\displaystyle y_p=A{t}^2+Bt+C,}$

Plugging this particular function into the original equation yields,

${\displaystyle 2At+B=t^2-(A{t}^2+Bt+C),}$

${\displaystyle 2At+B=(1-A)t^2-Bt-C,}$

${\displaystyle (A-1)t^2+(2A+B)t+(B+C)=0.}$

which gives:

${\displaystyle A-1=0,2A+B=0,B+C=0.}$

Solving for constants we get:

${\displaystyle y_p=t^2-2t+2}$

To solve for the general solution,

${\displaystyle y=y_p+y_c}$

where ${\displaystyle y_c}$ is the homogeneous solution ${\displaystyle y_c=c_1{e}^{-t}}$, therefore, the general solution is:

${\displaystyle y=t^2-2t+2+c_1{e}^{-t}}$