# Determine the particular solution to the equation 2\frac{dv(t)}{dt} + v(t) = 5t\sin(t)

## Question:

Determine the particular solution to the equation {eq}2\frac{dv(t)}{dt} + v(t) = 5t\sin(t) {/eq}

## Particular Solution:

Given a Linear constant coefficient differential equation with a forcing function in RHS we choose particular integral with following rules:

1) If the forcing function is :

{eq}e^{at} {/eq}

Then :

{eq}y_p=Ae^{at} {/eq}

If Forcing function is sinusoid , then :

{eq}y_p=C\sin(wt)+D\cos(wt) {/eq}

The given differential equation is:

{eq}2\frac{dv}{dt}+v=5t\sin t {/eq}

Since the forcing function is multiplication of linear term and sinusoid ,then particular integral is of the form:

{eq}v_p=(At+B)\sin t+(Ct+D)\cos t {/eq}

Differentiating both sides we get:

{eq}v_p'=(A-Ct-D)\sin t+(At+B+C)\cos t\\ 2v_p'=(2A-2Ct-2D)\sin t+(2At+2B+2C)\cos t {/eq}

Now Particular solution should satisfy the given differential equation:

{eq}2v_p'-+v_p=((A-2C)t+B-2A-2D)\sin t+((2A+C)t+2B+2C+D)\cos t=5t\sin t {/eq}

Comparing coefficients we get:

{eq}A-2C=5\\ 2A+C=0\\ 2B+2C+D=0\\ B+2A=2D {/eq}

Solving we get:

{eq}A=1\\ B=\frac{6}{5}\\ C=-2\\ D=\frac{8}{5} {/eq}

Hence the particular solution is:

{eq}v_p=t(\sin t-2\cos t)+\frac{2}{5}\left(3\sin t+4\cos t\right) {/eq}

Undetermined Coefficients: Method & Examples

from

Chapter 10 / Lesson 15
3.9K

The method of undetermined coefficients is used to solve a class of nonhomogeneous second order differential equations. This method makes use of the characteristic equation of the corresponding homogeneous differential equation.