# Find a particular solution to y'' - 8 y' + 16 y = {3.5 e^{4 t}} / {t^2 + 1}.

## Question:

Find a particular solution to {eq}\displaystyle y'' - 8 y' + 16 y = \frac{3.5 e^{4 t}}{t^2 + 1} {/eq}.

## Variation of Parameters:

Let {eq}\displaystyle y''+b(x)y'+c(x)y=f(x) {/eq} be a linear differential equations with {eq}\displaystyle y_c=C_1 u(x)+C_2v(x). {/eq} as the general solution to the associated homogeneous equation. Then a particular solution will have the form {eq}\displaystyle y_p=fu+gv, {/eq} and we find {eq}f {/eq} and {eq}g {/eq} using the following. Recall the Wronskian is

{eq}\displaystyle W=\begin{vmatrix} u & v\\ u{}' &v{}' \end{vmatrix}=uv{}'-vu{}' {/eq}

and

{eq}\displaystyle f=-\int \frac{vR}{W}\, dx\\ \displaystyle g=\int \frac{uR}{W}\, dx {/eq}

Note when using this formula we need to make sure the original differential equation has a leading coefficient of 1.

Consider the differential equation

{eq}\displaystyle y'' - 8 y' + 16 y = \frac{3.5 e^{4 t}}{t^2 + 1} {/eq}

Rewrite the differential equation as follows

{eq}\displaystyle (D^{2}-8D+16)y= \frac{3.5 e^{4 t}}{t^2 + 1} {/eq}

The auxiliary equation corresponding to the homogeneous differential equation is as follows

{eq}\displaystyle m^{2}-8m+16=0\\ \displaystyle (m-4)^2=0\\ \displaystyle \Rightarrow m=4,\:m=4 {/eq}

Therefore, the solution corresponding to the homogeneous differential equation is as follows

{eq}\displaystyle y_c=(a+bt)e^{4t} {/eq}

Therefore, the two L.I solutions of the differential equation are as follows

{eq}\displaystyle u=e^{4t},\, \, \, v=te^{4t},\, \, \, R= \frac{3.5 e^{4 t}}{t^2 + 1} {/eq}

The Wronskian of these two solutions is as follows

{eq}\displaystyle W=\begin{vmatrix} e^{4t}&te^{4t}\\ 4e^{4t}&e^{4t}+4e^{4t}t \end{vmatrix}\\ \displaystyle =e^{8t}+4e^{8t}t-4e^{8t}t\\ \displaystyle =e^{8t} {/eq}

The particular solution corresponding to the nonhomogeneous differential equation is as follows

{eq}\displaystyle y_p=fu+gv\\ \displaystyle f=-\int \frac{vR}{W}dt\\ \displaystyle =-\int \frac{te^{4t}}{e^{8t}}\cdot \frac{3.5 e^{4 t}}{t^2 + 1}\: dt\\ \displaystyle =-\int \frac{3.5 t}{t^2+1}\: dt\\ \displaystyle =-1.75\ln \left|t^2+1\right|\\ \displaystyle g=\int \frac{uR}{W}dt\\ \displaystyle =\int \frac{e^{4t}}{e^{8t}}\cdot \frac{3.5 e^{4 t}}{t^2 + 1}\: dt\\ \displaystyle =\int \frac{3.5 }{t^2+1}\: dt\\ \displaystyle \boxed{y_{p}=3.5\arctan \left(t\right)} {/eq}

Therefore,

{eq}\displaystyle y_p=fu+gv\\ \displaystyle y_p= -1.75e^{4t}\ln \left|t^2+1\right|+3.5te^{4t}\arctan \left(t\right) {/eq}

Hence the general solution of differential equation is as follows

{eq}\displaystyle y(t)=y_c+y_p\\ \displaystyle \boxed{y(t)=(a+bt)e^{4t}-1.75e^{4t}\ln \left|t^2+1\right|+3.5te^{4t}\arctan \left(t\right)}\\ {/eq}

First-Order Linear Differential Equations

from

Chapter 16 / Lesson 3
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In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.