Question

In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x

Answer 1

Answer:

Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

Step-by-step explanation:

Given Differential Equation,

$y"-2y'+y=xe^x$ ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

$\implies\:y(x)=y_c(x)+y_p(x)$

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

$y_c(x)=C_1e^x+c_2xe^x$

Now for particular solution,

let $y_p(x)=Ax^3e^x$

$y'=Ax^3e^x+3Ax^2e^x$

$y"=Ax^3e^x+6Ax^2e^x+6Axe^x$

Now putting these values in (1), we get

$Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x$

$Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x$

$6Axe^x=xe^x$

$6A=1$

$A=\frac{1}{6}$

$\implies\:y_p(x)=\frac{x^3e^x}{6}$

Therefore, Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$