Solve the Differential Equation by Variation of Parameters: y'' − 4y = 4xe^2x

We will be solving this by finding the homogeneous and complementary solutions of the equation.

Answer: The general solution of differential equation y'' − 4y = 4xe^2x is y = c\(_1\)e^2x + c\(_2\)e^-2x + (x²/2)e^2x - 4xe^2x + 16e^2x

Let's solve this step by step.

Explanation:

Given, differential equation: y'' − 4y = 4xe^2x

First solve the corresponding homogeneous differential equation: y'' − 4y = 0

The characteristic equation is r² - 4 = 0.

Let's find its roots by factorization.

(r - 2) (r + 2) = 0

r = 2, -2

Therefore, r = 2, -2 is a non repeated root thus one of the complimentary solutions y\(_c\) is y\(_c\) = c\(_1\)e^2x + c\(_2\)e^-2x 

Now, find the remaining particular solution y\(_p\).

y\(_1\) = e^2x , y\(_2\) = e^-2x, g(x) = 4xe^2x

W(y\(_1\), y\(_2\)) = y\(_1\) y\(_2\)' - y\(_1\)' y\(_2\) = e^2x × (-2e^-2x) - 2e^2x × e^-2x  = -4

\(y_p = -y_1\int \dfrac{y_2(x)g(x)}{W(x)}dx +y_2\int \dfrac{y_1(x)g(x)}{W(x)} dx\)

\(y_p = -e^{2x}\int \dfrac{e^{-2x}.4xe^{2x}}{-4}dx +e^{-2x}\int \dfrac{e^{2x}.4xe^{2x}}{-4}dx \)

\(y_p = e^{2x}\int xdx -e^{-2x}\int xe^{4x}dx \)

\(y_p = e^{2x}\dfrac{x^2}{2} -e^{-2x}[4xe^{4x}-16e^{4x}]\)

\(y_p = e^{2x}\dfrac{x^2}{2} -4xe^{2x}+16e^{2x}\)

Now, combine complementary solution and particular solution together to arrive at the general solution.

y = y\(_c\) + y\(_p\) 

y = c\(_1\)e^2x + c\(_2\)e^-2x + (x²/2)e^2x - 4xe^2x + 16e^2x

Thus, the general solution of differential equation y'' − 4y = 4xe^2x is y = c\(_1\)e^2x + c\(_2\)e^-2x + (x²/2)e^2x - 4xe^2x + 16e^2x​​​​​​​