# 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} + (x^{2}/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 r^{2} - 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} + (x^{2}/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} + (x^{2}/2)e^{2x} - 4xe^{2x} + 16e^{2x}

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