Solve the Differential Equation by Variation of Parameters: y'' − 4y = 4xe2x
We will be solving this by finding the homogeneous and complementary solutions of the equation.
Answer: The general solution of differential equation y'' − 4y = 4xe2x is y = c1e2x + c2e-2x + 1/2 x2e2x - 1/4 xe2x
Let's solve this step by step.
Given, differential equation: y'' − 4y = 4xe2x
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 yc is yc = c1e2x + c2e-2x
Now, find the remaining particular solution yp.
yp = Ax2e2x - Bx
y′p = 2Ax2e2x + 2Axe2x - B
y′′p = 4Ax2e2x + 4Axe2x + 4Axe2x + 2Ae2x = 4Ax2e2x + 8Axe2x + 2Ae2x
Substitute yp, y′p, and y′′p in given differential equation and solve for A and B.
(4Ax2e2x + 8Axe2x + 2Ae2x) − 4(Ax2e2x + B) = 4xe2x
4Ax2e2x + 8Axe2x + 2Ae2x − 4Ax2e2x - 4B = 4xe2x
8Axe2x + 2Ae2x − 4B = 4xe2x
On comparing, we get A = 1/2 and 2Ae2x − 4B = 0.
⇒B = e2x/4
Therefore, yp = 1/2 x2e2x - 1/4 xe2x
Now, combine complementary solution and particular solution together to arrive at the general solution.
y = yc + yp
y = c1e2x + c2e-2x + 1/2 x2e2x - 1/4 xe2x