160k views
4 votes
Consider the following differential equation to be solved by variation of parameters. 4y" - y = ex/2 9?

User Etki
by
8.6k points

1 Answer

1 vote

Final Answer:

The solution to the given differential equation 4y'' - y' = e^(x/2) + 9 is y(x) = C1e^(x/2) + C2e^(-x/4) - (1/4)e^(x/2) + 9/4, where C1 and C2 are constants.

Step-by-step explanation:

Certainly! Let's break down the explanation in a more mathematical form:

Given differential equation: 4y'' - y' = e^(x/2) + 9

1. Complementary Solution (Homogeneous Equation):

The associated homogeneous equation is 4y'' - y' = 0.

The characteristic equation is 4m^2 - m = 0.

Solving this equation gives the roots m = 1/4 and m = 0.

Therefore, the complementary solution is of the form yc = C1e^(x/2) + C2, where C1 and C2 are arbitrary constants.

2. Particular Solution (Non-Homogeneous Equation):

For the particular solution yp, assume yp = u1(x)e^(x/2), where u1(x) is the function to be determined.

Differentiate yp to find u1'(x) = 1/4 and u1''(x) = 0.

Substitute these into the original differential equation: 4(0) - 1/4 = e^(x/2) + 9.

Solving for u1(x) gives u1 = -1/4.

Therefore, the particular solution is yp = -1/4 e^(x/2).

3. General Solution and Final Answer:

The general solution is the sum of the complementary and particular solutions: y = yc + yp.

Hence, y(x) = C1e^(x/2) + C2 + yp = C1e^(x/2) + C2e^(-x/4) - (1/4)e^(x/2) + 9/4, where C1 and C2 are constants determined by initial conditions if provided.

User Dardisco
by
8.1k points