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Solve the given differential equation by undetermined

coefficients. y'' + 4y' + 4y = 3x + 4
y(x) =

1 Answer

2 votes

Answer:

The general solution is the sum of the particular solution and the complementary function: y((x) = (3/4)x + 1/2 + (C1 + C2x)e⁻²ˣ, where C1 and C2 are arbitrary constants.

To solve the given differential equation using the method of undetermined coefficients, assume a particular solution of the form:

y_p(x) = Ax + B

where A and B are constants to be determined.

First, let's find the derivatives of y_p(x):

y'_p(x) = A

y''_p(x) = 0

Now, substitute these derivatives into the original differential equation:

0 + 4(A) + 4(Ax + B) = 3x + 5

Simplifying this equation:

4Ax + 4B + 4A = 3x + 5

Now, equate the coefficients of like terms on both sides of the equation:

4A = 3 (coefficient of x on the right-hand side)

4B + 4A = 5 (constant term on the right-hand side)

Solving these equations simultaneously:

4A = 3

4B + 4A = 5

From the first equation, we find A = 3/4. Substituting this value into the second equation:

4B + 4(3/4) = 5

4B + 3 = 5

4B = 2

B = 1/2

Therefore, the particular solution is:

y_p(x) = (3/4)x + 1/2

To find the general solution, we also need the complementary function. The characteristic equation for the homogeneous equation y'' + 4y' + 4y = 0 is:

r² + 4r + 4 = 0

Factoring this equation, we have:

(r + 2)² = 0

The characteristic equation has a repeated root of -2. Therefore, the complementary function is:

y_c(x) = (C1 + C2x)e⁻²ˣ

where C1 and C2 are constants to be determined.

Hence, the general solution is the sum of the particular solution and the complementary function: y(x) = (3/4)x + 1/2 + (C1 + C2x)e⁻²ˣ , where C1 and C2 are arbitrary constants.

answered
User Wojtek Surowka
by
8.0k points
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