Question

Solve the following differential equation by variation of parameters. Fully evaluate all integrals. y" + 4y = sec(2x). a. Find the most general solution to the associated homogeneous differential equation. Use ci and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. b. Find a particular solution to the nonhomogeneous differential equation y" + 4y = sec(2x). c. Find the most general solution to the original nonhomogeneous differential equation. Use C1 and c2 in your answer to denote arbitrary constants.

Answer 1

a. To find the most general solution to the associated homogeneous differential equation y" + 4y = 0, we can assume a solution of the form y = e^(mx), where m is a constant.

By substituting this into the differential equation, we get:

(m^2)e^(mx) + 4e^(mx) = 0

Dividing both sides by e^(mx), we have:

m^2 + 4 = 0

Solving this quadratic equation, we find two complex roots: m = ±2i.

Therefore, the most general solution to the associated homogeneous differential equation is:

y = c1e^(2ix) + c2e^(-2ix)

where c1 and c2 are arbitrary constants.

b. To find a particular solution to the nonhomogeneous differential equation y" + 4y = sec(2x), we can use the variation of parameters method.

Assuming a particular solution of the form y = u1(x)e^(2ix), we can find u1(x) by substituting it into the differential equation:

[u1''(x) + 4u1(x)]e^(2ix) = sec(2x)

Using the formula for sec(2x) = 1/cos(2x), we can rewrite the equation as:

[u1''(x) + 4u1(x)]e^(2ix) = 1/cos(2x)

Now, we can equate the real and imaginary parts of the equation separately:

Real part: u1''(x) + 4u1(x) = 1/cos(2x)

Imaginary part: 2iu1''(x) + 8iu1(x) = 0

Solving the imaginary part, we find that u1(x) must be a constant, let's say A.

Substituting A back into the real part equation, we have:

A'' + 4A = 1/cos(2x)

To solve this, we can use the method of undetermined coefficients. We assume a particular solution of the form A = B/cos(2x), where B is a constant.

Differentiating and substituting back into the equation, we get:

-4B/cos^3(2x) + 4B/cos(2x) = 1/cos(2x)

Simplifying, we find that B = 1/4.

Therefore, a particular solution to the nonhomogeneous differential equation is:

y = (1/4)cos(2x)/cos(2x) + c1e^(2ix) + c2e^(-2ix)

Simplifying further, we have:

y = (1/4) + c1e^(2ix) + c2e^(-2ix)

c. Finally, to find the most general solution to the original nonhomogeneous differential equation, we combine the solutions from parts a and b.

The most general solution is given by:

y = c1e^(2ix) + c2e^(-2ix) + (1/4)cos(2x)/cos(2x)

Simplifying, we have:

y = c1e^(2ix) + c2e^(-2ix) + (1/4)