Sure, let's work through the differential equation [tex]\((D^2 + 4) y = 4 \tan(2x)\)[/tex] step by step.
1. Identify the Homogeneous Equation:
The first step is to identify the associated homogeneous differential equation:
[tex]\[
(D^2 + 4)y = 0
\][/tex]
where [tex]\(D\)[/tex] represents the derivative with respect to [tex]\(x\)[/tex].
2. Solve the Homogeneous Equation:
The characteristic equation for the homogeneous part is obtained by replacing [tex]\(D\)[/tex] with [tex]\(r\)[/tex]:
[tex]\[
r^2 + 4 = 0
\][/tex]
Solving for [tex]\(r\)[/tex], we get:
[tex]\[
r^2 = -4 \implies r = \pm 2i
\][/tex]
Thus, the solutions to the characteristic equation are complex: [tex]\(r = 2i\)[/tex] and [tex]\(r = -2i\)[/tex]. These solutions lead to the general solution of the homogeneous equation:
[tex]\[
y_h(x) = C_1 \cos(2x) + C_2 \sin(2x)
\][/tex]
where [tex]\(C_1\)[/tex] and [tex]\(C_2\)[/tex] are arbitrary constants.
3. Find a Particular Solution:
To find a particular solution [tex]\(y_p(x)\)[/tex] for the non-homogeneous differential equation [tex]\((D^2 + 4)y = 4 \tan(2x)\)[/tex], we can use the method of undetermined coefficients or variation of parameters. Given that [tex]\(\tan(2x)\)[/tex] typically suggests a non-standard function, variation of parameters is appropriate here.
4. Set Up Variation of Parameters:
The general solution [tex]\(y(x)\)[/tex] will be of the form:
[tex]\[
y(x) = y_h(x) + y_p(x)
\][/tex]
The particular solution [tex]\(y_p(x)\)[/tex] involves finding functions [tex]\(u_1(x)\)[/tex] and [tex]\(u_2(x)\)[/tex] such that:
[tex]\[
y_p(x) = u_1(x) \cos(2x) + u_2(x) \sin(2x)
\][/tex]
We then solve for [tex]\(u_1(x)\)[/tex] and [tex]\(u_2(x)\)[/tex] using the system derived from:
[tex]\[
u_1'(x) \cos(2x) + u_2'(x) \sin(2x) = 0
\][/tex]
[tex]\[
-2u_1'(x) \sin(2x) + 2u_2'(x) \cos(2x) = 4 \tan(2x)
\][/tex]
5. Solve for [tex]\(u_1(x)\)[/tex] and [tex]\(u_2(x)\)[/tex]:
After solving the system (e.g., by solving for the derivatives and integrating), we find:
[tex]\[
u_1(x) = \int \frac{2 \tan(2x) \sin(2x)}{\cos(2x)} \, dx = \frac{1}{2} \log|\sin(2x) + 1| - \frac{1}{2} \log|\sin(2x) - 1|
\][/tex]
and similarly:
[tex]\[
u_2(x) = 0
\][/tex]
6. Construct the Particular Solution:
Using [tex]\(u_1(x)\)[/tex] and [tex]\(u_2(x)\)[/tex], the particular solution is:
[tex]\[
y_p(x) = \left( \frac{1}{2} \log|\sin(2x) + 1| - \frac{1}{2} \log|\sin(2x) - 1| \right) \cos(2x)
\][/tex]
7. Form the General Solution:
Combine the homogeneous and particular solutions to get the general solution:
[tex]\[
y(x) = C_1 \cos(2x) + C_2 \sin(2x) + \left( \frac{1}{2} \log|\sin(2x) + 1| - \frac{1}{2} \log|\sin(2x) - 1| \right) \cos(2x)
\][/tex]
So, the detailed solution to the given differential equation is:
[tex]\[
\boxed{y(x) = C_1 \cos(2x) + C_2 \sin(2x) + \left( \frac{1}{2} \log(\sin(2x) + 1) - \frac{1}{2} \log(\sin(2x) - 1) \right) \cos(2x)}
\][/tex]
where [tex]\(C_1\)[/tex] and [tex]\(C_2\)[/tex] are constants determined by initial conditions.
