solve the initial value problem: y'+2y=20sin(t)=15cos(t)

To solve the initial value problem \( y' + 2y = 20\sin(t) + 15\cos(t) \), we'll follow these steps:

### Step 1: Solve the homogeneous equation
First, solve the homogeneous equation associated with the problem:

\[
y' + 2y = 0
\]

This is a first-order linear differential equation. To solve it:

1. Separate the variables:
   \[
   \frac{dy}{y} = -2dt
   \]
   
2. Integrate both sides:
   \[
   \ln|y| = -2t + C_1
   \]
   
3. Solve for \( y \):
   \[
   y(t) = C e^{-2t}
   \]
   where \( C = e^{C_1} \) is the constant of integration.

### Step 2: Find a particular solution
Now, find a particular solution to the non-homogeneous equation:

\[
y' + 2y = 20\sin(t) + 15\cos(t)
\]

We'll use the method of undetermined coefficients. Assume a particular solution of the form:

\[
y_p(t) = A\sin(t) + B\cos(t)
\]

Differentiate \( y_p(t) \):

\[
y_p'(t) = A\cos(t) - B\sin(t)
\]

Substitute \( y_p(t) \) and \( y_p'(t) \) into the original differential equation:

\[
(A\cos(t) - B\sin(t)) + 2(A\sin(t) + B\cos(t)) = 20\sin(t) + 15\cos(t)
\]

Group the terms involving \(\sin(t)\) and \(\cos(t)\):

\[
(A + 2B)\cos(t) + (-B + 2A)\sin(t) = 15\cos(t) + 20\sin(t)
\]

This gives us two equations:

\[
A + 2B = 15 \quad \text{(1)}
\]
\[
2A - B = 20 \quad \text{(2)}
\]

### Step 3: Solve the system of equations
Solve these simultaneous equations:

From equation (1):
\[
A = 15 - 2B \quad \text{(3)}
\]

Substitute (3) into equation (2):

\[
2(15 - 2B) - B = 20
\]

Simplify:

\[
30 - 4B - B = 20
\]
\[
30 - 5B = 20
\]
\[
5B = 10
\]
\[
B = 2
\]

Now, substitute \( B = 2 \) back into equation (3):

\[
A = 15 - 2(2) = 15 - 4 = 11
\]

So, the particular solution is:

\[
y_p(t) = 11\sin(t) + 2\cos(t)
\]

### Step 4: General solution
The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:

\[
y(t) = C e^{-2t} + 11\sin(t) + 2\cos(t)
\]

### Step 5: Apply the initial condition (if given)
If an initial condition \( y(t_0) = y_0 \) is given, substitute \( t_0 \) and \( y_0 \) into the general solution to solve for \( C \).

Let me know if you have an initial condition to apply, or if you want further details on any of these steps!

---

Would you like more details on any specific part? Here are 5 related questions:

1. How do you solve a homogeneous differential equation with different coefficients?
2. Can you explain the method of undetermined coefficients in more detail?
3. What if the right-hand side of the equation involved a different function, like an exponential or polynomial?
4. How do you solve for the constant of integration using an initial condition?
5. What happens if the particular solution has the same form as the homogeneous solution?

**Tip:** When solving differential equations, always double-check your particular solution by substituting it back into the original equation to ensure it satisfies the equation.

Learn how to solve the initial value problem y' + 2y = 20sin(t) + 15cos(t) using differential equations. This problem involves finding the general and particular solutions, applying the method of undetermined coefficients, and understanding the steps involved. Suitable for college-level students and those studying differential equations.

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