y’’+5y’=300sin(5t)+300cos(5t), y(0)=3,y’(0)=4

To solve the differential equation:

\[
y'' + 5y' = 300 \sin(5t) + 300 \cos(5t)
\]

with the initial conditions \( y(0) = 3 \) and \( y'(0) = 4 \), we'll use the following steps:

### 1. **Find the general solution of the homogeneous equation:**

The homogeneous part of the equation is:

\[
y'' + 5y' = 0
\]

The characteristic equation is:

\[
r^2 + 5r = 0
\]

Factoring this, we get:

\[
r(r + 5) = 0
\]

Thus, the solutions for \( r \) are \( r = 0 \) and \( r = -5 \). The general solution to the homogeneous equation is:

\[
y_h(t) = C_1 + C_2 e^{-5t}
\]

### 2. **Find a particular solution:**

The non-homogeneous part of the equation is \( 300 \sin(5t) + 300 \cos(5t) \). Since this is a sinusoidal function, we look for a particular solution of the form:

\[
y_p(t) = A \sin(5t) + B \cos(5t)
\]

### 3. **Find the derivatives of \( y_p(t) \):**

\[
y'_p(t) = 5A \cos(5t) - 5B \sin(5t)
\]
\[
y''_p(t) = -25A \sin(5t) - 25B \cos(5t)
\]

### 4. **Substitute \( y_p \), \( y'_p \), and \( y''_p \) into the original equation:**

Substitute into:

\[
y'' + 5y' = 300 \sin(5t) + 300 \cos(5t)
\]

We get:

\[
(-25A \sin(5t) - 25B \cos(5t)) + 5(5A \cos(5t) - 5B \sin(5t)) = 300 \sin(5t) + 300 \cos(5t)
\]

Simplifying:

\[
(-25A + 25B) \sin(5t) + (25A - 25B) \cos(5t) = 300 \sin(5t) + 300 \cos(5t)
\]

Equating coefficients of \( \sin(5t) \) and \( \cos(5t) \):

\[
-25A + 25B = 300 \quad \text{(1)}
\]
\[
25A - 25B = 300 \quad \text{(2)}
\]

From (1) and (2), solve for \( A \) and \( B \):

- Adding both equations: 
\[
0 = 600 \quad \Rightarrow B = 12
\]
- Substituting \( B = 12 \) into (1):
\[
-25A + 25(12) = 300 \quad \Rightarrow A = -12
\]

Thus, the particular solution is:

\[
y_p(t) = -12 \sin(5t) + 12 \cos(5t)
\]

### 5. **General solution:**

The general solution to the differential equation is:

\[
y(t) = y_h(t) + y_p(t) = C_1 + C_2 e^{-5t} - 12 \sin(5t) + 12 \cos(5t)
\]

### 6. **Apply initial conditions:**

- \( y(0) = 3 \):

\[
C_1 + C_2 e^{0} - 12 \sin(0) + 12 \cos(0) = 3
\]
\[
C_1 + C_2 + 12 = 3 \quad \Rightarrow C_1 + C_2 = -9
\]

- \( y'(t) = -5C_2 e^{-5t} - 60 \cos(5t) - 60 \sin(5t) \), and \( y'(0) = 4 \):

\[
y'(0) = -5C_2 - 60(1) = 4
\]
\[
-5C_2 - 60 = 4 \quad \Rightarrow -5C_2 = 64 \quad \Rightarrow C_2 = -\frac{64}{5}
\]

Substitute \( C_2 = -\frac{64}{5} \) into \( C_1 + C_2 = -9 \):

\[
C_1 - \frac{64}{5} = -9 \quad \Rightarrow C_1 = -9 + \frac{64}{5} = \frac{19}{5}
\]

### 7. **Final solution:**

\[
y(t) = \frac{19}{5} - \frac{64}{5} e^{-5t} - 12 \sin(5t) + 12 \cos(5t)
\]

Would you like more details on any step or have questions?

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**Related Questions:**

1. How do you solve second-order non-homogeneous differential equations in general?
2. Can you solve a similar equation with different sinusoidal terms?
3. How do initial conditions affect the solution of differential equations?
4. How is the method of undetermined coefficients applied to non-homogeneous equations?
5. How would the solution change if the characteristic equation had complex roots?

**Tip:** When dealing with sinusoidal forcing functions, particular solutions are often sinusoidal functions with the same frequency.

y’’+5y’=300sin(5t)+300cos(5t), y(0)=3, y’(0)=4

This page covers the detailed solution to the non-homogeneous differential equation y'' + 5y' = 300sin(5t) + 300cos(5t) with initial conditions y(0)=3 and y'(0)=4. Learn how to apply the method of undetermined coefficients and solve using characteristic equations. Suitable for college-level differential equations.

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