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This problem involves solving a second-order linear homogeneous differential equation with initial conditions. The equation is:

\[
y'' + 4y' - 5y = 0, \quad y(0) = 3, \quad y'(0) = -3.
\]

### Step 1: Finding the General Solution
1. **Form the characteristic equation** for the differential equation:
   
\[
r^2 + 4r - 5 = 0.
\]

2. **Solve the characteristic equation** using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -5 \):

\[
r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2}.
\]

This gives two distinct roots:

\[
r_1 = 1, \quad r_2 = -5.
\]

3. **General solution** for the differential equation:
   
\[
y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^t + C_2 e^{-5t}.
\]

### Step 2: Applying Initial Conditions
1. **Initial condition 1**: \( y(0) = 3 \).

\[
y(0) = C_1 e^0 + C_2 e^{-5 \cdot 0} = C_1 + C_2 = 3.
\]

2. **Initial condition 2**: \( y'(0) = -3 \).

First, compute the derivative of \( y(t) \):

\[
y'(t) = C_1 e^t - 5C_2 e^{-5t}.
\]

Now, apply the initial condition:

\[
y'(0) = C_1 e^0 - 5C_2 e^{-5 \cdot 0} = C_1 - 5C_2 = -3.
\]

### Step 3: Solving the System of Equations
We now have the following system of equations:

1. \( C_1 + C_2 = 3 \).
2. \( C_1 - 5C_2 = -3 \).

Solve these equations simultaneously:

From equation 1: \( C_1 = 3 - C_2 \).

Substitute \( C_1 \) into equation 2:

\[
(3 - C_2) - 5C_2 = -3 \Rightarrow 3 - 6C_2 = -3 \Rightarrow -6C_2 = -6 \Rightarrow C_2 = 1.
\]

Now, find \( C_1 \):

\[
C_1 = 3 - C_2 = 3 - 1 = 2.
\]

### Step 4: Final Solution
The solution to the differential equation is:

\[
y(t) = 2e^t + e^{-5t}.
\]

Would you like further details or have any questions on this solution?

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### Related Questions:
1. How would the solution change if the coefficients in the differential equation were different?
2. What happens if the initial conditions are altered?
3. How can we verify the solution using software tools?
4. How do we handle non-homogeneous differential equations?
5. What if the characteristic equation has complex roots?

### Tip:
When solving homogeneous linear differential equations, always start by finding the characteristic equation and solving for its roots to construct the general solution.

Solve y'' + 4y' - 5y = 0, y(0) = 3, y'(0) = -3

Learn how to solve the second-order linear homogeneous differential equation y'' + 4y' - 5y = 0 with initial conditions y(0) = 3 and y'(0) = -3. The solution involves finding the characteristic equation and applying the quadratic formula to find the general solution.

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