find the solution of y''+y'-2y=36e^4t with y(0)=7 and y'(0)=1.

We are tasked with solving the non-homogeneous second-order differential equation:

\[
y'' + y' - 2y = 36e^{4t}, \quad \text{with initial conditions} \ y(0) = 7 \ \text{and} \ y'(0) = 1.
\]

### Step 1: Solve the homogeneous equation
The homogeneous equation is:

\[
y'' + y' - 2y = 0.
\]

Assume the solution is of the form \( y_h = e^{rt} \). Substituting \( y_h = e^{rt} \) into the homogeneous equation gives the characteristic equation:

\[
r^2 + r - 2 = 0.
\]

Solve the characteristic equation:

\[
r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}.
\]

Thus, the roots are:

\[
r_1 = 1, \quad r_2 = -2.
\]

Therefore, the general solution to the homogeneous equation is:

\[
y_h(t) = c_1 e^{t} + c_2 e^{-2t}.
\]

### Step 2: Solve the particular solution
We need a particular solution \( y_p \) for the non-homogeneous equation. Since the non-homogeneous term is \( 36e^{4t} \), we try a particular solution of the form:

\[
y_p(t) = A e^{4t}.
\]

Substitute \( y_p(t) \) into the differential equation \( y'' + y' - 2y = 36e^{4t} \):

\[
(A \cdot 16 e^{4t}) + (A \cdot 4 e^{4t}) - 2A e^{4t} = 36e^{4t}.
\]

Factor out \( e^{4t} \):

\[
A(16 + 4 - 2) e^{4t} = 36e^{4t},
\]

which simplifies to:

\[
A(18) e^{4t} = 36e^{4t}.
\]

So, \( A = 2 \). Thus, the particular solution is:

\[
y_p(t) = 2e^{4t}.
\]

### Step 3: General solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(t) = y_h(t) + y_p(t) = c_1 e^{t} + c_2 e^{-2t} + 2e^{4t}.
\]

### Step 4: Apply initial conditions
Now, use the initial conditions \( y(0) = 7 \) and \( y'(0) = 1 \) to solve for \( c_1 \) and \( c_2 \).

1. At \( t = 0 \), we have \( y(0) = 7 \):

\[
c_1 e^{0} + c_2 e^{0} + 2e^{0} = 7,
\]
\[
c_1 + c_2 + 2 = 7,
\]
\[
c_1 + c_2 = 5. \quad \text{(Equation 1)}
\]

2. Differentiate \( y(t) \) to find \( y'(t) \):

\[
y'(t) = c_1 e^{t} - 2c_2 e^{-2t} + 8e^{4t}.
\]

At \( t = 0 \), we have \( y'(0) = 1 \):

\[
c_1 e^{0} - 2c_2 e^{0} + 8e^{0} = 1,
\]
\[
c_1 - 2c_2 + 8 = 1,
\]
\[
c_1 - 2c_2 = -7. \quad \text{(Equation 2)}
\]

### Step 5: Solve the system of equations
From Equation 1: \( c_1 + c_2 = 5 \).

From Equation 2: \( c_1 - 2c_2 = -7 \).

Solve this system of equations:

1. Add the two equations:
   \[
   (c_1 + c_2) + (c_1 - 2c_2) = 5 + (-7),
   \]
   \[
   2c_1 - c_2 = -2. \quad \text{(Equation 3)}
   \]

2. Substitute \( c_1 + c_2 = 5 \) into Equation 3:
   \[
   2(5 - c_2) - c_2 = -2,
   \]
   \[
   10 - 2c_2 - c_2 = -2,
   \]
   \[
   10 - 3c_2 = -2,
   \]
   \[
   -3c_2 = -12,
   \]
   \[
   c_2 = 4.
   \]

3. Substitute \( c_2 = 4 \) into \( c_1 + c_2 = 5 \):
   \[
   c_1 + 4 = 5,
   \]
   \[
   c_1 = 1.
   \]

### Step 6: Final solution
Thus, the final solution is:

\[
y(t) = e^{t} + 4e^{-2t} + 2e^{4t}.
\]

Would you like further details or clarifications on any part?

Here are 5 related questions you might find useful:
1. How do we handle non-homogeneous differential equations with different forcing functions?
2. What are the general steps for solving second-order linear differential equations?
3. How can initial conditions change the nature of a solution to a differential equation?
4. What are the differences between particular and homogeneous solutions in ODEs?
5. How can we verify that a solution to a differential equation is correct?

**Tip:** Always check if the forcing function resembles the terms of the homogeneous solution to avoid overlap when guessing the particular solution form.

This page explains how to solve the second-order non-homogeneous differential equation y'' + y' - 2y = 36e^4t with initial conditions y(0) = 7 and y'(0) = 1. It covers the step-by-step solution using the method of undetermined coefficients, solving both the homogeneous and particular solutions. Suitable for college-level students.

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