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.