The problem presented is a second-order linear homogeneous differential equation:

$4y'' = 3y'$

with initial conditions: $y(0) = 4, \quad y'(0) = 12.$

We are required to solve this using the superposition method.

Step 1: Rewrite the differential equation

Divide the entire equation by 4 to simplify:

$y'' = \frac{3}{4}y'.$

This is now a first-order differential equation in terms of $y'$, i.e.,

$y'' = \frac{3}{4} y'.$

Step 2: Solving the differential equation

Let $y' = v(x)$. Then, the equation becomes:

$v'(x) = \frac{3}{4} v(x).$

This is a separable differential equation. We can solve it by separating variables:

$\frac{dv}{v} = \frac{3}{4} dx.$

Integrating both sides:

$\ln|v| = \frac{3}{4} x + C_1,$

where $C_1$ is a constant of integration. Exponentiating both sides:

$v(x) = y'(x) = C_2 e^{\frac{3}{4}x},$

where $C_2 = e^{C_1}$ is a constant.

Step 3: Integrating to find $y(x)$

Now we integrate $y'(x) = C_2 e^{\frac{3}{4}x}$ to find $y(x)$:

$y(x) = \int C_2 e^{\frac{3}{4}x} dx.$

This gives:

$y(x) = \frac{4}{3} C_2 e^{\frac{3}{4}x} + C_3,$

where $C_3$ is another constant of integration.

Step 4: Apply initial conditions

We now use the initial conditions to solve for $C_2$ and $C_3$.

• $y(0) = 4$:

$4 = \frac{4}{3} C_2 + C_3.$

• $y'(0) = 12$:

$12 = C_2 e^{\frac{3}{4} \cdot 0} = C_2.$

Thus, $C_2 = 12$.

Now substitute $C_2 = 12$ into the first equation:

$4 = \frac{4}{3}(12) + C_3,$

$4 = 16 + C_3,$

$C_3 = 4 - 16 = -12.$

Final Solution:

Therefore, the solution to the differential equation is:

$y(x) = \frac{4}{3}(12)e^{\frac{3}{4}x} - 12,$

which simplifies to:

$y(x) = 16 e^{\frac{3}{4}x} - 12.$

Would you like further clarification or details on any of these steps?

Here are 5 related questions you might find interesting:

1. What is the significance of the superposition method in solving differential equations?
2. How do initial conditions affect the particular solution of a differential equation?
3. What is the role of the constant of integration in solving differential equations?
4. How would the solution differ if the right-hand side of the differential equation was non-homogeneous?
5. Can this method be extended to higher-order differential equations?

Tip: Always check that your solution satisfies both the differential equation and the given initial conditions.