Find the general solution of the given second-order differential equation: $y'' - y' - 12y = 0$

Neetesh Kumar | November 06, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec03 (Homework) Question - 1
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Step-by-Step-Solution:

We start by finding the characteristic equation to solve this second-order differential equation.

Step 1: Write the Characteristic Equation

The characteristic equation for the differential equation $y'' - y' - 12y = 0$ is obtained by assuming a solution of $y = e^{rx}$, where $r$ is a constant. Substituting this into the differential equation gives: $r^2 e^{rx} - r e^{rx} - 12 e^{rx} = 0$

Dividing by $e^{rx}$ (since $e^{rx} \neq 0$), we get: $r^2 - r - 12 = 0$

Step 2: Solve the Characteristic Equation

Now, we have a quadratic equation: $r^2 - r - 12 = 0$

We can solve this quadratic equation using the factoring:

$(r - 4)(r - 3) = 0$

This gives us two roots: $r_1 = 4$ and $r_2 = -3$

Step 3: Write the General Solution

Since we have two distinct real roots, $r_1 = 4$ and $r_2 = -3$, the general solution of the differential equation is:

$y(x) = C_1 e^{4x} + C_2 e^{-3x}$

where $C_1$ and $C_2$ are arbitrary constants.

Final Answer

The general solution is: $y(x) = \boxed{C_1 e^{4x} + C_2 e^{-3x}}$

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