Solving the Second-Order Linear Differential Equation: y'' - 4y' - 4y 0 with Initial Conditions

When dealing with second-order linear differential equations, it's crucial to understand the techniques and steps involved. This article will guide you through solving the differential equation y'' - 4y' - 4y 0, providing a detailed walkthrough with initial conditions y(0) 3 and y'(0) 7. By the end, you'll have a solid understanding of how to approach and solve such equations.

Determining the Type and Solving the Equation

This is a linear homogenous second order differential equation with constant coefficients. The equation can be rewritten in primes notation for clarity.

Rewrite the Equation in Primes Notation

In primes notation, the differential equation y'' - 4y' - 4y 0 can be expressed as:

Solution Steps

Find the Characteristic Equation: Substitute y e^{rx} into the differential equation, resulting in the characteristic equation: Solve the Characteristic Equation: Factor the characteristic equation to find its roots. In this case, we'll obtain a repeated root. Write the General Solution: For a repeated root, the general solution will include a term with x. Apply Initial Conditions: Use the initial conditions to determine the constants in the general solution. Final Solution: Substitute the constants back into the general solution.

Let's go through each step in detail.

Step 1: Find the Characteristic Equation

Substituting y e^{rx} into the equation y'' - 4y' - 4y 0 gives:

r^2 - 4r - 4  0

Step 2: Solve the Characteristic Equation

The characteristic equation can be factored, and we'll use the quadratic formula if necessary:

r^2 - 4r - 4  (r - 2)^2  0

This gives a repeated root:

r  -2

Step 3: Write the General Solution

With a repeated root, the general solution is:

y(x)  C_1 e^{-2x}   C_2 x e^{-2x}

Where C_1 and C_2 are constants determined by the initial conditions.

Step 4: Apply the Initial Conditions

The first initial condition is y(0) 3. Substituting x 0 in the general solution gives:

3  C_1 e^{0}   C_2 cdot 0 cdot e^{0}  C_1

So, C_1 3.

For the second initial condition, we need to take the first derivative of the general solution:

y'(x)  -2C_1 e^{-2x}   C_2 e^{-2x} - 2C_2 x e^{-2x}

Substituting x 0 and C_1 3 gives:

7  -2 cdot 3 cdot e^{0}   C_2 cdot e^{0}  -6   C_2

Therefore, C_2 13.

Step 5: Final Solution

Substituting C_1 3 and C_2 13 back into the general solution:

y(x)  3 e^{-2x}   13 x e^{-2x}

The solution to the differential equation with the given initial conditions is:

boxed{y(x)  3 e^{-2x}   13 x e^{-2x}}

Understanding Linear Differential Equations and Initial Conditions

For the readers to have a better understanding of the concepts involved, it's important to revisit the basics:

What is a Linear Differential Equation?

A linear differential equation is one where the dependent variable and its derivatives appear only to the first power and are not multiplied together.

What is a Linear Homogeneous Equation?

A linear homogeneous differential equation is one where the right-hand side is zero. The method of solving such an equation involves finding the characteristic equation and its roots.

Initial Conditions and Constants of Integration

Initial conditions are used to determine the specific solution by providing values of the function and its derivatives at specific points.

Conclusion

In summary, solving second-order linear differential equations involves several steps, including finding the characteristic equation, solving it for its roots, writing the general solution, and using initial conditions to find the constants. The example provided demonstrates these steps clearly, leading to the final solution for the given problem.