Problem 1:

We are tasked with finding the second solution $y_2(x)$ to the differential equation:

$y'' - 8y' + 16y = 0$ where $y_1 = e^{4x}$ is a given solution.

Solution Approach:

The general formula for finding the second solution $y_2(x)$ using reduction of order is:

$y_2 = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$

Where $P(x)$ is the coefficient of $y'$ in the standard form of the equation, which is:

$y'' + P(x) y' + Q(x) y = 0$

Step 1: Identify $P(x)$

Rewriting the given equation:

$y'' - 8y' + 16y = 0$

This is already in the standard form $y'' + P(x) y' + Q(x) y = 0$, where $P(x) = -8$ and $Q(x) = 16$.

Step 2: Apply the Reduction of Order Formula

Using the formula for the second solution:

$y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$

Substitute $P(x) = -8$ and $y_1(x) = e^{4x}$:

$y_2(x) = e^{4x} \int \frac{e^{-\int -8 dx}}{(e^{4x})^2} dx$

Now calculate the integrals:

$\int -8 dx = -8x$

Thus:

$y_2(x) = e^{4x} \int \frac{e^{8x}}{e^{8x}} dx = e^{4x} \int 1 dx$

The integral of 1 with respect to $x$ is $x$, so:

$y_2(x) = e^{4x} \cdot x = x e^{4x}$

Thus, the second solution is:

$y_2(x) = x e^{4x}$

Problem 2:

We are tasked with finding the second solution $y_2(x)$ to the differential equation:

$y'' + 2y' + y = 0$ where $y_1 = x e^{-x}$ is a given solution.

Step 1: Identify $P(x)$

Rewriting the given equation:

$y'' + 2y' + y = 0$

This is in the standard form $y'' + P(x) y' + Q(x) y = 0$, where $P(x) = 2$ and $Q(x) = 1$.

Step 2: Apply the Reduction of Order Formula

Using the formula for the second solution:

$y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$

Substitute $P(x) = 2$ and $y_1(x) = x e^{-x}$:

$y_2(x) = x e^{-x} \int \frac{e^{-\int 2 dx}}{(x e^{-x})^2} dx$

Now calculate the integrals:

$\int 2 dx = 2x$

Thus:

$y_2(x) = x e^{-x} \int \frac{e^{-2x}}{x^2 e^{-2x}} dx = x e^{-x} \int \frac{1}{x^2} dx$

The integral of $\frac{1}{x^2}$ is $-\frac{1}{x}$, so:

$y_2(x) = x e^{-x} \left( -\frac{1}{x} \right) = -e^{-x}$

Thus, the second solution is:

$y_2(x) = -e^{-x}$

Final Answers:

• For the first problem: $y_2(x) = x e^{4x}$
• For the second problem: $y_2(x) = -e^{-x}$

Would you like further details or clarifications on any part of the solution? Here are some related questions you may find interesting:

1. How can you apply the reduction of order method to higher-order linear differential equations?
2. What other methods can be used to find a second solution of a linear differential equation when one solution is already known?
3. How can we verify that the second solution $y_2(x)$ is independent of the first solution $y_1(x)$?
4. What is the general solution of a second-order linear homogeneous differential equation?
5. How does the concept of reduction of order relate to the method of undetermined coefficients?

Tip: Always double-check your integrals and ensure that the limits or indefinite integrals are handled correctly before substituting back into the formula