We are given the second-order linear ordinary differential equation (ODE):

$t^2 x'' + 4t x' + 2x = 0 \quad \text{for } t > 0$ We need to show that $x_1(t) = \frac{1}{t}$ and $x_2(t) = \frac{1}{t^2}$ are solutions to the ODE. Additionally, we need to show that the general solution is of the form:

$x(t) = \frac{c_1}{t} + \frac{c_2}{t^2} \quad \text{where } c_1, c_2 \in \mathbb{R}.$

Step 1: Verify that $x_1(t) = \frac{1}{t}$ is a solution

Let’s substitute $x_1(t) = \frac{1}{t}$ into the ODE.

• First, calculate $x_1'(t)$ and $x_1''(t)$: $x_1'(t) = -\frac{1}{t^2}, \quad x_1''(t) = \frac{2}{t^3}.$

• Now substitute $x_1(t)$, $x_1'(t)$, and $x_1''(t)$ into the ODE:

  $t^2 \cdot \frac{2}{t^3} + 4t \cdot \left(-\frac{1}{t^2}\right) + 2 \cdot \frac{1}{t} = 0.$

  Simplifying: $\frac{2}{t} - \frac{4}{t} + \frac{2}{t} = 0,$ which simplifies to: $0 = 0.$ Thus, $x_1(t) = \frac{1}{t}$ is indeed a solution to the ODE.

Step 2: Verify that $x_2(t) = \frac{1}{t^2}$ is a solution

Next, let’s substitute $x_2(t) = \frac{1}{t^2}$ into the ODE.

• First, calculate $x_2'(t)$ and $x_2''(t)$: $x_2'(t) = -\frac{2}{t^3}, \quad x_2''(t) = \frac{6}{t^4}.$

• Now substitute $x_2(t)$, $x_2'(t)$, and $x_2''(t)$ into the ODE:

  $t^2 \cdot \frac{6}{t^4} + 4t \cdot \left(-\frac{2}{t^3}\right) + 2 \cdot \frac{1}{t^2} = 0.$

  Simplifying: $\frac{6}{t^2} - \frac{8}{t^2} + \frac{2}{t^2} = 0,$ which simplifies to: $0 = 0.$ Thus, $x_2(t) = \frac{1}{t^2}$ is also a solution to the ODE.

Step 3: General solution

Since $x_1(t) = \frac{1}{t}$ and $x_2(t) = \frac{1}{t^2}$ are two linearly independent solutions to the homogeneous second-order linear differential equation, the general solution is a linear combination of these two solutions. That is, the general solution is:

$x(t) = c_1 x_1(t) + c_2 x_2(t) = c_1 \frac{1}{t} + c_2 \frac{1}{t^2},$ where $c_1$ and $c_2$ are arbitrary constants.

Thus, the general solution of the ODE is:

$x(t) = \frac{c_1}{t} + \frac{c_2}{t^2}.$

Conclusion

We have shown that $x_1(t) = \frac{1}{t}$ and $x_2(t) = \frac{1}{t^2}$ are solutions to the given ODE, and that the general solution is $x(t) = \frac{c_1}{t} + \frac{c_2}{t^2}$.

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Would you like more detailed explanations or have any questions? Here are some related questions for further exploration:

1. How do we check if two solutions are linearly independent?
2. What is the general form of a second-order linear homogeneous differential equation?
3. How does the method of undetermined coefficients work in solving ODEs?
4. What is the significance of finding two linearly independent solutions in solving second-order ODEs?
5. Can this method be applied to non-linear differential equations?

Tip: When solving linear ODEs with constant coefficients, you can often find the general solution by finding the roots of the characteristic equation.