One solution of t + 2 t =x² + t + 2 is x₁ (t) = t. Find a second independent X1 - x = 0 t² solution.
One solution of t + 2 t =x² + t + 2 is x₁ (t) = t. Find a second independent X1 - x = 0 t² solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please do number 5
![**4.** Consider the equation \( x'' - 2ax' + a^2x = 0 \), which has solution \( x = e^{at} \). Use reduction of order to find a second independent solution. (This shows the origin of the \( te^{at} \) solution in a second-order linear equation with constant coefficients, in the real, equal eigenvalue case.)
**5.** One solution of
\[
x'' - \frac{t+2}{t}x' + \frac{t+2}{t^2}x = 0
\]
is \( x_1(t) = t \). Find a second independent solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc10cce41-ef51-4545-a724-9736e236d3b7%2Fb5ff52bd-4624-4ae4-8624-0c365a9fb492%2Fgjlesod_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**4.** Consider the equation \( x'' - 2ax' + a^2x = 0 \), which has solution \( x = e^{at} \). Use reduction of order to find a second independent solution. (This shows the origin of the \( te^{at} \) solution in a second-order linear equation with constant coefficients, in the real, equal eigenvalue case.)
**5.** One solution of
\[
x'' - \frac{t+2}{t}x' + \frac{t+2}{t^2}x = 0
\]
is \( x_1(t) = t \). Find a second independent solution.
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