Chapter 4: Q43E (page 200)

Find a second linearly independent solution using reduction of order.
$tx'' - (t + 1) x' + x = 0, t > 0; f (t) = e^{t}$

Short Answer

Expert verified

The second linearly independent solution of the given equation

$tx'' - (t + 1) x' + x = 0, t > 0; f (t) = e^{t}$ is $y = c_{1} e^{t} - c_{2} \frac{t + 1}{e^{2 t}}$

Step by step solution

Finding y2

Given differential equation is $tx'' - (t + 1) x' + x = 0$ then the standard form of the solution is $x'' - \frac{t + 1}{t} x' + \frac{1}{t} x = 0$ and $f (t) = e^{t}$ is one of the solutions and role="math" localid="1664197935206" $p (t) = - \frac{t + 1}{t}$ . Then;

$\begin{array}{rcl} y_{2} & = & y_{1} (t) \int \frac{e^{- \int p (t)} dt}{y_{1}^{2} (t)} dt \\ = & e^{t} \int \frac{e^{\int \frac{1 + t}{t}} dt}{{(e^{t})}^{2}} dt \\ = & e^{t} \int \frac{e^{\ln (t^{1}) + t}}{e^{2 t}} \end{array}$

Simplification

Simplify the above

$= e^{t} \times \int \frac{t \times e^{t}}{e^{2 t}} = e^{t} \times \frac{(- t - 1) e^{- t}}{e^{2 t}} = - \frac{t + 1}{e^{2 t}}$

So, the solution is $y = c_{1} e^{t} - c_{2} \frac{t + 1}{e^{2 t}}$ .

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Find a second linearly independent solution using reduction of order.
$tx'' - (t + 1) x' + x = 0, t > 0; f (t) = e^{t}$

Short Answer

Step by step solution

Finding y2

Simplification

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Find a second linearly independent solution using reduction of order.tx''-(t+1)x'+x=0,t>0;f(t)=et

Short Answer

Step by step solution

Finding y2

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Probability and Statistics

Discrete Mathematics

Pure Maths

Decision Maths

Study anywhere. Anytime. Across all devices.

Find a second linearly independent solution using reduction of order.
$tx'' - (t + 1) x' + x = 0, t > 0; f (t) = e^{t}$