Chapter 4: Q42E (page 200)

A differential equation and a nontrivial solution fare given. Find a second linearly independent solution using reduction of order.
$t^{2} y'' + 6 ty' + 6 y = 0, t > 0; f (t) = t^{- 2}$

Short Answer

Expert verified

The second linearly independent solution of the given equation

$t^{2} y'' + 6 ty' + 6 y = 0, t > 0; f (t) = t^{- 2}$ is $y = c_{1} t^{- 2} + c_{2} t^{- 3}$ .

Step by step solution

Substitute the values

Given differential equation is $t^{2} y'' + 6 ty' + 6 y = 0$ then the standard form of the solution is $y'' + \frac{6}{t} y' + \frac{6}{t^{2}} y = 0$ and $f (t) = t^{- 2}$ is one of the solutions and $p (t) = \frac{6}{t}$ . Then

$\begin{array}{rcl} y_{2} & = & y_{1} (t) \int \frac{e^{- \int p (t)} dt}{y_{1}^{2} (t)} dt \\ = & t^{- 2} \int \frac{e^{- \int \frac{6}{t}} dt}{{(t^{- 2})}^{2}} dt \end{array}$

Simplification

Simplify the above and get;

$= t^{- 2} \int \frac{e^{\ln (t^{- 6})}}{t^{- 4}} = t^{- 2} \times \int \frac{t^{- 6}}{t^{- 4}} = t^{- 2} \times (- t^{- 1}) = - t^{- 3}$

So, the solution is $y = c_{1} t^{- 2} + c_{2} t^{- 3}$ .

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A differential equation and a nontrivial solution fare given. Find a second linearly independent solution using reduction of order.
$t^{2} y'' + 6 ty' + 6 y = 0, t > 0; f (t) = t^{- 2}$

Short Answer

Step by step solution

Substitute the values

Simplification

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A differential equation and a nontrivial solution fare given. Find a second linearly independent solution using reduction of order.t2y''+6ty'+6y=0,t>0;f(t)=t-2

Short Answer

Step by step solution

Substitute the values

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Applied Mathematics

Geometry

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

A differential equation and a nontrivial solution fare given. Find a second linearly independent solution using reduction of order.
$t^{2} y'' + 6 ty' + 6 y = 0, t > 0; f (t) = t^{- 2}$