Solve the initial value problem(t² – 5)a' – 2t.x = -2t(t² – 5), x(2) = 7.%3D

Answered: Solve the initial value problem (t2 –…

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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Question

Solve the initial value problem
(t² – 5)a' – 2t.x = -2t(t² – 5), x(2) = 7.
%3D

Expert Solution

Step 1

Given differential equation is $(t^{2} - 5) x^{'} - 2 t x = - 2 t (t^{2} - 5) . . . . . . (1)$ and $x (2) = 7$ .

We have to find the solution of the given initial value problem.

Now, simplify the equation $(1)$

$\begin{array}{l} \Rightarrow \frac{(t^{2} - 5)}{(t^{2} - 5)} x^{'} - \frac{2 t}{(t^{2} - 5)} x = - \frac{2 t (t^{2} - 5)}{(t^{2} - 5)} (Divide both sides by (t^{2} - 5)) \\ \Rightarrow \frac{d x}{d t} - \frac{2 t}{(t^{2} - 5)} x = - 2 t \end{array}$

Hence, equation $\frac{d x}{d t} - \frac{2 t}{(t^{2} - 5)} x = - 2 t$ is linear in $x$ and have the form $\frac{d x}{d t} + P (t) x = Q (t)$

Now, first find the integrating factor

$\begin{matrix} I . F = e^{\int P (t) d t} \\ = e^{\int - \frac{2 t}{(t^{2} - 5)} d t} \\ = e^{- \ln (t^{2} - 5)} \\ = e^{\ln \frac{1}{(t^{2} - 5)}} \\ = \frac{1}{(t^{2} - 5)} \end{matrix}$