Solve: y' + 3y = e':) By Method of Variation of Parameters :

Answered: Solve: y' + 3y = e' :) By Method of…

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: y' + 3y = e'
:) By Method of Variation of Parameters :

Expert Solution

Step 1

Given:

Given differential equation is

$y' + 3 y = e^{t}$

We solve this equation by variation of parameter method.

Step 2

Calculation:

First we find hommogeneous solution of the given differential equation

$y' + 3 y = e^{t}$

Associated homogeneous equation is

$y' + 3 y = 0$

Auxilliary equation is

$m + 3 = 0 \Rightarrow m = - 3$

Therefore hommogeneous solution is

$y_{c} = c e^{- 3 t} w h e r e y_{1} = e^{- 3 t}$

Now to find particular solution let

$y_{p} = y_{1} w (t) . . . . . . . . . . . . . . . . . . . . . . . (1) N o w,$

$\begin{array}{rcl} w' (t) & = & {y_{1}}^{- 1} g (t) \\ = & \frac{1}{e^{- 3 t}} \times e^{t} \\ = & e^{3 t} \times e^{t} \\ w' (t) & = & e^{4 t} \end{array}$

Therefore

$\begin{array}{rcl} w (t) & = & \int e^{4 t} d t \\ w (t) & = & \frac{e^{4 t}}{4} \end{array}$

Therefore ,

$\begin{array}{rcl} y_{p} & = & y_{1} w (t) \\ = & e^{- 3 t} \times \frac{e^{4 t}}{4} \\ y_{p} & = & \frac{e^{t}}{4} \end{array}$

Therefore the general solution of the differential equation $y' + 3 y = e^{t}$ is

$\begin{array}{rcl} y & = & y_{c} + y_{p} \\ y & = & c e^{- 3 t} + \frac{e^{t}}{4} \end{array}$

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