Finding The Solution of A Differential Equation

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
5

Homework Statement


y' + 3y = t + exp(2t)


Homework Equations





The Attempt at a Solution


My solution is y = t/3 + exp(-2t) -1/9 + c. The solution in the back of the book is y = cexp(-3t) + (t/3) - (1/9) + exp(-2t). How come the two solutions differ so vastly? I followed the procedure outlined in the textbook precisely.
 
Last edited:

Answers and Replies

  • #2
STEMucator
Homework Helper
2,075
140

Homework Statement


y' + 3y = t + exp(2t)


Homework Equations





The Attempt at a Solution


My solution is y = t/3 + exp(-2t) -1/9 + c. The solution in the back of the book is y = cexp(-3t) + (t/3) - (1/9) + exp(-2t).
The solution I'm getting is :

##y(t) = \frac{c}{e^{3 t}} + \frac{t}{3} + \frac{e^{2 t}}{5} - \frac{1}{9}##

You should check your work over again I think. You may have made a slight error.
 
Last edited:
  • #3
1,421
5
I got the exact same answer. I attached my work as a file. According to my text book, the solution to this sort of differential equation is:

[itex]y = \frac{1}{\mu (t)} \int \mu (t) g(t) dt + c[/itex], where [itex]\mu (t) = e^{\int p(t)dt}[/itex]
 

Attachments

  • #4
STEMucator
Homework Helper
2,075
140
I got the exact same answer. I attached my work as a file. According to my text book, the solution to this sort of differential equation is:

[itex]y = \frac{1}{\mu (t)} \int \mu (t) g(t) dt + c[/itex], where [itex]\mu (t) = e^{\int p(t)dt}[/itex]
Yes your book is 100% correct ( Barring I'd leave constants all the way until the end ), but don't neglect the ##e^{3t}## in the denominator. It will vary the value of ##c## if you're ever given an initial value problem.
 
Last edited:
  • #5
1,421
5
Oh, I see. So exp(-3t) would multiply the c, after I did the integration.
 
  • #6
STEMucator
Homework Helper
2,075
140
Oh, I see. So exp(-3t) would multiply the c, after I did the integration.
It will divide it actually. After evaluating the integrals I got this :

##e^{3t}y(t) = e^{3t} ( \frac{e^{2t}}{5} + \frac{t}{3} - \frac{1}{9} ) + c##
 

Related Threads on Finding The Solution of A Differential Equation

Replies
2
Views
3K
Replies
1
Views
2K
Replies
6
Views
11K
Replies
3
Views
1K
Replies
10
Views
5K
Replies
3
Views
812
Replies
9
Views
1K
Replies
2
Views
1K
Top