+0  
 
0
131
3
avatar+206 

As the title suggests, you are given two initial conditions, y(0) = 0 and y'(pi)=1

 

My initial idea was to perform solve this by laplace transform but one of the initial conditions was y'(pi) and not of zero and I don't know how to use that as the laplace transform only uses y'(0).

 

 

I assumed that y must be a function of some sort like y = A*sin ax + B*cos bx

 

After trying to put that into the equation and trying to simplify I ended up with contradicting information regarding the constants and couldn't progress. I tried doing it again and must have made another error somewhere. I would appreciate a walkthrough of this question. Thank you in advance.

Quazars  Apr 20, 2017
edited by Quazars  Apr 20, 2017
Sort: 

3+0 Answers

 #1
avatar
0

Did you try to see if Wolfram/Alpha will give you an answer?

Guest Apr 20, 2017
 #2
avatar+75323 
+3

y"  +  25y   =  3sin2x

 

First....let's find the solution to the homogenous equation 

 

y" + 25y  = 0

 

The characteristic  equation  for this is

r^2  +  25   =  0     →    r   = ± 5i

 

So, the complementary  solution to this homogenous equation is 

 

y  =  c1 cos(5x)  +  c2 sin (5x)      (1)

 

Now let's solve the non-homogenous equation

 

y"  +  25y   =  3sin2x

 

Guess that the solution  is

y  = Acos 2x  + Bsin2x

y' =  -2Asin2x  + 2Bcos2x

y"  =  -4Acos2x -4Bsin2x

 

So  we have

 

-4Acos2x - 4Bsin2x  + 25(Acos2x + Bsin2x)  =  3sin2x

(25A -4A)cos2x   + (25B - 4B)sin2x  =  3sin2x

21Acos2x   +  21Bsin2x  =  3sin2x

This implies that A  = 0       and  B  =   1/7

 

So......the solution to  this is that

y  =  (1/7)sin2x    (2)

 

Combining (1)  and (2)  we have this

 

y  =   c1 cos(5x)  +  c2 sin (5x)   + (1/7)sin(2x)

y'  = - 5c1 sin (5x) + 5c2 cos(5x) + (2/7)cos(2x)

 

Applying the initial conditions

y(0)  =  c1 (1) = 0    →  c1  = 0 

y' (pi)  = 1

5c2 (-1)  + (2/7)(1)  = 1

-5c2 + 2/7  = 1

-5c2 =  5/7

c2  =  -1/7

 

 

So.... the solution with the applied intial conditions is

 

 y  =  (-1/7)sin(5x) + (1/7) sin(2x)

And

y' =  (-5/7)cos(5x)  + (2/7)cos(2x)

 

Check

 

y(0)  = 0 ??

(-1/7)(0)  + (1/7)(0)  = 0

 

y'(pi)   =  1   ??

(-5/7)(-1)  + (2/7)(1)

5/7  +  2/7   =  1

  

 

cool cool cool

CPhill  Apr 20, 2017
edited by CPhill  Apr 21, 2017
 #3
avatar+206 
+1

I'm sure I say this every time but you are a life saver. Thank you :)

Quazars  Apr 21, 2017

6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details