Differential equation Task 7

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Task 7
Show that y=(1/4)tsin2t satisfies equation

d2y/dt2+4y=cos2t

Find the general solution and deduce the solution which satisfies y(0)=0 and y'(0)=0. What happens as t increases?

Solution

In the end I stay with:
y=Acos2t+Bsin2t+(1/4)tsin2t

dy/dx=-2Asin2t+2Bcos2t+(1/2)tcos2t+(1/4)sin2t

After applying boundary conditions I get:
0=0

What does it mean and what happens as 't' is increasing? I was asked to plot a graph.
 

Answers and Replies

  • #2
ehild
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Task 7
Show that y=(1/4)tsin2t satisfies equation

d2y/dt2+4y=cos2t

Find the general solution and deduce the solution which satisfies y(0)=0 and y'(0)=0. What happens as t increases?

Solution

In the end I stay with:
y=Acos2t+Bsin2t+(1/4)tsin2t

Yes, it is the general solution of the differential equation. Now you have to choose A and B so that both y and y' is zero at t=0.

dy/dx=-2Asin2t+2Bcos2t+(1/2)tcos2t+(1/4)sin2t

After applying boundary conditions I get:
0=0

What does it mean and what happens as 't' is increasing? I was asked to plot a graph.

0=0 does not mean anything. What are A and B if y(0)=0 and y'(0)=0?

Substitute zero for t in y=Acos2t+Bsin2t+(1/4)tsin2t, what do you get? Do the same for y'.

You have to plot the y(t) graph for t>0.


ehild
 
  • #3
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I get that
A=0
 
  • #5
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If B equals 0 as well then the graph will look like a Christmas tree.

But why would B equal 0?
 
  • #6
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But this is simple!
I had some brain freeze...
Thanks for help!
 
  • #7
ehild
Homework Helper
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Simple, isn't it? y'(0)=0, so -2Asin2t+2Bcos2t+(1/2)tcos2t+(1/4)sin2t=0 and A=0, 0=2Bcos(0) -->B=0.
Yes, the graph looks like a horizontal Christmas-tree:smile:

ehild
 

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