What is the Solution for Differential Equation Task 7?

[mathi85]

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.

 

[ehild]

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

 

[mathi85]

I get that
A=0

 

[mathi85]

How can I find B?

 

[mathi85]

If B equals 0 as well then the graph will look like a Christmas tree.

But why would B equal 0?

 

[mathi85]

But this is simple!
I had some brain freeze...
Thanks for help!

 

[ehild]

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

ehild