- #1

- 2

- 0

The problem: y'' - y = f(t), y(0) = 1, y'(0) = 0

f(t) = 1, 0 (less than equal to) t (less than) pi/2

f(t) = sin(t), t (greater than equal to) pi/2

Any help would be appreciated putting me in the right direction.

Thank you in advance.

--Julie

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- Thread starter bathtub2007
- Start date

- #1

- 2

- 0

The problem: y'' - y = f(t), y(0) = 1, y'(0) = 0

f(t) = 1, 0 (less than equal to) t (less than) pi/2

f(t) = sin(t), t (greater than equal to) pi/2

Any help would be appreciated putting me in the right direction.

Thank you in advance.

--Julie

- #2

- 9

- 0

I think the best way forward would be:

- #3

- 107

- 1

- #4

djeitnstine

Gold Member

- 614

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[tex]f(t)=1+H(t-\frac{\pi}{2})+\left(sin(t)-1\right)H(t-\frac{\pi}{2})[/tex]

Edit since latex isn't working....

f(t)=1+H(t-[pi/2])+[sin(t)-1]*H(t-[pi/2])

where H(t-[pi/2]) denotes the unit step function.

- #5

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Please advise to see whether his approach is correct or not.

Thanks

- #6

djeitnstine

Gold Member

- 614

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maths.tuition welcome to PF.

If you are speaking of doing the same thing that xfunctionx did for bathtub's differential equation that would be wrong

f(t) = 1, 0 (less than equal to) t (less than) pi/2

f(t) = sin(t), t (greater than equal to) pi/2

does not simply equal 1 for all t, please note I already stated the proper form of the function(in terms of the unit step function) to get the the laplace transform for f(t)

"**f(t)=1+H(t-[pi/2])+[sin(t)-1]*H(t-[pi/2])**"

If you are speaking of doing the same thing that xfunctionx did for bathtub's differential equation that would be wrong

f(t) = 1, 0 (less than equal to) t (less than) pi/2

f(t) = sin(t), t (greater than equal to) pi/2

does not simply equal 1 for all t, please note I already stated the proper form of the function(in terms of the unit step function) to get the the laplace transform for f(t)

"

Last edited:

- #7

- 279

- 1

maths.tuition welcome to PF.

If you are speaking of doing the same thing that xfunctionx did for bathtub's differential equation that would be wrong

f(t) = 1, 0 (less than equal to) t (less than) pi/2

f(t) = sin(t), t (greater than equal to) pi/2

does not simply equal 1 for all t, please note I already stated the proper form of the function(in terms of the unit step function) to get the the laplace transform for f(t)

"f(t)=1+H(t-[pi/2])+[sin(t)-1]*H(t-[pi/2])"

I think this last expression is wrong. Musn't this be:

[tex]L\left{H(t)-H\left(t-\frac{\pi}{2}\right)+sin(t)H\left(t-\frac{\pi}{2}\right)\right}= L\left{H(t)-H\left(t-\frac{\pi}{2}\right)+cos\left(t-\frac{\pi}{2}\right)H\left(t-\frac{\pi}{2}\right)\right}[/tex]

There seems to be something definitely wrong with the latex generation, it keeps posting things from older replies of other posts of me :grumpy:

coomast

- #8

djeitnstine

Gold Member

- 614

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one had to subtract 2 from sin and not 1 or else you would be left with 2-1+sin{t} = 1+sin(t)

thanks for catching that mistake

- #9

- 279

- 1

f(t)=1+H(t-[pi/2])+[sin(t)-2]*H(t-[pi/2])

one had to subtract 2 from sin and not 1 or else you would be left with 2-1+sin{t} = 1+sin(t)

thanks for catching that mistake

OK, next attempt for a good latex formula. I think it is still not OK, I got:

[tex]L\left[H(t)-H\left(t-\frac{\pi}{2}\right)+sin(t)H\left(t-\frac{\pi}{2}\right)\right]= L\left[H(t)-H\left(t-\frac{\pi}{2}\right)+cos\left(t-\frac{\pi}{2}\right)H\left(t-\frac{\pi}{2}\right)\right][/tex]

coomast

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