# Help me solve this differential equation

1. Apr 21, 2015

### grandpa2390

1. The problem statement, all variables and given/known data
find the inverse laplace transform.

1/(s^3 + 7s)

2. Relevant equations
sin(kt) = k/(s^2 + k^2)
cos(kt) = s/(s^2 + k^2)
3. The attempt at a solution

so this one is different from all the others I have done because it involves an imaginary number and I am not sure what the rule is that changes the problem.
somehow the answer becomes 1/7 - cos(sqrt(7)*t ) / 7

but I am getting positive sine instead of negative cosine and 7i in the denominator instead of 7. does the i reverse sine to cosine or something?

2. Apr 21, 2015

### Simon Bridge

... try taking the Laplace transform of the answer and see if it matches the question.

3. Apr 21, 2015

### grandpa2390

Well... considering it is the right answer... it should match the question. What is the method for getting that answer.

4. Apr 21, 2015

### Simon Bridge

It can be instructive to check that the "correct answer" matches the question ... did you try? Did you try the transform on your answer?

Going forward:
Did you try completing the square in s^2 in the denominator.

5. Apr 21, 2015

### grandpa2390

I don't know what that means.

I set up my partial fraction. A(s^2+7) + B(s) = 1
s=0 and sqrt(-7)
A= 1/7 and B= 1/sqrt(-7)

1/(7s) + 1/(sqrt(-7)*(s^2+7))

1/7 * 1/s = 1/7
1/sqrt(-7)sqrt(7) * sqrt(7)/(s^2 + 7) = 1/(7i) * sin(sqrt(7)t)

1/7 + sin(sqrt(7)t)/7i
but that is wrong I should get 1/7 - cos(sqrt(7)t)/7

Last edited: Apr 21, 2015
6. Apr 21, 2015

### Simon Bridge

Try getting the denominator in form: $(s^2 + a^2)^2$

7. Apr 21, 2015

### grandpa2390

there I posted my work.

8. Apr 21, 2015

### grandpa2390

so you are saying it is ok to multiply the initial problem by s/s and then I won't have to worry about i.

edit: no that didn't work

I do not have a clue.

Last edited: Apr 21, 2015
9. Apr 21, 2015

### Simon Bridge

... In the partial fractions decomposition: don't you need a Bs+C in the numerator of the quadratic denominator?
I found it helped to put a^2=7 and just write everything out in terms of a.

10. Apr 21, 2015

### grandpa2390

that might be it then. I should have done Bs+C. and that would put the s in the numerator required for the cosine function. but will it get rid of the i?

11. Apr 21, 2015

### Simon Bridge

If I read you correctly: the "i" comes from the imaginary root of the quadratic - nothing to do with the decomposition.
In general, the polynomial in the numerator should be, at most, one order less than the polynomial in the denominator.

12. Apr 21, 2015

### Ray Vickson

Use either (1) If $f(t) \leftrightarrow g(s)$ then $\int_0^t f(\tau) \, d\tau \leftrightarrow \frac{g(s)}{s}$; or (2) a partial-fraction expansion of $\frac{1}{s^3 + 7s}$.

13. Apr 23, 2015

### Simon Bridge

I had a think further - you tried a "partial fractions" of form:
$$\frac{A}{s}+\frac{B}{s^2+7}$$... and discovered the relations for A and B were: $$As^2 + 7A +Bs = 1\\ \implies As^2=0,\; 7A=1,\; Bs=0$$ ... the first tells you that A=0 while the second tells you that A=1/7 ... which is an inconsistency, which should have been a hint that you'd done something wrong.
By working out the roots of s^2+7 you were trying to find the values of s that would make the relations true(?) ... which is an OK strategy except that the decomposition is supposed to work for all values of s not just a few of them, so the strategy failed. How you incorporated the roots into the partial fractions is still beyond me - but I suspect it is enough to notice that you had not understood what the partial fractions method is supposed to do.

Last edited: Apr 24, 2015