# General Solution to the Diff. Eq. dy/dt = ty

1. Jan 11, 2009

### the7joker7

1. The problem statement, all variables and given/known data

Find the general solution of the differential equation specified.

$$\frac{dy}{dt}$$ = ty.

3. The attempt at a solution

I already know the answer to be ke$$^{t^{2}/2}$$, but can't figure out how it got here. I'm rusty with my integrals and am just really starting diff eqs, but I think I'm just supposed to take the integral of ty (which is (t^2y)/2 i believe) and add a + c to the end, c being some constant, so I'm confused...

2. Jan 11, 2009

### jgens

I think this might be a logical first step: dy/y = t dt. Now integrate that.

3. Jan 11, 2009

### the7joker7

dy/y = t dt

The integral of that would be...

dy = t^2/2

So

(t^2/2)/dt = ty

t^2/2 = ty dt

Where do I go from here?

4. Jan 11, 2009

### jgens

No. You might want to brush up on your integrals. The integral of dy/y is not dy but rather log(y). Therefore, your equation should be?

5. Jan 11, 2009

### the7joker7

log(y) = t^2/2

So y = e^(t^2/2)

Alright. And the k is because it's a general equation and k needs to be there so it can be solved for any initial condition. But why is the k being multiplied and not added?

6. Jan 11, 2009

### jgens

Well, the equation should read: log(y) = t^2/2 + C; therefore, y = e^(t^2/2 + C) = e^(t^2/2)e^C. We let e^C = k and the equation reads: y = ke^(t^2/2)

7. Jan 12, 2009

### the7joker7

Just to be sure, is the basic starting step for all problems of this type getting dy/y?

Like if I have the same question with dy/dt = t/(t^2y + y), I would try to get dy/y on one side?

8. Jan 12, 2009

### Staff: Mentor

The basic goal is to get y and dy on one side, and t and dt on the other, which is called separation. Your example differential equation is not separable, if I understand what you wrote. Is the first term in the denominator t^(2y) or is it t^2 * y? If it's the first, your example isn't separable.

9. Jan 12, 2009

### the7joker7

Oops, my bad. It's the 2nd one.

dy/dt = t/((t^2)y + y)

10. Jan 12, 2009

### the7joker7

Lesse

dy/dt = t/((t^2)y + y)

dy/y = t/((t^2)y^2 + y^2) dt

dy/y * y^2 + y^2 = t/(t^2) dt

dy/y * 2(y^2) = 1/t

ln(y) * 2(y^3/3) = 1/t

Ugh...I dunno, something in that ballpark? I'm confused...

11. Jan 12, 2009

### jgens

I don't believe that's quite right. Your initial equation is: dy/dt = t/(y(t^2 + 1) correct? Then we claim that y dy = (t dt)/(t^2 + 1) which we then integrate.

12. Jan 12, 2009

### NoMoreExams

These are called SEPARABLE DEs. Separable because we can separate them into f(y) dy = g(t) dt. And then integrate both sides. This is also why DE requires you to do well in Calculus classes so you don't have to think about how to integrate :) This is also why separable DEs are introduced in Calc 2(I guess), to give you a flavor of what to come.

13. Jan 12, 2009

### the7joker7

y dy = (t dt)/(t^2 + 1)

y^2/2 = (t dt/t^2) + t dt (If I understand correctly, this is separating?)

y^2/2 = ((t^2)/2)/t^2 + t^2/2

y^2/2 = 1/2 + t^2/2

y^2 = 1/4 + t^2/4

y = sqrt(k(.25 + t^2/4))

Eh?

14. Jan 12, 2009

### NoMoreExams

No the separating was the process of writing dy/dt = g(t)/f(y) as f(y) dy = g(t) dt.

a/(b+c) is definitely not a/b + a/c, you should remember that from algebra.

You should know how to integrate t dt/(t^2 + 1) from calculus.

15. Jan 12, 2009

### jgens

In the case that you're uncertain how to integrate t dt/(t^2 + 1) use the substitution u = t^2 + 1. However, you really should learn how to do these steps and properly apply elementary algebra - especially is you're working with DEs.