Question

find the solution. dy/dt = (ty)^2

Answer 1

SO you want to integrate?

Answer 2

well separate variables and integrate solve for y

Answer 3

Do you have the answers on an answer sheet or something?

Answer 4

I just need check if I got it right.

Answer 5

need to*

Answer 6

yea y(t) = -3/(t^3 + k)

Answer 7

So where is the problem, you need the steps?

Answer 8

Don't misunderstand my question please, but I think you mentioned everything that's necessary already above, this is an ODE, it is separable.

Answer 9

i just need the steps. i dont know how you get that answer

Answer 10

\[dy/dt = (ty)^2\]\[y^{-2}*dy = t^2dt\]\[\int y^{-2}dy = \int t^2 dt\]\[-y^{-1} + c_1 = \frac{1}{3}t^3 + c_2\]\[y^{-1} = -\frac{1}{3}t^3 + c\]Take reciprocal of both sides
\[y = - \frac{1}{\frac{1}{3}t^3 + c}\]\[y = -\frac{3}{t^3+c}\]

Answer 11

I don't get your last line @whpalmer4 How can you get a three on the numerator with a 1/3t^3 +c in the denominator?

Answer 12

(1/3)

Answer 13

Ah, I just factor out the 1/3 and turn it into 3 in the numerator. The values of c obviously aren't identical, but one constant is as good as another, I always say :-)

Answer 14

Yeah. so you can just make 3c into c, correct? Cause multiplying the constant woulds till give you a constant but the constant would be much larger?

Answer 15

Right. What I'm really doing is

\[y = -\frac{1}{\frac{1}{3}t^3 + c} = -\frac{1}{\frac{1}{3}(t^3 + 3c)} = -\frac{3}{t^3+3c} = -\frac{3}{t^3+k}\]

Answer 16

\[k = 3c\]

Answer 17

thank you whpalmer4. that was very helpful

Answer 18

Glad I could still remember how to do it! It's been a while :-)

Answer 19

As a little exercise, take the derivative of the result and plug it back into the original to make sure it works (it does, but it's good practice).