Question

Solve differential equation- 2 parts.
Calculus II

Answer 1

Hmm ok so this is separable.
Separating the t's and y's gives us,\[\large y^{-4}\;dy=t^{5/2}\;dt\]

What part are you having trouble with? :O

Answer 2

well when I solve for y = all I get is 0.

Answer 3

Hmmm that's strange +_+
Lemme show ya a few steps, maybe you can see where you went wrong.

Answer 4

Integrating both sides, we'll apply the `Power Rule for Integration`,\[\large \int\limits y^{-4}\;dy=\int\limits t^{5/2}\;dt\]Giving us,\[\large -\frac{1}{3}y^{-3}=\frac{2}{7}t^{7/2}+C\]

Answer 5

Any part of that super confusing?

Answer 6

no not yet lol

Answer 7

Multiplying both sides by -3 gives us,\[\large y^{-3}=C-\frac{6}{7}t^{7/2}\]And thennnn we want to take the reciprocal of both sides to deal with the negative exponent,\[\large y^3=\frac{1}{C-\dfrac{6}{7}t^{7/2}}\]

Answer 8

Then take the cube root of both sides, and tada! We have our general solution.

I'm not sure how you're coming up with zero +_+ Hmm

Answer 9

for y=?

Answer 10

for a constant solution

Answer 11

Constant solutions...? :\

Answer 12

hmm...

Answer 13

Well we figured out part B at least.
Hmm

Answer 14

Constant solutions mean \(y=C\) i.e. \(\frac{dy}{dt}=0\). Observe this reduces our equation to:$$0=t^{5/2}y^4$$since \(t\) is allowed to vary we conclude \(y^4=0\) and therefore \(y=0\) is our only constant solution.