Question

Calculus Help
Fan + Medal

Answer 1

dy/dx = xy/3 , y>0
Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4

Answer 2

\[\frac{ dy }{ dx } = \frac{ xy }{ 3 } , y>0\]

Answer 3

I think we can re-arrange and integrate both sides.

\[\frac{ dy }{ dx } = \frac{ xy }{ 3 }\]

\[\frac{ 3~dy }{ y } = x*dx \]

Answer 4

I feel the next step would probably be to just integrate both sides.

\[3*\int\limits \frac{ dy }{ y } = \int\limits x*dx \]

Answer 5

How would you integrate dy/y ?

Answer 6

ln |y| ?

Answer 7

\[\int\limits \frac{ 1 }{ x } dx = \ln|x|+C\]

Answer 8

you familiar with that identity right?

Answer 9

Yes. I figured it out. It would be \[\ln |y| = \frac{ x^2 }{ 2 } + C\]

Answer 10

don't forget the 3 outside the integral

Answer 11

\[3\ln(y) = \frac{ x^{2} }{ 2 } + C \]

Answer 12

\[\ln|y|= \frac{ x^{2} }{ 6 } + \frac{ C }{ 3 }\]

Answer 13

I'm wondering though if we can just get rid of the ln and just take both sides to the e^

Answer 14

I dont see why not.

Answer 15

It might get kinda messy, tho right?

Answer 16

\[e^{\ln|y|} = e^{\frac{ x^{2} }{ 6 }} + e^{\frac{ C }{ 3 }}\]

Answer 17

Yup we could.

Answer 18

dunno lol I just don't like ln that's why

Answer 19

I guess this would make our lives a bit easier.

\[y = e^{\frac{ x^{2} }{ 6 }} + e^{\frac{ C }{ 3 }}\]

now we plug in f(0) and try to solve

Answer 20

\[y = e^{\frac{ 0 }{ 6 }} + e^{\frac{ c }{ 3 }}\]

\[4 = 1+e^{\frac{ c }{ 3 }}\]

Answer 21

I was jjust typing that out. awesome

Answer 22

:)

Answer 23

\[3 = e^{\frac{ c }{ 3 }}\]

so what now?

Answer 24

solve for C.

Answer 25

C= 3ln(3) or 3.296

Answer 26

@Photon336

Answer 27

yep let's put it together.

Answer 28

so \[y = e^{\frac{ x^{2} }{ 6 }} + e^{\frac{ 3\ln(3) }{ 3 }}\]

Answer 29

\[(e^{0} = 1)+e^{\frac{ \cancel\3\ln(3) }{ \cancel\3 }} ~ e^{\ln(3)} = 3 + 1 = 4 \]

Answer 30

Yay. We got it. Thank you!!

Answer 31

Most importantly always check the result when you can.

Answer 32

the idea was just using algebra and rearranging in the first part so that we could integrate both sides and build our function.