- #1

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dy/dt = t/(y + 1) with initial condition y(2) = 0

here's what i have done so far

(y + 1) dy = t dt

y

^{2}/2 + y = t

^{2}/2 + C

i am confused on how to isolate y.

this is the solution

- Thread starter magnifik
- Start date

- #1

- 360

- 0

dy/dt = t/(y + 1) with initial condition y(2) = 0

here's what i have done so far

(y + 1) dy = t dt

y

i am confused on how to isolate y.

this is the solution

- #2

Char. Limit

Gold Member

- 1,204

- 14

First, multiply both sides by 2. Then complete the square on the LHS.

- #3

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

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Don't forget, use y(2)=0, to evaluate C. (or c_{1}).

- #4

- #5

HallsofIvy

Science Advisor

Homework Helper

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Once you have multiplied both sides by 2, you will still have [itex]y^2+ 2y[/itex] on the left side and just "taking the square root" won't give you y. Complete the square as Char. Limit said in the very first response to this thread.

- #6

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Thanks for the help!

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