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## Denebel 3 years ago use separation of variables to solve the initial value problem. dy/dx = (y+2)(x+2) and y=1 when x=0

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1. Denebel

*oops I meant (y+5)

2. Denebel

dy/dx= (y+5)(x+2) and y=1 when x=0

3. Roachie

dy/(y+5) = (x+2) dx integrate both sides

4. Denebel

I did, but I don't know what to do after.. I did this? |dw:1327548662810:dw|

5. Denebel

Can you show me your steps?

6. Roachie

ok, take your ln ( y+5) = 1/2x^2+2x + C and raise expotentiate to get rid of the ln y+5 = Ce^(1/2x^2+2x ) then y = Ce^(1/2x^2+2x ) - 5

7. Roachie

y=6e^(.5x^2+2x) + 5

8. Denebel

Why can't I plug in the numbers before putting in e?

9. Roachie

Because this is how I was taught :) lemme try it that way

10. Roachie

that gives ln(6) = C... 1.7917594692280550008124773583807022727229906921830047058553

11. Denebel

Oh ok. thanks :) How did you get your C to become a coefficient?

12. Denebel

Referring to this: y+5 = Ce^(1/2x^2+2x )

13. Denebel

Is this a rule?

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