Question

Use seperation of variables to solve the initial value problem:
dy/dx= (y+5)(x+2) and y=1 when x=0

Answer 1

\[\frac{dy}{dx}=\frac{y+5}{x+2}\]\[\frac{dy}{y+5}=\frac{dx}{x+2}\]\[\int\limits_{}^{}\frac{dy}{y+5}=\int\limits_{}^{}\frac{dx}{x+2}\]Does that help?

Answer 2

Its actually multiplication, not division :(

Answer 3

(y+5)*(x+2)

Answer 4

Ahh, my bad.

So it would instead be:\[\int\limits_{}^{}\frac{dy}{y+5}=\int\limits_{}^{}(x+2)dx\]

Answer 5

Yeah I got that, would it then be ln(y+5)= 1/2x^(2) + 2x +C ?

Answer 6

Yes.

Answer 7

Does c = ln6?

Answer 8

Yes.

Answer 9

Okay Awesome! The answer showed an equation. Do you know how I get that?

Answer 10

You want an equation for y in terms of x?

Answer 11

Yep

Answer 12

\[\ln(y+5)=\frac{x^{2}}{2}+2x+\ln(6)\]\[e^{\ln(y+5)}=e^{ \frac{x^{2}}{2}+2x+\ln(6)}\]\[y+5=e^{ \frac{x^{2}}{2}+2x} \times e^{\ln(6)}\]\[y=6 \times e^{ \frac{x^{2}}{2}+2x} -5\]

Answer 13

Will you Marry me?

Answer 14

I'm already engaged to someone else. :P

Answer 15

You Jerk, Huge thanks though!

Answer 16

You're welcome! :P