Question

solve y' = xe^(y-x^(2)), when x=0, y=0?

Answer 1

dy/dx right?

Answer 2

\[\frac{ dy }{ dx }= xe^{y-x^2}\]

Answer 3

\[xe^{y-x^2} = xe^y \times e^{-x^2}\]

Answer 4

Might be able to separate the variables now and solve?

Answer 5

\[\frac{ dy }{ dx } = xe^y \times e^{-x^2}\]
If we divide both sides by e^(y) and multiply both sides by (dx)
\[\frac{ dy }{ e^y } = xe^{-x^2} dx\]

Answer 6

Integrate both sides

Answer 7

why integrate?

Answer 8

To eliminate the dy and dx \[\int\limits_{}^{}\frac{ dy }{ e^y } = \int\limits_{}^{}xe^{-x^2} dx\] and so he can plug in the initial conditions to solve for C

Answer 9

ohh i thought that will be
dy=x(e^y/ e^x^2)

Answer 10

It is the same thing since there is a negative sign in the exponent of the e^-(x^2)

Answer 11

C=-1/2 ?

Answer 12

Have not attempted the problem but if you show your work I can verify

Answer 13

\[\int\limits\limits \frac{ dy }{ e^y } - \int\limits\limits \frac{ xdx }{e ^{x ^{2}} } =0\]
\[\int\limits \frac{ du }{ e^u } - \frac{ 1 }{ 2 } \int\limits \frac{ du }{ e ^{u} } = \int\limits 0\]
\[\int\limits e ^{-u} du - \frac{ 1 }{ 2 } \int\limits e ^{-u} du = C\]
\[\frac{ -1 }{ e ^{y} } + \frac{ 1}{ 2ex ^{2}} = C\]
\[\frac{ 1 }{ 2 } e ^{-x ^{2}} = C + e ^{-y}\]
\[\frac{ 1 }{ 2 } e ^{0} = C + e ^{0}\]
\[1/2 = C+1 \]
C=-1/2
???

Answer 14

@Johnbc

Answer 15

Great Job, that is correct. Using you're value of C plug it back into your final equation before you valued X and Y so that you have the full answer of y'(x).