partial derivatives

[cremer39]

Does anyone know how to solve the following problem?

If dy/dx=e^-2y and y=0 when x=5, find x when y=3

 

[Daniel_Feldman]

dy/dx=e^(-2y)

e^(2y)dy=dx

Integrating both sides

(1/2)e^(2y)=x+c

y=0 when x=5 so

(1/2)=5+c

c=-9/2

(1/2)e^(2y)=x-9/2

y=3

x=(1/2)e^6+9/2


That's my thinking...hopefully someone can confirm or change it....

 

[tkhunny]

dy/dx=e^(-2y)
e^(2y) dy = dx
Introduce the integration
½e^(2y) = x + C
e^(2y) = 2x + D
2y = log(2x+D)
y = ½log(2x+D)

We have 0 = ½log(2(5)+D) ==> 2(5)+D = 1 ==> D = -9

y = ½log(2x-9)

Find 'x': 3 = ½log(2x-9)

 

[Gene]

Looks good except log should be ln