Asked by Kyle
                Solve for the function y.
dy/dx = xe^y, when y(1)=0
I got as far as:
e^-y dy = x dx
But then how do you continue from there?
            
        dy/dx = xe^y, when y(1)=0
I got as far as:
e^-y dy = x dx
But then how do you continue from there?
Answers
                    Answered by
            Damon
            
    dy e^-y = x dx
- e^-y = .5 x^2 + c
when x = 1, y = 0
- e^0 = .5 * 1 + c
-1 = .5 + c
c = -1.5
so
e^-y = -.5 x^2 + 1.5
    
- e^-y = .5 x^2 + c
when x = 1, y = 0
- e^0 = .5 * 1 + c
-1 = .5 + c
c = -1.5
so
e^-y = -.5 x^2 + 1.5
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