Initial Value Problem (or IVP)

Question

Find the implicit IVP solution

y'(x) = - (e^x ⁄ e^y+ye^y)

y(0) = 0

What can we say about the existence of the explicit solution? If there is, can we say who it is?

Answer 1

SEparate variables: (e^y + ye^y)dy = - e^xdx;

We integrate now both sides: ∫(e^y + ye^y)dy = - ∫e^xdx; e^y + ye^y - e^y = - e^x + C; ye^y = - e^x + C.

y(0) = 0; 0·e⁰ = - 1 + C; C = 0.

So, ye^y = - e^x + 1 is implicit solution of this IVP

Answer 2

y' = -e^x / ( e^y + ye^y)

First cross mulriplication

y' ( e^y + ye^y) = -e^x



y' ( e^y + ye^y) + e^x = 0

y'e^y( 1 + y) + e^x = 0

 you know that integral of e^x is e^x

 also you know the rule of product (u.v)' = u'v + u.v'

so ( y.e^y)' = y'.e^y + y (y'e^y) =after factor of  y'.e^y you will have:

             y'.e^y ( 1 + y)

as a result anti derivatve of y'e^y( 1 + y) + e^x = 0 equal: y.e^y + e^x + c = 0 by using y(0) =0----> c= -1

for prove:

fine derivative of y.e^y + e^x - 1 = 0  

                  y' e^y + y.y'e^y + e^x = 0

                  factor of y'

                  y'( e^y + ye^y ) = - e^x

both side divided by e^y + ye^y you will have

y' = -e^x / ( e^y + ye^y)

I hope it is useful,

MInoo