If #y'(x)+y(x)g'(x)=g(x)g'(x)" ; y(0)=0 ; g(0)=g(2)=0 " where " x in RR# then #y(2)=?#

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Answer 1 · 2016-12-27T12:41:42

#y(2)=-2#

This equation is of type

#y'+P(x)y=Q(x)#

Making #u(x)=int_(x_0)^xP(xi)d xi# we have

#d/(dx)(e^(u(x)) y)=e^(u(x))(y'+P(x) y) = e^(u(x)) Q(x)# so

#y = e^(-u(x))int_(x_0)^x e^(u(xi))Q(xi)d xi#

Here #P(x)= g'# so #u(x) = g(x)# and

#y=e^(-g(x))int_(x_0)^x g(xi) d/(d xi)(e^(g(xi)))d xi =e^(-g(x))(( g(x)-1)e^(g(x))+C)# or

#y = g(x)-1+Ce^(-g(x))#

Now, using the initial conditions, #y(0)=g(0)= 0# we have

#0=-1+C# so #C = 1# and

#y = g(x)-1+e^(-g(x))# and also

#y(2)=g(2)-1+e^(-g(2)) = 0#