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

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

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