# 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)=?#

This equation is of type

#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

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To solve the given differential equation, we can use the method of integrating factors. Given ( y'(x) + y(x)g'(x) = g(x)g''(x) ), we'll first rewrite the equation as ( y'(x) + y(x)g'(x) - g(x)g''(x) = 0 ).

Now, we'll find the integrating factor, which is given by ( e^{\int P(x)dx} ), where ( P(x) ) is the coefficient of ( y(x) ), which is ( g'(x) ) in this case.

So, the integrating factor, ( I(x) = e^{\int g'(x)dx} ).

Integrating ( g'(x) ), we get ( g(x) ).

Therefore, the integrating factor becomes ( I(x) = e^{g(x)} ).

Now, we'll multiply both sides of the differential equation by the integrating factor:

( e^{g(x)}[y'(x) + y(x)g'(x) - g(x)g''(x)] = 0 ).

This simplifies to:

( e^{g(x)}[y'(x)e^{g(x)} + y(x)g'(x)e^{g(x)} - g(x)g''(x)e^{g(x)}] = 0 ).

Now, we recognize that the left-hand side can be written as the derivative of a product:

( \frac{d}{dx}[y(x)e^{g(x)}] = 0 ).

Integrating both sides with respect to ( x ), we get:

( \int \frac{d}{dx}[y(x)e^{g(x)}] dx = \int 0 dx ).

This simplifies to:

( y(x)e^{g(x)} = C ), where ( C ) is the constant of integration.

Given ( y(0) = 0 ), we have:

( y(0)e^{g(0)} = C ).

Since ( g(0) = 0 ), ( e^{g(0)} = 1 ), so ( C = 0 ).

Therefore, ( y(x)e^{g(x)} = 0 ).

Given ( g(2) = 0 ), we have:

( y(2)e^{g(2)} = 0 ).

Since ( g(2) = 0 ), ( e^{g(2)} = 1 ).

So, ( y(2) = 0 ).

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