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Let y^(prime)(x)+y(x)g^(prime)(x)=g(x)g^...

Let `y^(prime)(x)+y(x)g^(prime)(x)=g(x)g^(prime)(x),y(0),x in R ,` where `f^(prime)(x)` denotes `(dy(x))/(dx),` and `g(x)` is a given non-constant differentiable function on `R` with `g(0)=g(2)=0.` Then the value of `y(2)` is______

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`dy/dx +y cdot g'(x) =g(x)g'(x)`
`IF=e^(intg'(x)dx)=e^g(x)`
`therefore` Solution is `y(e^(g(x)))=int g(x) cdot g'(x)cdote^(g(x))dx+C`
Put `g(x)=t,g'(x)dx=dt`
`y(e^(g(x)))=int t cdot e^(t) dt +C`
`=t cdot e^(t)-int1 cdot e^(t) dt+C =t cdot e^(t) - e^(t) +C`
`ye^(g(x))=(g(x)-1) e^(g(x))+C … (i)`
Given, `y(0)=0,g(0)=g(2)=0`
`therefore` Eq. (i) becomes,
`y(0) cdot e^(g(0))= (g(0)-1) cdot e^(g(x))+C`
`rArr 0=(-1)cdot1+C rArr C=1`
`therefore y(x) cdot e^(g(x))= (g(x)-1) e^(g(x))+1`
`rArr y(2) cdot e^(g(2))= (g(2)-1)e^(g(2))+1,` where g(2)=0
`rArr y(2) cdot 1 =(-1) cdot1+1`
`y(2)=0`
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