Discussion Forum

Let  $y'(x)+y(x)g'(x)=g(x)g'(x)y(0)=0, x \epsilon R,$ where f'(x)  denoted $\frac{d f(x)}{dx}$ 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 .....

Answer:

Explanation:

$\frac{dy}{dx}+y.g'(x)=g(x)g'(x)$

 IF= $e^{\int g'(x)dx}= e^{g(x)}$

 $\therefore$ solution is 

              $y(e^{g(x)})=\int g(x).g'(x).e^{g(x)} dx+C$

Put   $g(x)=t, g'(x)dx=dt$

 $y(e^{g(x)})=\int t.e^{t} dt+C$

  =$t.e^{t}-\int 1.e^{t} dt+C=t.e^{t}-e^{t}+C$

 $y e^{g(x)}=(g(x)-1)e^{g(x)}+C$........(i)$

 Given , y(0)=0,g(0)=g(2)=0

$\therefore$  Eq.(i) becomes 

        $y(0).e^{g(0)}=(g(0)-1).e^{g(0)}+C$

 $\Rightarrow$   $0= (-1).1+C \Rightarrow C=1$

 $\therefore$  $y(x).e^{g(x)}=(g(x)-1)e^{g(x)}+1$

$\Rightarrow$ $y(2).e^{g(2)}=(g(2)-1)e^{g(2)}+1$

 where g(2)=0

 $\Rightarrow$ $y(2).1= (-1).1+1 \Rightarrow y(2)=0$