Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x) y(0)=0, x \in R$, where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ 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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Aicharya · Accepted Answer

$\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) g^{\prime}(x)$ $ \mathrm{IF}=e^{\int g^{\prime}(x) d x}=e^{g(x)} $ $	herefore$ Solution is $ y\left(e^{g(x)}ight)=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+C $ Put $g(x)=t, g^{\prime}(x) d x=d t$ $ \begin{aligned} & y\left(e^{g(x)}ight)=\int t \cdot e^t d t+C \ = & t \cdot e^t-\int 1 \cdot e^t d t+C=t \cdot e^t-e^t+C \ & y e^{g(x)}=(g(x)-1) e^{g(x)}+C \quad \ldots(\mathrm{i}) \end{aligned} $ Given, $y(0)=0, g(0)=g(2)=0$ $	herefore$ Eq. (i) becomes, $ \begin{aligned} & \quad y(0) \cdot e^{g(0)}=(g(0)-1) \cdot e^{g(0)}+C \ & \Rightarrow \quad 0=(-1) \cdot 1+C \Rightarrow C=1 \ & 	herefore \quad y(x) \cdot e^{g(x)}=(g(x)-1) e^{g(x)}+1 \ & \Rightarrow \quad y(2) \cdot e^{g(2)}=(g(2)-1) e^{g(2)}+1 \ & 	ext { where, } \quad g(2)=0 \ & \Rightarrow \quad y(2) \cdot 1=(-1) \cdot 1+1 \Rightarrow y(2)=0 \end{aligned} $

Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x) y(0)=0, x \in R$, where $f^{\prime}(x)$ denotes…

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