Differential Equations 2 Question 24
25. 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 ……
(2011)
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Solution:
- $\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) g^{\prime}(x)$
$$ IF=e^{\int g^{\prime}(x) d x}=e^{g(x)} $$
$\therefore$ Solution is $y\left(e^{g(x)}\right)=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+C$
$$ \begin{aligned} & \text { Put } \begin{aligned} g(x) & =t, g^{\prime}(x) d x=d t \\ y\left(e^{g(x)}\right) & =\int t \cdot e^{t} d t+C \\ =t \cdot e^{t}-\int 1 \cdot e^{t} d t+C & =t \cdot e^{t}-e^{t}+C \\ y e^{g(x)} & =(g(x)-1) e^{g(x)}+C \\ y(0) & =0, g(0)=g(2)=0 \end{aligned} \end{aligned} $$
$\therefore$ Eq. (i) becomes,
$$ \begin{array}{rlrl} & & y(0) \cdot e^{g(0)} & =(g(0)-1) \cdot e^{g(0)}+C \\ \Rightarrow & & 0 & =(-1) \cdot 1+C \Rightarrow \quad C=1 \\ & \therefore & y(x) \cdot e^{g(x)} & =(g(x)-1) e^{g(x)}+1 \\ \Rightarrow & y(2) \cdot e^{g(2)} & =(g(2)-1) e^{g(2)}+1, \text { where } g(2)=0 \\ \Rightarrow & & y(2) \cdot 1 & =(-1) \cdot 1+1 \\ & & y(2) & =0 \end{array} $$
