JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 13)
Let y = y(x), x > 1, be the solution of the differential equation $$(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}}$$, with $$y(2) = {{1 + {e^4}} \over {2{e^4}}}$$. If $$y(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}}$$, then the value of $$\alpha + \beta $$ is equal to _________.
Answer
14
Explanation
$\frac{d y}{d x}+y\left(\frac{2 x}{x-1}\right)=\frac{1}{(x-1)^{2}}$
$$ \text { I.F} =e^{\int \frac{2 x}{x-1} d x} $$
$$ \begin{aligned} & =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\ & =e^{2 x+2 \ln (x-1)} \\\\ & =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2} \end{aligned} $$
$\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$
$\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$
$\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$
$\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$
$$ \downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}} $$
$\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$
$\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$
$$ \text { I.F} =e^{\int \frac{2 x}{x-1} d x} $$
$$ \begin{aligned} & =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\ & =e^{2 x+2 \ln (x-1)} \\\\ & =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2} \end{aligned} $$
$\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$
$\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$
$\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$
$\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$
$$ \downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}} $$
$\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$
$\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$
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