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Let $\mathrm{y=y(x), x>1}$ be the solution of the diffrential equation $\mathrm{(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}, \text { with } y(2)=\frac{1+e^{4}}{2 e^{4}}}$ . If $\mathrm{y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}}$ then the value of $\mathrm{\alpha +\beta }$ is equal to
Option: 1
14

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{\frac{d y}{d x}+\left(\frac{2 x}{x-1}\right) y=\frac{1}{(x-1)^{2}}} \\$

$\mathrm{I \cdot F \cdot=e^{\int \frac{2 x}{x-1} d x}}$

$\mathrm{\text { Now } \int \frac{2 x}{x-1} d x }\\$

$\mathrm{= 2 \left[\int \frac{x-1}{x-1} d x+\int \frac{1}{x-1} d x\right]} \\$

$\mathrm{= 2[x+\log (x-1)]}$

$\mathrm{\therefore I. F =e^{2 x+\log (x-1)^{2}}} \\$

$\mathrm{=(x-1)^{2} \cdot e^{2 x}} \\$

$\mathrm{\therefore y \cdot(x-1)^{2} e^{2 x}=\int \frac{1}{(x-1)^{2}} \cdot(x-1)^{2} e^{2 x} d x}$

$\mathrm{\Rightarrow y(x-1)^{2} e^{2 x}=\frac{e^{2 x}}{2}+c} \\$

$\mathrm{\Rightarrow y=\frac{1}{2(x-1)^{2}}+\frac{c}{e^{2 x} \cdot(x-1)^{2}}}$

$\mathrm{As\: y(2)=\frac{1+e^{4}}{2 e^{4}}}$

$\mathrm{\Rightarrow \frac{1+e^{4}}{2 e^{4}}=\frac{1}{2}+\frac{c}{e^{4}}} \\$

$\mathrm{\Rightarrow \frac{1+e^{4}}{2 e^{4}}=\frac{e^{4}+2 c}{2 e^{4}}} \\$

$\mathrm{\Rightarrow c=\frac{1}{2}}$

Putting $\mathrm{x=3}$

$\mathrm{y(3) =\frac{1}{2 \cdot 4}+\frac{1}{2 e^{6} \cdot(4)}} \\$

$\mathrm{=\frac{e^{6}+1}{8 e^{6}}} \\$

$\mathrm{\therefore \quad \alpha=6, \,\, \beta=8 \Rightarrow \alpha+\beta=14 }$

Hence answer is $14$

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