Verified Solution • 2025–26 Session moderate

QJEEM22Ma4576: JEE Main 2022 Mathematics - Class 12 JEEM22Ma4576 Solution: Let y = y(x), x > 1, be the solution of the differential…

Let y = y(x), x > 1, be the solution of the differential equation (x−1)dydx+2xy=1x−1(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}} , with y(2)=1+e42e4y(2) = {{1 + {e^4}} \over {2{e^4}}} . If y(3)=eα+1βeαy(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}} , then the value of α+β\alpha + \beta is equal to _________.

Step-by-Step Verified Solution

dydx+y(2xx−1)=1(x−1)2\frac{d y}{d x}+y\left(\frac{2 x}{x-1}\right)=\frac{1}{(x-1)^{2}}

 I.F=e∫2xx−1dx\text { I.F} =e^{\int \frac{2 x}{x-1} d x}

amp;=e2∫(x−1x−1+1x−1)dxamp;=e2x+2ln⁡(x−1)amp;=e2x(x−1)2\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}

⇒∫d(y⋅e2x(x−1)2)=∫e2xdx\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x

⇒y⋅e2x(x−1)2=e2x2+c\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c

↓y(2)=1+e42e4\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}

1+e42e4⋅e4=e42+c\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c

⇒c=e42(1+e4−e4e4)=12\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}

⇒y⋅e2x(x−1)2=e2x+12\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}

↓y(3)=eα+1βeα\downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}

⇒eα+1βeα⋅e6⋅4=e6+12\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}

⇒α=6\Rightarrow \alpha=6 and β=8⇒α+β=14\beta=8 \Rightarrow \alpha+\beta=14

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