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(B) Given differential equation: $\frac{dy}{dx} = \frac{x+y-2}{x-y}$.Let $x = X+h$ and $y = Y+k$. Substituting these,we get $\frac{dY}{dX} = \frac{X+Y

Vedclass · Accepted Answer

$11$

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(B) Given differential equation: $\frac{dy}{dx} = \frac{x+y-2}{x-y}$.Let $x = X+h$ and $y = Y+k$. Substituting these,we get $\frac{dY}{dX} = \frac{X+Y+(h+k-2)}{X-Y+(h-k)}$.For the equation to be homogeneous,we set $h+k-2=0$ and $h-k=0$. Solving these gives $h=1$ and $k=1$.Thus,$\frac{dY}{dX} = \frac{X+Y}{X-Y}$. Dividing numerator and denominator by $X$,we get $\frac{dY}{dX} = \frac{1+(Y/X)}{1-(Y/X)}$.Let $Y/X = v$,then $Y = vX$ and $\frac{dY}{dX} = v + X\frac{dv}{dX}$.$v + X\frac{dv}{dX} = \frac{1+v}{1-v} \Rightarrow X\frac{dv}{dX} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}$.Separating variables: $\frac{1-v}{1+v^2} dv = \frac{dX}{X}$.Integrating both sides: $\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dX}{X}$.$	an^{-1}(v) - \frac{1}{2} \ln(1+v^2) = \ln|X| + C$.Substituting $v = \frac{y-1}{x-1}$ and $X = x-1$: $	an^{-1}\left(\frac{y-1}{x-1}ight) - \frac{1}{2} \ln\left(1 + \left(\frac{y-1}{x-1}ight)^2ight) = \ln|x-1| + C$.Since the curve passes through $(2,1)$,$	an^{-1}(0) - \frac{1}{2} \ln(1+0) = \ln|1| + C \Rightarrow C = 0$.Comparing with the given form,$\alpha = 1$ and $\beta = 2$.Therefore,$5\beta + \alpha = 5(2) + 1 = 11$.

If the solution curve of the differential equation $\frac{dy}{dx} = \frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is $\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{\beta} \log_e\left(\alpha + \left(\frac{y-1}{x-1}\right)^2\right) = \log_e|x-1|$,then $5\beta + \alpha$ is equal to

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