JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 7)
If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
$$\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
$$2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)$$
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)$$
Explanation
$\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$
Let $x-1=X, y-1=Y$
$$ \frac{d Y}{d X}=\frac{X+Y}{X-Y} $$
Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$
$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$
$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$
$\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$
$$ \begin{aligned} &\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\ &\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c \end{aligned} $$
Curve passes through $(2,1)$
$0-0=0+c \Rightarrow c=0$
If $(k+1,2)$ also satisfies the curve
$$ \begin{aligned} &\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\ &2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right) \end{aligned} $$
Let $x-1=X, y-1=Y$
$$ \frac{d Y}{d X}=\frac{X+Y}{X-Y} $$
Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$
$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$
$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$
$\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$
$$ \begin{aligned} &\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\ &\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c \end{aligned} $$
Curve passes through $(2,1)$
$0-0=0+c \Rightarrow c=0$
If $(k+1,2)$ also satisfies the curve
$$ \begin{aligned} &\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\ &2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right) \end{aligned} $$
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