If the solution curve of the differential equation <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>−</mo><mn>2</mn></mrow><mrow><mi>x</mi><mo>−</mo><mi>y</mi></mrow></mfrac></mrow></math> passes through the points (2, 1) and (k + 1, 2), k > 0, then

dydx=x+y−2x−yLet x – 1 = X, y – 1 = Ythen DE: dYdX=X+YX−Y=1+YX1−YXPut y = vxthen dYdX=V+XdVdXV + XdVdX=1+V1−VXdVdX=1+V1−V−V=1+V21−V1−V1+V2dV=dXXV−1V2+1dV+dXX=012ln|V2+1|−tan−1V+ln|X|=clnV2+1⋅X−tan−1V=cln(1+(Y−1X−)2⋅|X−1|)tan−1Y−1X−1=c(2,1)⇒ln(1+0⋅1)−0=cc = 0ln(X−1)2+(Y−1)2=tan−1Y−1X−1point (k + 1, 2) lnk2+1=tan−11k⇒12ln(k2+1)=tan−11k

If the solution curve of the differential equation&nbsp;<math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>−</mo><mn>2</mn></mrow><mrow><mi>x</mi><mo>−</mo><mi>y</mi></mrow></mfrac></mrow></math>&nbsp;passes through the points (2, 1) and (k + 1, 2), k &gt; 0, then

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First Order Differential Equation Differential Equations Maths Class 12th

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 (k + 1, 2), k > 0, then

Option 1 -

$2 t a n^{- 1} (\frac{1}{k}) = l o g_{e} (k^{2} + 1)$

Option 2 -

$t a n^{- 1} (\frac{1}{k}) l o g_{e} (k^{2} + 1)$

Option 3 -

$2 t a n^{- 1} (\frac{1}{k + 1}) l o g_{e} (k^{2} + 2 k + 2)$

Option 4 -

$2 t a n^{- 1} (\frac{1}{k}) l o g_{e} (\frac{k^{2} + 1}{k^{2}})$

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$\frac{d y}{d x} = \frac{x + y - 2}{x - y}$

Let x – 1 = X, y – 1 = Y

then DE: $\frac{d Y}{d X} = \frac{X + Y}{X - Y} = \frac{1 + \frac{Y}{X}}{1 - \frac{Y}{X}}$

Put y = vx

then $\frac{d Y}{d X} = V + X \frac{d V}{d X}$

V + $X \frac{d V}{d X} = \frac{1 + V}{1 - V}$

$X \frac{d V}{d X} = \frac{1 + V}{1 - V} - V$

$= \frac{1 + V^{2}}{1 - V}$

$\frac{1 - V}{1 + V^{2}} d V = \frac{d X}{X}$

$\frac{V - 1}{V^{2} + 1} d V + \frac{d X}{X} = 0$

$\frac{1}{2} l n | V^{2} + 1 | - t a n^{- 1} V + l n | X | = c$

$l n \sqrt[]{V^{2} + 1} \cdot X - t a n^{- 1} V = c$

$l n (\sqrt[]{1 + {(\frac{Y - 1}{X -})}^{2}} \cdot | X - 1 |) t a n^{- 1} \frac{Y - 1}{X - 1} = c$

$(2, 1) \Rightarrow l n (\sqrt[]{1 + 0} \cdot 1) - 0 = c$

c = 0

$l n \sqrt[]{{(X - 1)}^{2} + {(Y - 1)}^{2}} = t a n^{- 1} \frac{Y - 1}{X - 1}$

point (k + 1, 2) $l n \sqrt[]{k^{2} + 1} = t a n^{- 1} \frac{1}{k}$

$\Rightarrow \frac{1}{2} l n (k^{2} + 1) = t a n^{- 1} \frac{1}{k}$

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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 (k + 1, 2), k > 0, then

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If the solution curve of the differential equation dydx=x+y−2x−y passes through the points (2, 1) and (k + 1, 2), k > 0, then

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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 (k + 1, 2), k > 0, then