If the solution curve of the differential equation <math><mrow><mrow><mo>(</mo><mrow><mrow><mo>(</mo><mrow><mi>t</mi><mi>a</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mi>y</mi></mrow><mo>)</mo></mrow><mo>−</mo><mi>x</mi></mrow><mo>)</mo></mrow><mi>d</mi><mi>y</mi><mo>=</mo><mrow><mo>(</mo><mrow><mn>1</mn><mo>+</mo><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math> passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

(tan−1y)−x)dy=(1+y2)dxdxdy+x1+y2=tan−1y1+y2l.F=e∫11+y2dy=etan−1yx.etan−1y∫etan−1ytan−1y1+y2dyLetetan−1y=t=xetan−1y=etan−1yy−etan−1y+c...(i)? It passes through (1, 0) = c = 2Now put y = tan 1, thenex = e – e + 2⇒x=26

If the solution curve of the differential equation&nbsp;<math><mrow><mrow><mo>(</mo><mrow><mrow><mo>(</mo><mrow><mi>t</mi><mi>a</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mi>y</mi></mrow><mo>)</mo></mrow><mo>−</mo><mi>x</mi></mrow><mo>)</mo></mrow><mi>d</mi><mi>y</mi><mo>=</mo><mrow><mo>(</mo><mrow><mn>1</mn><mo>+</mo><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math>&nbsp;passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine Differential Equations

If the solution curve of the differential equation $((t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

Option 1 -

Option 2 -

$\frac{2}{e}$

Option 3 -

Option 4 -

$\frac{1}{e}$

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Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$(t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$

$\frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{t a n^{- 1} y}{1 + y^{2}}$

$l . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}$

$x . e^{t a n^{- 1} y} \int \frac{e^{t a n^{- 1} y} t a n^{- 1} y}{1 + y^{2}} d y$

Let

$e^{t a n^{- 1} y} = t$

$= x e^{t a n^{- 1} y} = e^{t a n^{- 1} y} y - e^{t a n^{- 1} y} + c . . . (i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x = \frac{2}{6}$

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If the solution curve of the differential equation $((t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

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If the solution curve of the differential equation ((tan−1y)−x)dy=(1+y2)dx passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

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If the solution curve of the differential equation $((t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is