Let <math> <mi>y</mi> <mo>=</mo> <mi>y</mi> <mo>(</mo> <mi>x</mi> <mo>)</mo> </math> be the solution curve of the differential equation, <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mi>x</mi> </mrow> </mrow> </mfenced> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </math> , satisfying <math> <mi>y</mi> <mo>(</mo> <mn>0</mn> <mo>)</mo> <mo>=</mo> <mn>1</mn> </math> . This curve intersects the <math> <mi>x</mi> <mo>-</mo> </math> axis at a point whose abscissa is:

f ' ( x ) = a ( x + 1 ) ( x - 1 ) x 2  f ' ( x ) = a x 4 - x 2 f  f ( x ) = a x 5 5 - a x 3 5 + C ? f ( 0 ) = 0 ⇒ c = 0  l i m x → 0 ? f ( x ) x 3 = 2  ⇒ a = - 6 Minima at ∴ f ' ( x ) = - 6 x 2 - 1 x 2 Maxima at x = - 1

Let <math> <mi>y</mi> <mo>=</mo> <mi>y</mi> <mo>(</mo> <mi>x</mi> <mo>)</mo> </math> be the solution curve of the differential equation, <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mi>x</mi> </mrow> </mrow> </mfenced> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </math> , satisfying <math> <mi>y</mi> <mo>(</mo> <mn>0</mn> <mo>)</mo> <mo>=</mo> <mn>1</mn> </math> . This curve intersects the <math> <mi>x</mi> <mo>-</mo> </math> &nbsp;axis at a point whose abscissa is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Maths Class 12th Differential Equations

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x -$ axis at a point whose abscissa is:

Option 1 -

$2 + e$

Option 2 -

$- e$

Option 3 -

2 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

$f^{'} (x) = a (x + 1) (x - 1) x^{2}$

$f^{'} (x) = a x^{4} - x^{2} f$

$f (x) = \frac{a x^{5}}{5} - \frac{a x^{3}}{5} + C$
$? f (0) = 0 \Rightarrow c = 0$

${l i m}_{x \to 0} ? \frac{f (x)}{x^{3}} = 2$

$\Rightarrow a = - 6$

Minima at $∴ f^{'} (x) = - 6 (x^{2} - 1) (x^{2})$

Maxima at $x = - 1$

<math> <mrow> <mi mathvariant="normal"> </mi> </mrow> <msup> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mo>(</mo> <mi>x</mi> <mo>)</mo> <mo>=</mo> <mi>a</mi> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math>  <math> <msup> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mo>(</mo> <mi>x</mi> <mo>)</mo> <mo>=</mo> <mi>a</mi> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>-</mo> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>f</mi> </math>  <math> <mi>f</mi> <mo>(</mo> <mi>x</mi> <mo>)</mo> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>a</mi> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>a</mi> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mi>C</mi> </math> <math> <mo>?</mo> <mi>f</mi> <mo>(</mo> <mn>0</mn> <mo>)</mo> <mo>=</mo> <mn>0</mn> <mo>⇒</mo> <mi>c</mi> <mo>=</mo> <mn>0</mn> </math>  <math> <msub> <mrow> <mrow> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mi>x</mi> <mo>→</mo> <mn>0</mn> </mrow> </mrow> </msub> <mo>?</mo> <mfrac> <mrow> <mrow> <mi>f</mi> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> </math>  <math> <mo>⇒</mo> <mi mathvariant="normal">a</mi> <mo>=</mo> <mo>-</mo> <mn>6</mn> </math> Minima at <math> <mo>∴</mo> <msup> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mo>(</mo> <mi>x</mi> <mo>)</mo> <mo>=</mo> <mo>-</mo> <mn>6</mn> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> Maxima at <math> <mi>x</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </math>

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Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x -$ axis at a point whose abscissa is:

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Let y = y ( x ) be the solution curve of the differential equation, y 2 - x d y d x = 1 , satisfying y ( 0 ) = 1 . This curve intersects the x - axis at a point whose abscissa is:

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Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x -$ axis at a point whose abscissa is: