Le the solution curve y = f(x) of the differential equation <math><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>y</mi></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>x</mi><mi>y</mi></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>4</mn></mrow></msup><mo>+</mo><mn>2</mn><mi>x</mi></mrow><mrow><mroot><mrow><mn>1</mn></mrow><mrow></mrow></mroot><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>,</mo><mi>x</mi><mo>∈</mo><mrow><mo>(</mo><mrow><mo>−</mo><mn>1</mn><mo>,</mo><mn>1</mn></mrow><mo>)</mo></mrow></mrow></math> pass through the origin. Then <math><mrow><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mo>−</mo><mfrac><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mrow><mfrac><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></munderover><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></mrow></mstyle></mrow></math> is equal to

dydx+xyx2−1=x4+2x1−x2, I.F.e∫xdxx2−1=|x2−1|=1−x2 (?x∈(−1,1))Solution of differential equation is y1−x2=∫(x4+2x)dx=x55+x2+cCurve is passing through origin, c = 0 y=x5+5x251−x2∴∫−3232x5+5x251−x2dx=π3−34

Le the solution curve y = f(x) of the differential equation&nbsp;<math><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>y</mi></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>x</mi><mi>y</mi></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>4</mn></mrow></msup><mo>+</mo><mn>2</mn><mi>x</mi></mrow><mrow><mroot><mrow><mn>1</mn></mrow><mrow></mrow></mroot><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>,</mo><mi>x</mi><mo>∈</mo><mrow><mo>(</mo><mrow><mo>−</mo><mn>1</mn><mo>,</mo><mn>1</mn></mrow><mo>)</mo></mrow></mrow></math>&nbsp;pass through the origin. Then&nbsp;<math><mrow><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mo>−</mo><mfrac><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mrow><mfrac><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></munderover><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></mrow></mstyle></mrow></math>&nbsp;is equal to

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

Le the solution curve y = f(x) of the differential equation $\frac{d x}{d y} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1} - x^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} f (x) d x$ is equal to

Option 1 -

$\frac{π}{3} - \frac{1}{4}$

Option 2 -

$\frac{π}{3} - \frac{\sqrt[]{3}}{4}$

Option 3 -

$\frac{π}{6} - \frac{\sqrt[]{3}}{4}$

Option 4 -

$\frac{π}{6} - \frac{\sqrt[]{3}}{2}$

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1 - x^{2}}},$ $I . F . e^{\int \frac{x d x}{x^{2} - 1}} = \sqrt[]{| x^{2} - 1 |} = \sqrt[]{1 - x^{2}}$ $(? x \in (- 1, 1))$

Solution of differential equation is $y \sqrt[]{1 - x^{2}} = \int (x^{4} + 2 x) d x = \frac{x^{5}}{5} + x^{2} + c$

Curve is passing through origin, c = 0 $y = \frac{x^{5} + 5 x^{2}}{5 \sqrt[]{1 - x^{2}}}$

$∴ \int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} \frac{x^{5} + 5 x^{2}}{5 \sqrt[]{1 - x^{2}}} d x = \frac{π}{3} - \frac{\sqrt[]{3}}{4}$

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Le the solution curve y = f(x) of the differential equation $\frac{d x}{d y} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1} - x^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} f (x) d x$ is equal to

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Le the solution curve y = f(x) of the differential equation dxdy+xyx2−1=x4+2x1−x2,x∈(−1,1) pass through the origin. Then ∫−3232f(x)dx is equal to

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Le the solution curve y = f(x) of the differential equation $\frac{d x}{d y} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1} - x^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} f (x) d x$ is equal to