Let y = y(x) be 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><mn>1</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mi>y</mi><mo>=</mo><msup><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mi>x</mi><mo>−</mo><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup><mo>,</mo><mi>x</mi><mo>></mo><mn>1</mn></mrow></math> passing through the point <math><mrow><mrow><mo>(</mo><mrow><mn>2</mn><mo>,</mo><mroot><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mrow></mrow></mroot></mrow><mo>)</mo></mrow><mo>.</mo></mrow></math> Then <math><mrow><mroot><mrow><mn>7</mn></mrow><mrow></mrow></mroot><mi>y</mi><mrow><mo>(</mo><mrow><mn>8</mn></mrow><mo>)</mo></mrow></mrow></math> is equal to:

dydx+1x2−1y= (x−1x+1)1/2dydx+py=QI.F=e∫Pdx= (x−1x+)12x−2loge|x+1|+CCurves passes through (2, 13)⇒C=2loge3−53at x=8, 7y (8)=19−6loge3

Let y = y(x) be 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><mn>1</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mi>y</mi><mo>=</mo><msup><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mi>x</mi><mo>−</mo><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup><mo>,</mo><mi>x</mi><mo>&gt;</mo><mn>1</mn></mrow></math>&nbsp;passing through the point&nbsp;<math><mrow><mrow><mo>(</mo><mrow><mn>2</mn><mo>,</mo><mroot><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mrow></mrow></mroot></mrow><mo>)</mo></mrow><mo>.</mo></mrow></math>&nbsp;Then&nbsp;<math><mrow><mroot><mrow><mn>7</mn></mrow><mrow></mrow></mroot><mi>y</mi><mrow><mo>(</mo><mrow><mn>8</mn></mrow><mo>)</mo></mrow></mrow></math>&nbsp;is equal to:

19 – 6 loge 3

11 + 6 loge 3

12 – 2loge 3

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

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt[]{\frac{1}{3}}) .$ Then $\sqrt[]{7} y (8)$ is equal to:

Option 1 -

11 + 6 log_e 3

Option 2 -

Option 3 -

12 – 2log_e 3

Option 4 -

19 – 6 log_e 3

3 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}$

$\frac{d y}{d x} + p y = Q$

$I . F = e^{\int P d x} = {(\frac{x - 1}{x +})}^{\frac{1}{2}}$

$x - 2 l o g_{e} | x + 1 | + C$

Curves passes through $(2, \frac{1}{\sqrt[]{3}})$

$\Rightarrow C = 2 l o g_{e} 3 - \frac{5}{3}$

$a t x = 8, \sqrt[]{7} y (8) = 19 - 6 l o g_{e} 3$

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Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt[]{\frac{1}{3}}) .$ Then $\sqrt[]{7} y (8)$ is equal to:

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Let y = y(x) be the solution curve of the differential equation dydx+1x2−1y=(x−1x+1)1/2,x>1 passing through the point (2,13). Then 7y(8) is equal to:

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Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt[]{\frac{1}{3}}) .$ Then $\sqrt[]{7} y (8)$ is equal to: