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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

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Vishal Baghel | Contributor-Level 10

4 months ago

The given D.E. is

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

$\frac{d x}{d y} + \frac{y}{1 - y^{2}} \times x = \frac{\propto y}{1 - y^{2}}$ which is of form

$\begin{array}{l} \frac{d x}{d y} + P x = Q & P = \frac{y}{1 - y^{2}} & Q = \frac{\propto y}{1 - y^{2}} \\ \int p d x = \int \frac{y}{1 - y^{2}} d x = - \frac{1}{2} \int \frac{- 2 y}{1 - y^{2} d x} \end{array}$

$\begin{array}{l} = - \frac{1}{2} l o g | 1 - y^{2} | \\ = l o g {[1 - y^{2}]}^{- \frac{1}{2}} \end{array}$

$∴ I . F = e^{\int P d x} = e^{l o g {[1 - y^{2}]}^{- \frac{1}{2}}} = {[1 - y^{2}]}^{- \frac{1}{2}}$

$∴$ option (D ) is correct.

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95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

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95. Choose the correct answer: The integrating factor of the differential equation (1−y2)dxdy+yx=ay(−1<<1) .

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95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .