Let c be the arbitrary constant- then the solution of the differential equation e x y dx -1- e x cosec2 y dy -0 isA- 1- e x y - cB- e x -1 cosec2 y - cC- e x -1 y - cD- e x -1- c y

The correct option is C (e^x 1) y = c e^xdxe^x 1 = cosec^2ydy y Integrating , we get log(e^x 1) + log y = log c ⇒ (e^x 1) y = c ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

Let c be the ...

Question

Let $c$ be the arbitrary constant, then the solution of the differential equation $e^{x} cot y d x + (1 - e^{x}) {cosec}^{2} y d y = 0$ is

$(1 - e^{x}) cot y = c$

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$(e^{x} - 1) {cosec}^{2} y = c$

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$(e^{x} - 1) cot y = c$

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$(e^{x} - 1) = c (cot y)$

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Solution

The correct option is C
$(e^{x} - 1) cot y = c$

$\frac{e^{x} d x}{e^{x} - 1} = \frac{{cosec}^{2} y d y}{cot y}$

Integrating , we get

$log (e^{x} - 1) + log cot y = log c$

$\Rightarrow (e^{x} - 1) cot y = c$

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Let c be the arbitrary constant, then the solution of the differential equation ex coty dx+(1–ex)cosec2y dy=0 is

The correct option is C (ex−1)coty=c exdxex−1=cosec2ydycoty Integrating , we get log(ex−1)+logcoty=logc ⇒(ex−1)coty=c

Let $c$ be the arbitrary constant, then the solution of the differential equation $e^{x} cot y d x + (1 - e^{x}) {cosec}^{2} y d y = 0$ is

The correct option is C
$(e^{x} - 1) cot y = c$

$\frac{e^{x} d x}{e^{x} - 1} = \frac{{cosec}^{2} y d y}{cot y}$

Integrating , we get

$log (e^{x} - 1) + log cot y = log c$

$\Rightarrow (e^{x} - 1) cot y = c$