Let c be the arbitrary constant- then the solution of the differential equation e-x- y-dx - -1 - e-x- cosec-2y- dy - 0 is -1-e-x- y - c-e-x-1- cosec-2 nbsp-y - c-e-x-1- y - c-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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

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

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

Suggest Corrections

Similar questions

Q. The general solution of the differential equation

\frac{d y}{d x} = e^{x + y}

, is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

Q. Let

f (x) = \frac{1}{x} I n (\frac{x}{e^{x}})

then its primitive with respect to x is

Q. The solution of the differential equation

[(cos x) d x - d y] (1 + x^{2}) + e^{x} [(1 + x^{2}) {tan}^{- 1} x + 1] d x = 0

is
(where

C

is arbitrary constant)

Let I $= \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x . J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x,$ Then, for an arbitrary constant c, the value of J-I equals

Solution of differential equation $\frac{d y}{d x} + \frac{tan y}{x} = x e^{x} sec y$ is

Join BYJU'S Learning Program

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

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

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