The integrating factor of the differential equation d y d x = y t a n x − y 2 s e c x is [MP PET 1995; Pb. CET 2002] Options: A . &nbsp tan x B . &nbsp secx C . &nbsp -sec x D . &nbsp cot x

Answer: Option B : B The differential equation is d y d x − y t a n x = − y 2 s e c x I . F . = e − ∫ t a n x d x This is Bernoulli's equation i.e. reducible to linear equation. Dividing the equation by y 2 , we get 1 y 2 d y d x − 1 y t a n x = − s e c x . . . . . . . . . . . . ( i ) Put 1 y = y ⇒ − 1 y 2 d y d x = d Y d x Equation (i) reduces to − d y d x = − y t a n x = − s e c x ⇒ d Y d x + Y t a n x = s e c x , Which is a linear equation Hence I . F . = e − ∫ t a n x d x = s e c x .

The integrating factor of the differential equation dydx=y tan x

Discussion Forum : Differential Equations

Question -

The integrating factor of the differential equation $\frac{d y}{d x} = y t a n x - y^{2} s e c x$ is

[MP PET 1995; Pb. CET 2002]

Options:

A . tan x

B . secx

C . -sec x

D . cot x

Answer: Option B
:
B

The differential equation
is $\frac{d y}{d x} - y t a n x = - y^{2} s e c x I . F . = e^{- \int t a n x d x}$
This is Bernoulli's equation i.e. reducible to
linear equation.
Dividing the equation by $y^{2},$ we get
$\frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x . . . . . . . . . . . . (i)$
Put $\frac{1}{y} = y \Rightarrow - \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d Y}{d x}$
Equation (i) reduces to $- \frac{d y}{d x} = - y t a n x = - s e c x \Rightarrow \frac{d Y}{d x} + Y t a n x = s e c x,$ Which is a linear equation
Hence $I . F . = e^{- \int t a n x d x} = s e c x .$

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