Solution of the differential equation x dy = y( x - y) dx, 0

Correct Answer is : sec x = (tan x + C ) y

Correct Answer is : sec x = (tan x + C ) y

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Army Group X Practice Set 1

Question : 46 of 70

Marks: +1, -0

\cos x dy = y(\sin x - y) dx,

0 < x < \frac{\pi}{2} \text{ is }

Solution:

\cos x dy = y(\sin x - y) dx

⇒

\frac{dy}{dx} = y \sin x - \frac{y^2}{\cos x}

⇒

\frac{1}{y^2} \frac{dy}{dx} = \frac{1}{y} \tan x - \sec x

⇒

-\frac{1}{y^2} \frac{dy}{dx} + \frac{1}{y} \tan x = \sec x \ \dots\dots (i)

\text{Put } \frac{1}{y} = t \text{ in Eq.(i),}

-\frac{1}{y^2} \frac{dy}{dx} = \frac{dt}{dx} \ \dots\dots (ii)

From Eqs. (i) and ( ii), we get

\frac{dt}{dx} + t \cdot \tan x = \sec x

It is a linear differential equation.

\therefore \text{IF} = e^{\int \tan x \, dx} = e^{\log |\sec x|} = \sec x

\therefore

Solution of differential equation

t \cdot \sec x = \int \sec x \sec x \, dx + C

⇒

\frac{1}{y} \sec x = \tan x + C

\sec x = y (\tan x + C)

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