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BITSAT 2019 Paper

Section: Mathematics

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Question : 83 of 150

Marks: +1, -0

In order to solve the differential equation

\frac{dy}{dx}

+ y(x sinx + cosx) = 1

the integrating factor is:

x cos x
x sec x
x sin x
x cosec x

Solution:

(b) Given differential equation is :

x cosx dy / dx + y (x sinx + cosx) = 1

Dividing both the sides by x cos x,

⇒

\frac{dy}{dx}

+

\frac{xy\sin x}{x\cos x}

+

\frac{y\cos x}{x\cos x}

=

\frac{1}{x\cos x}

⇒

\frac{dy}{dx}

+ y tan x +

\frac{y}{x}

=

\frac{\sec x}{x}

⇒

\frac{dy}{dx}

+

\left(\tan x + \frac{y}{x}\right)

y =

\frac{\sec x}{x}

which is of the form

\frac{dy}{dx}

+ P y = Q

Here,P = tan x +

\frac{1}{x}

and Q =

\frac{\sec x}{x}

Integrating factor = e

^{\int P\,dx}

= e

^{\int \tan x + \frac{1}{x}\,dx}

= e

^{(\log \sec x + \log x)}

= e

^{\log (\sec x \cdot x)}

= x sec x

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