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BITSAT 2013 Paper

Section: Mathematics

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Question : 117 of 150

Marks: +1, -0

The solution to the differential equation

\frac{dy}{dx}

\frac{yf'(x)-y^2}{f(x)}

where f (x) is a given function

f(x) = y(x + c)
f(x) = cxy
f(x) = c(x + y)
yf(x) = cx

Solution:

We have

\frac{dy}{dx}

=

\frac{f'(x)}{f(x)}

y -

\frac{y^2}{f(x)}

⇒

\frac{dy}{dx}

-

\frac{f'(x)}{f(x)}

y = -

\frac{y^2}{f(x)}

Divide by

y^2

:

y^{-2} \frac{dy}{dx}

-

y^{-1} \frac{f'(x)}{f(x)}

= -

\frac{1}{f(x)}

Put

y^{-1}

= z ⇒

-y^{-2} \frac{dy}{dx}

=

\frac{dz}{dx}

-

\frac{dz}{dx}

-

\frac{f'(x)}{f(x)}

(z) =

-\frac{1}{f(x)}

⇒

\frac{dz}{dx}

\frac{dz}{dx}

+

\frac{f'(x)}{f(x)}

(z) =

\frac{1}{f(x)}

I.F. =

e^{\int \frac{f'(x)}{f(x)} dx}

=

e^{\log f(x)}

= f (x)

∴ The solution is

z 9f(x)) = ∫

\frac{1}{f(x)}

(f(x)) dx + c

⇒

y^{-1}

(f(x)) = x + c ⇒ f (x) = y (x + c)

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