The solution of differential equation (y x - 1) y dx = x dy is

Correct Answer is : None of these

Correct Answer is : None of these

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CGPET 2015 Solved Paper

Section: Mathematics

Question : 111 of 150

Marks: +1, -0

(y \log x - 1) y dx = x dy

Solution:

Given differential equation is

(y \log x - 1) y dx = x dy

⇒

\frac{dy}{dx} = \frac{(y \log x -1)y}{x}

\frac{y^2 \log x}{x} - \frac{y}{x}

⇒

\frac{dy}{dx} + \frac{y}{x} = \frac{y^2 \log x}{x}

⇒

\frac{1}{y^2} \frac{dy}{dx} + \frac{y^{-1}}{x} = \frac{\log x}{x}

.... (i)

Put

y^{-1} = v \Rightarrow -y^{-2} \frac{dy}{dx} = \frac{dv}{dx}

⇒

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

From Eq. (i), we have

- \frac{dv}{dx} + \frac{v}{x} = \frac{\log x}{x}

⇒

\frac{dv}{dx} - \frac{v}{x} = - \frac{\log x}{x}

...(ii)

This is linear differential equation.

Here,

\text{IF} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}

So, solution is

v \cdot \text{IF} = \int \text{IF} \cdot Q \, dx + C

v \cdot \frac{1}{x} = \int \frac{1}{x} \left( -\frac{\log x}{x} \right) dx + C

⇒

v \cdot \frac{1}{x} = - \int \frac{\log x}{x^2} dx + C

-\left[ \log x \left(-\frac{1}{x}\right) + \int \frac{1}{x} \cdot \frac{1}{x} dx \right] + C

⇒

\frac{1}{xy} = \frac{\log x}{x} + \frac{1}{x} + C \quad \left[\because v = \frac{1}{y}\right]

⇒

1 = y \left[ \log x + 1 + Cx \right]

⇒

1 = y \left[ \log x + \log e + Cx \right] \quad [\because 1 = \log e]

⇒

1 = y \left[ \log(e x) + Cx \right]

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