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

Correct Answer is : None of these

Correct Answer is : None of these

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Manipal Entrance Test 2020 Math Paper

Question : 44 of 70

Marks: +1, -0

(y \log x-1) y d x = x d y

Solution:

Given differential equation is

(y \log x-1) y d x = x d y

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

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

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

\Rightarrow \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}

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

From Eq. (i), we have

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

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

.....(ii)

This is linear differential equation.

Here,

I F = e^{\int\left(-\frac{1}{x}\right) dx} = e^{-\log x} = \frac{1}{x}

So, solution is

v \cdot I F = \int IF \cdot Q \, dx + C

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

\Rightarrow 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

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

\Rightarrow 1 = y[\log x + 1 + C x]

\Rightarrow 1 = y[\log x + \log e + C x] \quad [\because 1 = \log e]

\Rightarrow 1 = y[\log e \cdot x + C x]

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