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Differential Equations Part 2

Section: Mathematics

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Question : 100 of 100

Marks: +1, -0

If

y \frac{dy}{dx} = x \left[ \frac{y^{2}}{x^{2}} + \frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi'\left(\frac{y^{2}}{x^{2}}\right)} \right], x > 0, \phi > 0

y(1) = -1

\phi\left(\frac{y^{2}}{4}\right)

[31 Aug 2021 Shift 2]

4 ϕ (2)
4 ϕ (1)
2 ϕ (1)
ϕ (1)

Solution:

Given,

y \frac{dy}{dx} = x \left[ \frac{y^{2}}{x^{2}} + \frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi'\left(\frac{y^{2}}{x^{2}}\right)} \right]

...(i)

Let

t = \frac{y}{x}

\Rightarrow y = x t

\Rightarrow \frac{dy}{dx} = t + x \frac{dt}{dx}

∴ Eq. (i) becomes

t \left( t + x \frac{dt}{dx} \right) = \left( t^{2} + \frac{\phi(t^{2})}{\phi'(t^{2})} \right)

\Rightarrow x t \frac{dt}{dx} = \frac{\phi(t^{2})}{\phi'(t^{2})}

\Rightarrow \frac{t \phi'(t^{2})}{\phi(t^{2})} dt = \frac{dx}{x}

Integrating both sides

\int \frac{t \phi'(t^{2})}{\phi(t^{2})} dt = \int \frac{dx}{x}

Let

\phi(t^{2}) = u

\Rightarrow t \phi'(t^{2}) dt = \frac{du}{2}

\therefore \frac{1}{2} \int \frac{du}{u} = \int \frac{dx}{x}

\Rightarrow \frac{1}{2} \ln u = \ln x + C

\Rightarrow \frac{1}{2} \ln \phi(t^{2}) = \ln x + C

\Rightarrow \frac{1}{2} \ln \left( \phi\left(\frac{y^{2}}{x^{2}}\right) \right) = \ln x + C

If

x = 1, y = -1

, then

C = \frac{1}{2} \ln ( \phi(1) )

\therefore \frac{1}{2} \ln \left( \phi\left(\frac{y^{2}}{x^{2}}\right) \right) = \ln x + 2 \ln ( \phi(1) )

If

x = 2

, then

\ln \left( \phi\left(\frac{y^{2}}{4}\right) \right) = \ln 4 + \ln [ \phi(1) ]

Or

\phi\left(\frac{y^{2}}{4}\right) = 4 \phi(1)

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