The general solution of the equation dy/dx = y² - x/2y(x+1) is

Correct Answer is : y2=(1+x)log⁡c1+x−1y² = (1+x) c/1+x - 1y2=(1+x)log1+xc−1

Correct Answer is : y2=(1+x)log⁡c1+x−1y² = (1+x) c/1+x - 1y2=(1+x)log1+xc−1

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UPSEE 2017 Solved Paper

Question : 106 of 150

Marks: +1, -0

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

Solution:

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

........(i)

\Rightarrow y\frac{dy}{dx} = \frac{y^2 - x}{2(x+1)}

\Rightarrow y\frac{dy}{dx} - \frac{y^2}{2(x+1)} = -\frac{x}{2(x+1)}

.......(ii)

Put

y^2 = t

By differentiating both side w.r.t. '

x

', we get

2y\frac{dy}{dx} = \frac{dt}{dx}

\therefore

Eq. (ii) reduces to

\frac{1}{2}\frac{dt}{dx} - \frac{t}{2(x+1)} = \frac{-x}{2(x+1)}

⇒

\frac{dt}{dx} - \frac{1}{x+1}t = \frac{-x}{x+1}

This is linear differential equation with

P = \frac{-1}{x+1} \text{ and } Q = \frac{-x}{x+1}

\therefore IF = e^{\int P \, dx}

= e^{\int \frac{-1}{x+1} \, dx}

= e^{-\log(x+1)} = e^{\log \frac{1}{x+1}}

= \frac{1}{x+1}

\therefore

Required solution will be

IF = \int Q(IF) \, dx + \log_C [

Here,

\log c

is constant

]

y^2 \cdot \frac{1}{1+x} = \int \left( \frac{-x}{x+1} \times \frac{1}{x+1} \right) dx

+ \log c

\Rightarrow \frac{y^2}{1+x} = -\int \frac{(x+1)-1}{(x+1)^2} \, dx

+ \log c

\Rightarrow \frac{y^2}{1+x} = -\left[ \int \frac{1}{1+x} \, dx - \int \frac{1}{(1+x)^2} \, dx \right]

+ \log c

\Rightarrow \frac{y^2}{1+x} = -\left[ \log(1+x) + \frac{1}{1+x} \right]

+ \log c

\Rightarrow \frac{y^2}{1+x} = \log \frac{c}{1+x} - \frac{1}{1+x}

\Rightarrow y^2 = (1+x) \log \frac{c}{1+x} - 1

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