If the curve satisfying x(x + 1) - y = x(x + 1), passes through (1, 0), then the value of at x = 4, is.

JEE MainMathematicsToughDifferential equation

If the curve satisfying x(x + 1) $\frac{dy}{dx}$ - y = x(x +
1), passes through (1, 0), then the value of
$\frac{5}{2}\left( \frac{5y}{4} - \log_{e}4 \right)$ at x = 4, is.

Show answer and solution

x(x+1)

\frac{dy}{dx}

- y = x (x + 1)

\Rightarrow \frac{dy}{dx} + \left( \frac{- 1}{x(x + 1} \right)

y = 1
Here,

P(x) = \frac{- 1}{x(x + 1)},Q(x) = 1

And integrating factor

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

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

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

So the solution is

y.(\frac{x + 1}{x}) = \int 1.\frac{x + 1}{x}dx + c

y \cdot (\frac{x + 1}{x}) = x + \log_{e}x + C

(x = 1,y = 0 \Rightarrow C = - 1)

\Rightarrow y(\frac{x + 1}{x}) = (x - 1) + \log_{e}x

y(4)(\frac{5}{4}) = 3 + \log_{e}4

\frac{5}{4}y(4) - \log_{e}4 = 3

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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