The solution of the differential equation satisfying the condition y

JEE MainMathematicsModerateDifferential equation

The solution of the differential equation
$\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition y

Show answer and solution

Given equation can be rewritten as

\frac{dy}{dx} - \frac{1}{x} \cdot y = 1

Now, I.F.

= e^{- \int\frac{1}{x}dx} = e^{\mathcal{- l}nx} = \frac{1}{x}

Required solution is

y(\frac{1}{x}) = \int\frac{1}{x}dx\mathcal{= l}nx + c

Since, y(1) = 1

\Rightarrow 1 = \mathcal{l}n1 + c \Rightarrow c = 1

\therefore

y = x\ \mathcal{l}xnx + x

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

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