Solve the differential equation $x^{2} d y + y (x + y) d x = 0$ , given that $y = 1$ when $x = 1$ .

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Solution

$x^{2} d y + (y x + y^{2}) d x = 0$
$\frac{d y}{d x} + \frac{y}{x} + (\frac{y}{x})^{2} = 0$ $\frac{y}{x} = x$
$v + \frac{x d v}{d x} + v + v^{2} = 0$ $\frac{d y}{d x} = v + \frac{x d v}{d x}$
$\frac{x d v}{d x} = - (v^{2} + 2 v)$
$\int \frac{d v}{(v + 2)} = \int \frac{- d x}{x} \Rightarrow \frac{1}{2} \int \frac{v + 2 - v}{(v + 2) v} = \int \frac{- d x}{x}$
$\frac{1}{2} \int (\frac{1}{v} - \frac{1}{v + 2}) d v = \int \frac{- d x}{x} \Rightarrow \frac{1}{2} l n (\frac{v}{v + 2}) = - l m x + c$
$\Rightarrow \frac{1}{2} l n (\frac{y}{y + 2 x}) = - l n x + c$

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Similar questions

The solution of differential equation x²dy + y(x + y)dx = 0 when x = 1, y = 1 is:

x^{2} d y + y (x + y) d x = 0

x^{2} \frac{d y}{d x} = y^{2} + 2 x y

y = 1

x = 1

({tan}^{- 1} x - y) d x = (1 + x^{2}) d y

x^{2} d y + (x y + y^{2}) d x = 0

x = 1

y = 1

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