First slide

Methods of solving first order first degree differential equations

Question

The solution of the differential equation $x d y - y d x = (\sqrt{x^{2} + y^{2}}) d x$ is

Moderate

A

$\frac{x}{y} = \sqrt{x^{2} + y^{2}} + c$
B

$y + \sqrt{x^{2} + y^{2}} = c x^{2}$
C

$x + \sqrt{x^{2} + y^{2}} = c y^{2}$
D

none

Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{y + \sqrt{x^{2} + y^{2}}}{x} = \frac{y}{x} + \frac{\sqrt{x^{2} + y^{2}}}{x} \\ \frac{d y}{d x} = \frac{y}{x} + \frac{x \sqrt{1 + {(\frac{y}{x})}^{2}}}{x} \\ \frac{d y}{d x} = \frac{y}{x} + \sqrt{1 + {(\frac{y}{x})}^{2}} \\ p u t y = v x \\ \frac{d y}{d x} = v + x \frac{d v}{d x} \\ v + x \frac{d v}{d x} = v + \sqrt{1 + v^{2}} \\ x \frac{d v}{d x} = \sqrt{1 + {(v)}^{2}} \\ \Rightarrow \int \frac{d v}{\sqrt{1 + v^{2}}} = \int \frac{d x}{x} \end{array}$
$\begin{array}{l} \Rightarrow \ln (v + \sqrt{v^{2} + 1}) = \log x + \log c \\ \Rightarrow \ln (\frac{y}{x} + \sqrt{{(\frac{y}{x})}^{2} + 1}) = \ln | c x | \\ \Rightarrow y + \sqrt{x^{2} + y^{2}} = c x^{2} \end{array}$

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