Let the solution curve of the differential equation $xdy - ydx = \sqrt{x^{2} + y^{2}}dx,x 0,y(1

Question

Let the solution curve of the differential equation $xdy - ydx = \sqrt{x^{2} + y^{2}}dx,x > 0,y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to

Accepted Answer

$\frac{xdy - ydx}{x^{2}} = \frac{\sqrt{x^{2} + y^{2}}}{x^{2}}dx$

$${d\left( \frac{y}{x} \right) = \sqrt{1 + \frac{y^{2}}{x^{2}}} \cdot \frac{1}{x}dx }{\int\frac{d\left( \frac{y}{x} \right)}{\sqrt{1 + \left( \frac{y}{x} \right)^{2}}} = \int\frac{1}{x}dx }{\Rightarrow ln\left( \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} \right) = lnx + lnk = lnkx }{\Rightarrow y + \sqrt{x^{2} + y^{2}} = kx^{2} }{0 + 1 = k }{\Rightarrow y + \sqrt{x^{2} + y^{2}} = x^{2} }{y + \sqrt{9 + y^{2}} = 9 }{y = 4}$$

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