Question

(x^2+y^2)dx−2xydy=0
a. x+y/x=c
b. x-y/x=c
c. x-y^2/x=c
d. x+y^2/x=c

Answer 1

Answer on Question #79650 – Math – Differential Equations

**Question**

$$(x^2 + y^2)dx - 2xydy = 0.$$

a. $x + \frac{y}{x} = c$

b. $x - \frac{y}{x} = c$

c. $x - \frac{y^2}{x} = c$

d. $x + \frac{y^2}{x} = c$

**Solution**

This differential equation is homogeneous.

Replace $y = ux, dy = udx + xdu$. Here $u = u(x)$.

$$(x^2 + (ux)^2)dx - 2xux(udx + xdu) = 0$$ $$x^2(1 + u^2)dx - 2x^2u^2dx - 2x^3udu = 0$$ $$x^2(1 - u^2)dx - 2x^3udu = 0$$ $$\frac{x^2}{x^3}dx = \frac{2u}{1 - u^2}du$$ $$\frac{1}{x}dx = -\frac{1}{1 - u^2}d(1 - u^2)$$ $$\ln(x) = -\ln(1 - u^2) + \ln(c)$$

Replace: $u = \frac{y}{x}$

$$x\left(1 - \frac{y^2}{x^2}\right) = c$$ $$x - \frac{y^2}{x} = c$$

Answer: c. $x - \frac{y^2}{x} = c$.

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