Question

Solve the differential equation (x2+y2)dx−2xydy=0

Answer 1

Answer on Question #69635 - Math - Differential Equations

Question

Solve the differential equation

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

Solution

Rewrite the differential equation as

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

This is Euler homogeneous equation, since

$$f (c x, c y) = \frac {c ^ {2} x ^ {2} + c ^ {2} y ^ {2}}{2 (c x) (c y)} = \frac {c ^ {2} (x ^ {2} + y ^ {2})}{c ^ {2} (2 x y)} = \frac {x ^ {2} + y ^ {2}}{2 x y} = f (x, y).$$

Since the numerator and denominator are homogeneous "degree 2" we multiply the right-hand side of the equation by "1" in the form $(1 / x^{2}) / (1 / x^{2})$ :

$$y ^ {\prime} = \frac {x ^ {2} + y ^ {2}}{2 x y} \frac {\left(\frac {1}{x ^ {2}}\right)}{\left(\frac {1}{x ^ {2}}\right)} \Rightarrow y ^ {\prime} = \frac {1 + \left(\frac {y}{x}\right) ^ {2}}{2 \left(\frac {y}{x}\right)}.$$

Now we introduce the change of functions $v = y / x$ ,

$$y ^ {\prime} = \frac {1 + v ^ {2}}{2 v}.$$

Since $y = xv$ , then $y' = v + xv'$ , which implies

$$v + x v ^ {\prime} = \frac {1 + v ^ {2}}{2 v} \Rightarrow x v ^ {\prime} = \frac {1 + v ^ {2}}{2 v} - v = \frac {1 + v ^ {2} - 2 v ^ {2}}{2 v} = \frac {1 - v ^ {2}}{2 v}.$$

We obtained the separable equation

$$v ^ {\prime} = \frac {1}{x} \left(\frac {1 - v ^ {2}}{2 v}\right).$$

We rewrite and integrate it,

$$- \frac {2 v}{v ^ {2} - 1} v ^ {\prime} = \frac {1}{x} \Rightarrow - \int \frac {2 v}{v ^ {2} - 1} d v = \int \frac {1}{x} d x + C _ {0}.$$

The substitution $u = v^2(x) - 1$ implies $du = 2v(x)dv = 2v(x)v'(x)dx$ , so

$$- \int \frac {d u}{u} = \int \frac {d x}{x} + C _ {0} \Rightarrow - \ln u = \ln x + C _ {0} \Rightarrow u = \frac {1}{e ^ {\ln x + C _ {0}}}.$$

But $u = 1 / e^{\ln x}e^{C_0}$ , so denoting $C_1 = e^{C_0}$ , then $u = 1 / (C_1x)$ . So, we get

$$v ^ {2} - 1 = \frac {1}{C _ {1} x} \Rightarrow \left(\frac {y}{x}\right) ^ {2} - 1 = \frac {1}{C _ {1} x} \Rightarrow \left(\frac {y}{x}\right) ^ {2} = \frac {1 + C _ {1} x}{C _ {1} x} \Rightarrow y ^ {2} = \frac {1}{C _ {1}} x + x ^ {2}$$

or

$$y = \pm \sqrt {C _ {2} x + x ^ {2}}. (1)$$

If $v = 1$ , then $y = x$ , $y' = 1$ and

$$y ^ {\prime} = \frac {x ^ {2} + y ^ {2}}{2 x y} \Rightarrow 1 = 1,$$

Which is true, hence $y = x$ is a solution that can be obtained from the general solution (1) if $C_2 = 0$ .

If $v = -1$ , then $y = -x$ , $y' = -1$ and

$$y ^ {\prime} = \frac {x ^ {2} + y ^ {2}}{2 x y} \Rightarrow - 1 = - 1,$$

which is true, hence $y = -x$ is a solution that can be obtained from the general solution (1) if $C_2 = 0$ .

Answer: $y = \pm \sqrt{C_2x + x^2}$ .

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