Question

Solve the equation(x2+y2)dx−2xydy=0

Answer 1

Answer on Question #69637 - Math - Differential Equations

Question

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

Solution

$$(x ^ {2} + y ^ {2}) d x - 2 x y d y = 0$$ $$x ^ {2} + y ^ {2} - 2 x y y ^ {\prime} = 0$$ $$y y ^ {\prime} = \frac {x ^ {2} + y ^ {2}}{2 x}$$ $$y ^ {\prime} = \frac {x ^ {2} + y ^ {2}}{2 x y}$$ $$y ^ {\prime} = \frac {x}{2 y} + \frac {y}{2 x}$$ $$y ^ {\prime} - \frac {y}{2 x} = \frac {x}{2} \cdot \frac {1}{y}$$

Method 1

$$y y ^ {\prime} - \frac {1}{2 x} y ^ {2} = \frac {x}{2}$$ $$z = y ^ {2} \quad (1)$$ $$\frac {1}{2} z ^ {\prime} - \frac {1}{2 x} z = \frac {x}{2}$$ $$z ^ {\prime} - \frac {1}{x} z = x$$

The integrating factor of the differential equation (2) is

$$e ^ {\int - \frac {1}{x} d x} = e ^ {- \ln x} = \frac {1}{x}$$

Multiplying both sides of (2) by $\frac{1}{x}$

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

Integrating both sides with respect to $x$

$$\frac {z}{x} = x + C$$ $$z = x (x + C)$$

Get back to substitution (1) and replace the variable in the previous equality

$$y ^ {2} = x (x + C)$$ $$y = \pm \sqrt {x \cdot (x + C)}$$

Method 2

$$y' - \frac{y}{2x} = \frac{x}{2}y^{-1}$$

$y' + P(x)y = Q(x)y^n$, $n \neq 1$, is the Bernoulli differential equation:

$$z = \frac{1}{y^{n-1}},$$ $$z = e^{-\int P_1(x)dx} \cdot \left(\int Q_1(x) \cdot e^{\int P_1(x)dx} dx + C\right),$$

where

$$Q_1(x) = -(n-1)Q(x), \quad P_1(x) = -(n-1)P(x),$$ $$P(x) = -\frac{1}{2x}, \quad Q(x) = \frac{x}{2}, \quad n = -1,$$ $$P_1(x) = -\frac{1}{x}, \quad Q_1(x) = x,$$ $$z = e^{-\int\left(-\frac{1}{x}\right)dx} \cdot \left(\int x \cdot e^{\int -\frac{1}{x} dx} dx + C\right),$$ $$e^{-\int\left(-\frac{1}{x}\right)dx} = e^{inx} = x,$$ $$\int x \cdot e^{\int -\frac{1}{x} dx} dx = \int x \cdot e^{in\frac{1}{x}} dx = \int x \cdot \frac{1}{x} dx = x,$$ $$z = x \cdot (x + C),$$ $$z = \frac{1}{y^{-2}}, \quad y^2 = z,$$ $$y^2 = x \cdot (x + C),$$ $$y = \pm \sqrt{x \cdot (x + C)}.$$

Answer: $y = \pm \sqrt{x \cdot (x + C)}$

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