Problem 2 - Differential Equations

Solve each of the following differential equations:

1.

$\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$;

2.

$y\frac{dy}{dx} + \frac{d^2 y}{dx^2} = 0$.

$$\frac{dy}{dx} = \frac{xy}{x^2 + y^2} = \frac{u}{1 + u^2} \; \hbox{with} \; u=\frac{y}{x}$$

$$\frac{du}{dx} = \frac{y'}{x} - \frac{y}{x^2} = \frac{y'}{x} - \frac{u}{x} \; \hbox{and therefore} \; y' = x\frac{du}{dx} + u$$

$$x\frac{du}{dx} = \frac{u}{1+u^2} - u$$

$$\frac{du}{\frac{-u^3}{1+u^2}} = \frac{dx}{x}$$

$$- \int_{x_0}^{x} \frac{1}{u^3}du - \int_{x_0}^{x} \frac{1}{u}du = \int_{x_0}^x \frac{1}{x}dx$$

$$\left[ \frac{1}{2u^2} - \ln u \right]_{x_0}^x = \ln t - \ln t_0$$

$$\frac{x^2}{2y(x)^2} - \ln \frac{y(x)}{x} = \ln x + \ln k_0$$

$$\frac{x^2}{2y(x)^2} = \ln y(x) + \ln k_0$$

$$y\frac{dy}{dx} + \frac{d^2 y}{dx^2} = 0$$

$$\int_{x_0}^x y\frac{dy}{dx}dx + \int_{x_0}^x \frac{d^2 y}{dx^2}dx = \int_{x_0}^x 0 dx$$

$$\left[ \frac{y(x)^2}{2} \right]_{x_0}^x + \left[ \frac{dy}{dx} \right]_{x_0}^x = 0$$

$$\frac{y(t)^2}{2} + \frac{dy}{dx}(t) = k_0$$

$$2 k_0 - y^2 = 2 \frac{dy}{dx}$$

$$\frac{2dy}{2k_0 - y^2} = dx$$

1.

if $k_0 > 0$ then

$$\frac{1}{\sqrt{2k_0}}\left( \frac{1}{\sqrt{2k_0} - y} + \frac{1}{\sqrt{2k_0} + y} \right) = dx$$

$$\frac{1}{\sqrt{2k_0}} \left( \int_{x_1}^x \frac{1}{\sqrt{2k_0} - y}dx + \int_{x_1}^x \frac{1}{\sqrt{2k_0} + y}dx \right) = \int_{x_1}^x dx$$

$$\frac{1}{\sqrt{2k_0}} \left[ -\ln (\sqrt{2k_0} - y) +\ln (\sqrt{2k_0} + y) \right]_{x_1}^x = [ x ]_{x_1}^x$$

$$\frac{k_1}{\sqrt{2k_0}}\ln \frac{\sqrt{2k_0} + y}{\sqrt{2k_0} - y} = x$$

2.

if $k_0 = 0$ then

$$-\frac{2dy}{y^2} = dx$$

$$\frac{2}{y} = x + k_1$$

3.

if $k_0 < 0$ then

$$-2\frac{dy}{(\sqrt{2k_0})^2 + y^2} = dx$$

$$-\frac{2}{\sqrt{2 k_0}} \arctan \frac{y}{\sqrt{2k_0}} = x + k_1$$