Author Topic: Tut0301-Quiz1  (Read 436 times)

Siyan Chen

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Tut0301-Quiz1
« on: September 27, 2019, 02:13:45 PM »
$$
\frac{dy}{dx} = \frac {x^2-3y^2}{2xy}
$$
It’s not homogeneous, since: $$\frac{dy}{dx}*2xy - x^2 + 3y^2 = 0 $$

Using substitution to solve the equation:
Let $$u = \frac {y}{x}  \\then, y = ux$$
$$\frac {dy}{dx} = \frac {du}{dx} * x + u$$
and,
$$\frac {dy}{dx} = \frac {x^2-3y^2}{2xy} \\ = \frac {1-3(y/x)^2}{2(y/x)}  \\ = \frac {1-3(u^2)}{2u} $$
then,
$$\frac {1-3(u^2)}{2u} = \frac {du}{dx} * x + u \\
\frac {1-3(u^2)}{2u} - u = \frac {du}{dx} * x \\
\frac {1-5(u^2)}{2u} = \frac {du}{dx} * x \\
\frac{2u}{1-5(u^2)} du = \frac {1}{x} dx $$
integrating both sides:
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx$$
Using substitution again: $$let \ a = 1-5(u^2),$$
$$\frac{da}{du} = -10u => du = \frac{da}{-10u}$$
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{-5a} da \\= -\frac{1}{5}\ln{|a|}+C $$
$$ => \int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx \\
     -\frac{1}{5}\ln|1-5(u^2)| = \ln{x}+C \\
     -\frac{1}{5}\ln|1-5(\frac{y}{x})^2| = \ln|x|+C
$$
« Last Edit: September 27, 2019, 05:35:21 PM by Siyan Chen »