Find the General Solution of xdx + (sin²(y/x)(ydx-xdy)) = 0

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xdx + (sin²(y/x)(ydx-xdy)) = 0

Tags
homogenous equations

fitzmerl duron Tue, 08/22/2017 - 16:59

Here's mine:

I recognize that the differential equation above is a homogenous type, so we need the substitutions $x = vy$ and $dx = vdy + ydv$ to get the solutions to these type of differential equations.

Armed with these hints, the differential equation $x\space dx + (sin \left(\frac{y}{x}\right))^2(y\space dx - x \space dy) = 0$ becomes:

$$x\space dx + \left(sin \left(\frac{y}{x}\right)\right)^2(y\space dx - x \space dy) = 0$$ $$x\space dx + y \left(sin \left(\frac{y}{x}\right)\right)^2 \space dx - x \left(sin \left(\frac{y}{x}\right)\right)^2 \space dy) = 0$$ $$vy\space (vdy + ydv) + y \left(sin \left(\frac{y}{vy}\right)\right)^2 \space (vdy + ydv) - vy \left(sin \left(\frac{y}{vy}\right)\right)^2 \space dy = 0$$ $$v^2y\space dy + vy^2\space dv + vy\left(sin\left(\frac{1}{v} \right) \right)^2 dy+ y^2\left(sin\left(\frac{1}{v} \right) \right)^2 dv - vy\left(sin\left(\frac{1}{v} \right) \right)^2 dy = 0$$ $$v^2y\space dy + vy^2\space dv + y^2\left(sin\left(\frac{1}{v} \right) \right)^2 dv = 0$$

It is apparent that we can put $v$'s on left side and $y$'s on right side. It looks like this:

$$v^2y\space dy + vy^2\space dv + y^2\left(sin\left(\frac{1}{v} \right) \right)^2 dv = 0$$ $$ vy^2\space dv + y^2\left(sin\left(\frac{1}{v} \right) \right)^2 dv = -v^2y\space dy$$

Divide both left and right sides of the equation by $y^2$, it becomes:
$$ v\space dv + \left(sin\left(\frac{1}{v} \right) \right)^2 dv = -\frac{v^2}{y}\space dy$$

Divide both left and right sides of the equation by $v^2$, it becomes:
$$ \frac{v}{dv} + \frac{\left(sin\left(\frac{1}{v} \right) \right)^2}{v^2} dv = -\frac{1}{y}\space dy$$

Get the integrals of the terms of the differential equation above, it becomes:

$$ \int \frac{v}{dv} + \int \frac{\left(sin\left(\frac{1}{v} \right) \right)^2}{v^2} dv = \int -\frac{1}{y}\space dy$$ $$ln \space v + \left(-\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4}\right) = - ln \space y + c$$

Then, rearranging the expression above to make sense:
$$ln \space v + \left(-\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4}\right) = - ln \space y + c$$ $$ -\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4} = -ln \space v - ln \space y + c$$ $$ -\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4} = -ln \space v - ln \space y + ln \space c$$ $$ -\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4} = ln \space v^{-1} + ln \space y^{-1} + ln \space c$$ $$ -\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4} = ln \left(\frac{c}{vy} \right)$$

Then do this:

$$ -\frac{1}{2v} + \frac{sin \left(\frac{2}{v}\right)}{4} = ln \left(\frac{c}{vy} \right)$$ $$\frac{-4 + 2v \left( sin \left(\frac{2}{v}\right) \right) }{8v} = ln \left( \frac{c}{vy} \right)$$

Returning the original values, it becomes:

$$\frac{-4 + 2\left(\frac{x}{y}\right) \left( sin \left(\frac{2}{\frac{x}{y}}\right) \right) }{8 \left( \frac{x}{y} \right)} = ln \left( \frac{c}{\left(\frac{x}{y} \right) y} \right)$$ $$\left(\frac{y}{8x}\right) \left( -4 + \frac{2x}{y} sin \left( \frac{2y}{x} \right) \right) = ln \left( \frac{c}{x}\right) $$ $$\frac{-y}{2x} + \frac{1}{4} sin \left( \frac{2y}{x} \right) = ln \left( \frac{c}{x}\right) $$

Ultimately, the solution of the differential equation I got is:

$$xe^{\left(\frac{-y}{2x} + \frac{1}{4} sin \left( \frac{2y}{x} \right) \right)} = c$$

Hope it helps....

Jhun Vert Sat, 08/26/2017 - 09:41

You may check this out.

$x \, dx + \left[ \sin^2 \left( \dfrac{y}{x} \right) \right] (y \, dx - x \, dy) = 0$

$\dfrac{x \, dx}{x^2} - \left[ \sin^2 \left( \dfrac{y}{x} \right) \right] \left( \dfrac{x \, dy - y \, dx}{x^2} \right) = 0$

From double angle formulas:
$\cos 2\theta = 1 - 2 \sin^2 \theta$

$\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)$

$\dfrac{dx}{x} - \dfrac{1}{2} \left[ 1 - \cos \left( \dfrac{2y}{x} \right) \right] d\left( \dfrac{y}{x} \right) = 0$

$\dfrac{dx}{x} - \dfrac{1}{2} d\left( \dfrac{y}{x} \right) + \dfrac{1}{2} \cos \left( \dfrac{2y}{x} \right) \, d\left( \dfrac{y}{x} \right) = 0$

$\dfrac{dx}{x} - \dfrac{1}{2} d\left( \dfrac{y}{x} \right) + \dfrac{1}{4} \cos \left( \dfrac{2y}{x} \right) \, d\left( \dfrac{2y}{x} \right) = 0$

$\displaystyle \int \dfrac{dx}{x} - \dfrac{1}{2} \int d\left( \dfrac{y}{x} \right) + \dfrac{1}{4} \int \cos \left( \dfrac{2y}{x} \right) \, d\left( \dfrac{2y}{x} \right) = 0$

$\ln x - \dfrac{y}{2x} + \dfrac{1}{4} \sin \left( \dfrac{2y}{x} \right) = c$
 

Basically, this answer is similar to the answer in #3 but expressed in different form.