Give the correct answer and solve the given equation [x-y arctan(frac{y}{x})]dx+x arctan (frac{y}{x})dy=0

Give the correct answer and solve the given equation $\left[x-y\mathrm{arctan}\left(\frac{y}{x}\right)\right]dx+x\mathrm{arctan}\left(\frac{y}{x}\right)dy=0$

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Arnold Odonnell

We will first write this equation as
$x\mathrm{arctan}\left(\frac{y}{x}\right)\cdot \frac{dy}{dx}=y\mathrm{arctan}\left(\frac{y}{x}\right)-x$ (1)
Notice that $\mathrm{arctan}\left(\frac{y}{x}\right)=0$
means that However, this is
not a solution of the given equation.
Thus, $\mathrm{arctan}\left(\frac{y}{x}\right)\ne 0,$ so we can divide the equation (1)
by $x\mathrm{arctan}\left(\frac{y}{x}\right)$
$\frac{dy}{dx}=\frac{y}{x}-\frac{1}{\mathrm{arctan}\left(\frac{y}{x}\right)}$ (2)
Make the substitution Then
$\frac{dy}{dx}=\frac{d}{dx}\left(ux\right)=x\frac{dy}{dx}+u$ (3)
(use the Chain Rule). Furthermore, (2) becomes
$\frac{dy}{dx}=u-\frac{1}{\mathrm{arctan}\left(u\right)}$ (4)
Combining (3) and (4), we get $x\frac{du}{dx}+u=u-\frac{1}{\mathrm{arctan}\left(u\right)}⇒x\frac{du}{dx}=-\frac{1}{\mathrm{arctan}\left(u\right)}$
Notice that this is a separable equation! It can be written as
$\mathrm{arctan}\left(u\right)du=-\frac{1}{x}dx$
Integrate both sides:
$\int \mathrm{arctan}\left(u\right)du=-\int \frac{dx}{d}$ (5)
Clearly,
$-\int \frac{dx}{d}=-\mathrm{ln}|x|+{C}_{2},$
where ${C}_{2}$ is some constant.
For the other integral, we use the integration by parts:
$\int \mathrm{arctan}\left(u\right)du=\left\{\left(t=\mathrm{arctan}\left(u\right)dv=du\right),\left(dt=\frac{du}{1+{u}^{2}}v=u\right)\right\}$
$=u\mathrm{arctan}\left(u\right)-\int \frac{udu}{1+{u}^{2}}du$
$=u\mathrm{arctan}\left(u\right)-\frac{1}{2}\mathrm{ln}|1+{u}^{2}|+{C}_{1}$
$=u\mathrm{arctan}\left(u\right)-\mathrm{ln}\sqrt{1+{u}^{2}}+{C}_{1}$
where ${C}_{1}$ is some constant.
(In last equality we need used that
and the property of the logarithmic function: $x\mathrm{ln}y=\mathrm{ln}{y}^{x}.\right)$
Finnaly. (5) becomes
$=u\mathrm{arctan}\left(u\right)-\mathrm{ln}\sqrt{1+{u}^{2}}=-\mathrm{ln}|x|+C,$
where $C={C}_{2}-{C}_{1}$