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 [x-y arctan(frac{y}{x})]dx+x arctan (frac{y}{x})dy=0

Question
Integrals
asked 2021-03-06

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

Answers (1)

2021-03-07

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

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