Give the correct answer and solve the given equation [x−yarctan⁡(yx)]dx+xarctan⁡(yx)dy=0

Question

Arnold Odonnell · Accepted Answer

We will first write this equation as xarctan⁡(yx)⋅dydx=yarctan⁡(yx)−x (1) Notice that arctan⁡(yx)=0 means that yx=0, so y=0. However, this is not a solution of the given equation. Thus, arctan⁡(yx)≠0, so we can divide the equation (1) by xarctan⁡(yx) dydx=yx−1arctan⁡(yx) (2) Make the substitution u=yx or y=ux. Then dydx=ddx(ux)=xdydx+u (3) (use the Chain Rule). Furthermore, (2) becomes dydx=u−1arctan⁡(u) (4) Combining (3) and (4), we get xdudx+u=u−1arctan⁡(u)⇒xdudx=−1arctan⁡(u) Notice that this is a separable equation! It can be written as arctan⁡(u)du=−1xdx Integrate both sides: ∫arctan⁡(u)du=−∫dxd (5) Clearly, −∫dxd=−ln⁡|x|+C2, where C2 is some constant. For the other integral, we use the integration by parts: ∫arctan⁡(u)du={(t=arctan⁡(u)dv=du),(dt=du1+u2v=u)} =uarctan⁡(u)−∫udu1+u2du =uarctan⁡(u)−12ln⁡|1+u2|+C1 =uarctan⁡(u)−ln⁡1+u2+C1 where C1 is some constant. (In last equality we need used that 1+u2&amp;gt;0, so |1+u2|=1+u2, and the property of the logarithmic function: xln⁡y=ln⁡yx.) Finnaly. (5) becomes =uarctan⁡(u)−ln⁡1+u2=−ln⁡|x|+C, where C=C2−C1

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

Answered question

Answer & Explanation

New Questions in Integral Calculus