Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$

Question

Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$

i am trying to solving integrating combination

since given that

$\cot (x^2+y^2)dy+xdx+ydy=0$

then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem

Answer 1

$$2\cot(x^2+y^2)dy+2xdx+2ydy=0$$ $$2\cot(x^2+y^2)dy+d(x^2+y^2)=0$$ $$\dfrac{d(x^2+y^2)}{\cot(x^2+y^2)}=-2dy$$ $$-\ln\cos(x^2+y^2)=-2y+C$$

Answer 2

Dividing both sides of $\cot (x^2+y^2)dy+xdx+ydy=0$ by $dy$ we get $\cot (x^2+y^2)+x\frac {dx}{dy}+y=0$

Then,

$$x\frac {dx}{dy}=-y-\cot (x^2+y^2) \iff \frac {dx}{dy}=\frac{-y-\cot (x^2+y^2)}{x} \iff \frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$$

Now you can integrate this: $\frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$

Answer 3

Using polar coordinates, we see that \begin{align} dx =&\ \cos\theta dr -r\sin\theta d\theta,\\ dy =&\ \sin\theta dr + r\cos\theta d\theta \end{align} then we see that \begin{align} \cot(x^2+y^2)dy +xdx+ydy = (\cot(r^2)\sin\theta+r) dr+r\cot(r^2)\cos\theta d\theta =0. \end{align} Hence it follows \begin{align} \frac{d\theta}{dr} = - \frac{\cot(r^2)\sin\theta+ r}{r\cot(r^2)\cos\theta} = -\frac{\tan\theta}{r}-\tan(r^2)\sec\theta. \end{align} Lastly, using the fact that \begin{align} \frac{d (\sin\theta)}{dr} = \cos\theta\frac{d\theta}{dr} = -\frac{\sin\theta}{r}-\tan(r^2) \end{align} then we have obtain a linear equation \begin{align} u'+\frac{1}{r}u=-\tan(r^2) \ \ &\implies\ \ \frac{d}{dr}(ru) = -r\tan(r^2)\\ &\implies\ u= \frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r}. \end{align} Finally, we have that \begin{align} \sin \theta =\frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r} \ \implies \ y = \frac{1}{2}\log|\cos(x^2+y^2)|+C \end{align}

Answer 4

$$\cot (x^2+y^2)dy+xdx+ydy=0$$ $$\cot (x^2+y^2)y'+x+yy'=0$$ Susbstitute $u=x^2+y^2$ and $u'=2x+2yy'$ $$\cot(u)y'+\frac {u'}2=0$$ $$\cot(u)y'=-\frac {u'}2$$ Multiply by $dx$ $$\cot(u)dy=-\frac {du}2$$ $$-2dy=\frac {du}{\cot(u)}$$ Integrate $$-2y+K=\int\frac {du}{\cot(u)}$$ $$-2y+K=\int\frac {\sin(u)du}{\cos(u)}=-\ln|\cos(u)|$$ $$\boxed{y=\frac 12 \ln|\cos(x^2+y^2)|+K}$$