Solve the differential equation (x^2+y^2)(x dx+y dy) =…

Solve the differential equation $(x^2+y^2)(x\,dx+y\,dy) = (x^2+y^2+x)(x\,dy-y\,dx)$.

Explanation of Method

Converting to polar coordinates $x=r\\cos\\theta,\\ y=r\\sin\\theta$ exploits the identities $x\\,dx+y\\,dy=r\\,dr$ and $x\\,dy-y\\,dx=r^2\\,d\\theta$, collapsing the equation to a first-order linear ODE for $r(\\theta)$, solved by an integrating factor.

Convert to polar form

Substituting $x\\,dx+y\\,dy=r\\,dr$ and $x\\,dy-y\\,dx=r^2\\,d\\theta$ turns the equation into $r^3\\,dr=r^2(r^2+r\\cos\\theta)\\,d\\theta$. Dividing by $r^2$ (for $r\\ne0$) gives $r\\,dr = r(r+\\cos\\theta)\\,d\\theta$, and dividing again by $r$,\n\\\[ \\frac{dr}{d\\theta} - r = \\cos\\theta. \\\]

Apply the integrating factor

This is linear in $r$ with $P(\\theta)=-1$, so $\\mu=e^{-\\theta}$. Multiplying through,\n\\\[ \\frac{d}{d\\theta}\\big(re^{-\\theta}\\big) = e^{-\\theta}\\cos\\theta. \\\]

Evaluate the integral

Using $\\int e^{-\\theta}\\cos\\theta\\,d\\theta = \\frac{e^{-\\theta}(\\sin\\theta-\\cos\\theta)}{2}+C$ (confirmed directly by computer-algebra integration and by differentiating the claimed antiderivative), \n\\\[ re^{-\\theta} = \\frac{e^{-\\theta}(\\sin\\theta-\\cos\\theta)}{2} + C. \\\]

Solve for $r(\\theta)$

Multiplying by $e^{\\theta}$ gives $r = \\dfrac{\\sin\\theta-\\cos\\theta}{2} + Ce^{\\theta}$.

Independent verification

Substituting $r(\\theta)=\\frac{\\sin\\theta-\\cos\\theta}{2}+Ce^\\theta$ back into $\\frac{dr}{d\\theta}-r=\\cos\\theta$ reduces it to $0=0$ identically. Going further, parametrizing $x(\\theta)=r(\\theta)\\cos\\theta$, $y(\\theta)=r(\\theta)\\sin\\theta$ and substituting into the original Cartesian equation also collapses it to $0=0$ identically in $\\theta$ and $C$ — a full end-to-end check.

Result

Final answer:

Verified — the source's answer is correct. The general solution in polar form is \[ \\boxed{r = \\frac{\\sin\\theta - \\cos\\theta}{2} + Ce^{\\theta}} \] where $r=\\sqrt{x^2+y^2}$, $\\theta=\\arctan(y/x)$, and $C$ is an arbitrary constant; equivalently in Cartesian form, \[ \\sqrt{x^2+y^2} = \\frac{y-x}{2\\sqrt{x^2+y^2}} + Ce^{\\arctan(y/x)}. \]