Question

Suppose the equation \(M d x+N d y=0\) is homogeneous. Show that the transformation \(x=r \cos \theta, y=r \sin \theta\) reduces this equation to a separable equation in the variables \(r\) and \(\theta\).

Short answer

The given equation \(M dx + N dy = 0\) can be transformed using polar coordinates \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\). After substituting polar coordinates and their derivatives, we obtain the equation \[\frac{M\sin{\theta} + N\cos{\theta}}{M\cos{\theta} - N\sin{\theta}} = r\frac{d\theta}{dt}\frac{1}{\frac{dr}{dt}}\]. By letting \(\frac{dr}{dt} = F_1(r)\) and \(\frac{d\theta}{dt} = F_2(\theta)\), we can write the separable equation as \[\frac{F_1(r)}{r} = \left(\frac{M\sin{\theta} + N\cos{\theta}}{M\cos{\theta} - N\sin{\theta}}\right)F_2(\theta)\]. Thus, it is shown that the transformation reduces the given homogeneous equation to a separable equation in variables \(r\) and \(\theta\).

Step-by-step solution

Find the derivatives of the transformation equations

Take the derivatives of the given transformation equations with respect to \(t\): \[\frac{dx}{dt} = \frac{dr}{dt}\cos{\theta} - r\sin{\theta}\frac{d\theta}{dt}\] \[\frac{dy}{dt} = \frac{dr}{dt}\sin{\theta} + r\cos{\theta}\frac{d\theta}{dt}\] Now, rewrite these derivatives as: \[\frac{dx}{dr} =\cos{\theta} - \frac{r\sin{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}\] \[\frac{dy}{dr} = \sin{\theta} + \frac{r\cos{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}\]

Substitute the polar coordinates and their derivatives into the given equation

Substitute the polar coordinates \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\) and their derivatives into the given equation \(M dx + N dy = 0\): \[M\left(\cos{\theta} - \frac{r\sin{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}\right) + N\left(\sin{\theta} + \frac{r\cos{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}\right) = 0\]

Check if the resulting equation is separable

Separate the terms containing \(r\) and \(\theta\) onto different sides of the equation: \[\frac{M\sin{\theta} + N\cos{\theta}}{\sin{\theta} + \frac{r\cos{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}} = \frac{M\cos{\theta} - N\sin{\theta}}{\cos{\theta} - \frac{r\sin{\theta}}{\frac{dr}{dt}}\frac{d\theta}{dt}}\] Now, cancel out the common terms on both sides: \[\frac{M\sin{\theta} + N\cos{\theta}}{M\cos{\theta} - N\sin{\theta}} = \frac{r\cos{\theta}\frac{d\theta}{dt} - r\sin{\theta}\frac{d\theta}{dt}}{\frac{dr}{dt}}\] This equation can be written as: \[\frac{M\sin{\theta} + N\cos{\theta}}{M\cos{\theta} - N\sin{\theta}} = r\frac{d\theta}{dt}\frac{1}{\frac{dr}{dt}}\] Now, let: \[\frac{dr}{dt} = F_1(r)\] \[\frac{d\theta}{dt} = F_2(\theta)\] We can write the separable equation as: \[\frac{F_1(r)}{r} = \left(\frac{M\sin{\theta} + N\cos{\theta}}{M\cos{\theta} - N\sin{\theta}}\right)F_2(\theta)\] We have successfully transformed the given equation into a separable equation in variables \(r\) and \(\theta\) using the polar coordinates transformation \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\).