solve partial differantial equation x((∂f(x,y))/∂x)+y((∂f(x,y))/∂y)=f(x,y)ln(x^2 +y^2 ) ((∂^2 f(x,y))/∂x^2 )+((∂^2 f(x,y))/∂y^2 )=0


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 214678 by issac last updated on 16/Dec/24

$solve$ $partial differantial equation$ $x \frac{\partial f (x, y)}{\partial x} + y \frac{\partial f (x, y)}{\partial y} = f (x, y) \ln (x^{2} + y^{2})$ $\frac{\partial^{2} f (x, y)}{\partial x^{2}} + \frac{\partial^{2} f (x, y)}{\partial y^{2}} = 0$
	Answered by MrGaster last updated on 09/Feb/25

	$x = r \cos θ, y = r \sin θ$ $f (x, y) = f (r, \cos θ, r \sin θ) = g (r, θ)$ $\Rightarrow \frac{\partial f}{\partial x} = \frac{\partial g}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial g}{\partial θ} \frac{\partial θ}{\partial x} = \cos θ \frac{\partial g}{\partial r} - \frac{\sin θ}{r} \frac{\partial g}{\partial θ}$ $\Rightarrow \frac{\partial f}{\partial y} = \frac{\partial g}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial g}{\partial θ} \frac{\partial θ}{\partial y} = \sin θ \frac{\partial g}{\partial r} + \frac{\cos θ}{r} \frac{\partial g}{\partial θ}$ $\Rightarrow \frac{\partial^{2} f}{\partial x^{2}} = \cos^{2} θ \frac{\partial^{2} g}{\partial r^{2}} - \frac{2 \cos θ \sin θ}{r} \frac{\partial^{2} g}{\partial r \partial θ} + \frac{\cos^{2} θ}{r} \frac{\partial g}{\partial r} - \frac{2 \cos θ \sin θ}{r^{2}} \frac{\partial g}{\partial θ}$ $\Rightarrow x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = r \cos θ (\cos θ \frac{\partial g}{\partial r} - \frac{\sin θ}{r} \frac{\partial g}{\partial θ}) + r \sin θ (\sin θ \frac{\partial g}{\partial r} + \frac{\cos θ}{r} \frac{\partial g}{\partial θ})$ $= r \cos^{2} θ \frac{\partial g}{\partial r} - \cos θ \sin θ \frac{\partial g}{\partial θ} + r \sin^{2} θ \frac{\partial g}{\partial r} + \cos θ \sin θ \frac{\partial g}{\partial θ}$ $= r \frac{\partial g}{\partial r}$ $= g (r, θ) \ln (r^{2})$ $\Rightarrow r \frac{\partial g}{\partial r} = g (r, θ) \ln (r^{2})$ $r \frac{\partial g}{\partial r} = g (r, θ) \ln (r^{2})$ $\frac{\partial g}{g} = \frac{\ln (r^{2})}{r} d r$ $\ln g = \int \frac{2 \ln r}{r} d r = \ln^{2} r + C (θ)$ $g (r, θ) = e^{C (θ)} r^{2 \ln r}$ $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = 0$ $\Rightarrow \frac{\partial^{2} g}{\partial r^{2}} + \frac{1}{r^{2}} \frac{\partial^{2} g}{\partial θ^{2}} + \frac{1}{r} \frac{\partial g}{\partial r} = 0$ $Substitute g (r, θ) = e^{C (θ)} r^{2 \ln r} :$ $\frac{\partial g}{\partial r} = e^{C (θ)} (2 \ln r + 2) r^{2 \ln r - 1}$ $\frac{\partial^{2} g}{\partial r^{2}} = e^{C (0)} (2 \ln r + 2) (2 \ln r - 1) r^{2 \ln r - 2}$ $\frac{\partial^{2} g}{\partial θ^{2}} = e^{C (θ)} \frac{d^{2} C (θ)}{\partial θ^{2}} r^{2} \ln r$ $\Rightarrow e^{C (θ)} (2 \ln r + 2) (2 \ln r - 1) r^{2 \ln r - 2} + \frac{1}{r^{2}} e^{C (θ)} \frac{d^{2} C (θ)}{d θ^{2}} r^{2 \ln r} + \frac{1}{r} e^{C (θ)} (2 \ln r + 2) r^{2 \ln r - 1} = 0$ $e^{C (θ)} r^{2 \ln r - 2} [(2 \ln r + 2) (2 \ln r - 1) + \frac{1}{r^{2}} \frac{d^{2} C (θ)}{d θ^{2}} + \frac{1}{r} (2 \ln r + 2)] = 0$ $∵ e^{C (θ)} r^{2 \ln r - 2} \neq 0, have :$ $(2 \ln r + 2) (2 \ln r - 1) + \frac{1}{r^{2}} \frac{d^{2} C (θ)}{d θ^{2}} + \frac{1}{r} (2 \ln r + 2) = 0$ $This implies :$ $\frac{d^{2} C (θ)}{d θ^{2}} = - r^{2} [(2 \ln r + 2) (2 \ln r - 1) + \frac{1}{r} (2 \ln r + 2)$ $Since the left side is independent of r, the right$ $side must be constant . Thefore, C (θ) is a$ $liear function of θ :$ $C (θ) = C_{1} θ + C_{2}$ $Thus, the solution is :$ $g (r, θ) = e^{C_{1} θ + C_{2}} r^{2 \ln r}$ $f (x, y) = e^{C_{1} θ + C_{2}} r^{2 \ln r}$ $\begin{matrix} f (x, y) = C_{1} e^{C_{2}} {(x^{2} + y^{2})}^{\ln (x^{2} + y^{2})} \end{matrix}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum