Solve Differential Equations using Polar Coordinates
Type 2: Some times transformation to the polar coordinates facilitates separation of variables. In this connection it is convenient to remember the following differentials. If x = r cosq ; y = r sinq then ,
(i) x dx + y dy = rdr
(ii) dx2 + dy2 = dr2 + r2 dq2
(iii) x dy – ydx = r2 dq
If x = r secq & y = r tanq then x dx – y dy = r dr and x dy – y dx = r2 secq dq.
Note : x = r cos q ; y = r sinq Þ x2 + y2 = r2 and tanq = y/x
Hence x dx + y dy = r dr and x dy – y dx = x2 sec2q dq
s dy – ydx = r2 dq
|||ly x = r secq and y = r tanq x2 – y2 = r2 and y/x = sinq
Hence x dx – ydy = r dr and x dy – y dx = x2 cosq dq = r2 secq dq Examples :
(i) (x dx –y dy)/(x dy –y dx ) =Ö[(1+x2-y2)/(x2-y2)] (T/S)
(ii) [(x+y (dy/dx))/(x(dy/dx)-y)] = Ö[(1-x2 -y2)/(x2+y2)] (T/S)
(iii) x dx + y dy = x (x dy – y dx)
[Solution : (iii) x = r cosq; y = r sinq ; Tanq =(y/x)
Þ x dx + y dy = r dr ; x dy – y dy = r2 dq
r dr = r cosq r2 dq
∫ (dr/r2) =∫ cosqdq
Þ -(1/r) = sinq +c
(1/Ö(x2 +y2)) +(y/(Ö(x2 +y2)) = c
y+ 1 =c Ö(x2+y2)
(y+1)2 =c2 (x2 + y2)
(iv) (x dx + ydy) /(Ö(x2+y2)) = ((y dx –x dy)/x2)
[Solution : Put x = r cosq ; y = r sinq
Þ x2 + y2 = r2 Þ x dx + y dy = r dr
& y/x = tanq Þ ((x dy –ydx)/x2) = sec2 q dq
∫ (r dr/r) =- ∫ sec2 q dq
R=- tanq+c
Ö(x2+y2) +(y/x) =C
Ans]
