Determine the singular points of the following differential equation and classify each singular point as regular or irregular.
1 answer:
Let be p(x)=<span>(x^{2} -1 q(x)=</span><span>3(x+1) r(x)=1 the three coefficients of the equation a is a singular point of the equation if lim p(x) =0 x------>a so let's find a </span> lim p(x) = lim x²-1=a²-1=0 x------>a x------>a a²-1=0 implies a=+ or -1 so the sigular points are a= -1 or a=1 case 1 for a= -1 lim (x-(-1)) q(x)/p(x)=lim (x+1) 3(x+1)/x²-1=lim3(x+1)/x-1= 0/-2=0 x------> -1 x------> -1 x------> -1 lim (x-(-1))² r(x)/p(x)= lim(x+1)²/x²-1= 0/-2=0 x------> -1 x------> -1 lim (x-(-1)) q(x)/p(x) and lim (x-(-1))² r(x)/p(x) are finite so -1 is regular x------> -1 x------> -1 singular point case 2 a=1 lim (x-1)) q(x)/p(x)=lim (x-1) 3(x+1)/x²-1=lim3(x+1)/x+1= 3 x------> 1 x------> 1 x------> 1 lim (x-1))² r(x)/p(x)= lim(x-1)²/x²-1= =0 x------> 1 x------> 1 1 is also a regular singular point
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<h2>From:</h2><h2 /><h2>kenny</h2>
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