Consider the line y=8x-2 find the equation of the line that is parallel to this line and passes through the point (-5,2)
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Answer:
Step-by-step explanation:
The given differential equation is:
the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.
So, for a regular singular point ; is located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.
Let first convert it to standard form by dividing through with x³
The standard form of the differential equation is :
Thus;
The zeros of is 0
Therefore , the singular points of above given differential equation is 0
Classify each singular point as regular or irregular.
Let p(x) = xP(x) and q(x) = x²Q(x)
p(x) = xP(x)
p(x) =
p(x) = 2
q(x) = x²Q(x)
q(x) =
q(x) =
The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.
Thus ; from above; we can say that q(x) is not analytic at x = 0
do not satisfy the condition,at most to the second power in the denominator of Q(x).
Thus, the point x =0 is an irregular singular point