Answer:
r = 1/2
c1 = 3/4
c2 = 27/64
c3 = 405/1408
Step-by-step explanation:
Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)
r = 1/2
c1 = 3/4
c2 = 27/64
c3 = 405/1408
The solution is attached.
Answer:
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Answer:
C
Step-by-step explanation:
See pic below for why...
Standard form of a quadratic f(x) = ax^2 + bx + c
In a quadratic function you find the vertex using the formula x = -b/2a; once you find the x value, substitute it into the equation and solve for y.
To know if a parabola opens up or down you look at the coefficient of the x^2 term. If it is positive, the graph opens up; if it is negative, the graph opens down.
