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An IVP or Initial Value Problem is a differential equation with an appropriate number of initial conditions.
Illustration 3: The subsequent is an IVP.
4x2 y'' + 12y' + 3y = 0; y(4) = 1/8; y'4(-3/64);
Illustration 4: Here's the other IVP.
2ty' + 4y = 3
y(1) = -4.
As I noticed previously the number of initial conditions needed, such will depend on the order of the differential equation.
Any point on parabola, (k 2 ,k) Perpendicular distance formula: D=(k-k 2 -1)/2 1/2 Differentiating and putting =0 1-2k=0 k=1/2 Therefore the point is (1/4, 1/2) D=3/(32 1/2
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