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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.
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i dont know how to do it
1 1/3:2/3:1/6
what is the value of zero to the power raised to zero?
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