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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.
If the diameter of a right cylinder is doubled and the height is tripled, its volume is a. multiplied by 12. b. multiplied by 2. c. multiplied by 6 d. multiplied by 3.
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from 0->1: Int sqrt(1-x^2) Solution) I=∫sqrt(1-x 2 )dx = sqrt(1-x 2 )∫dx - ∫{(-2x)/2sqrt(1-x 2 )}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x 2 ) - ∫-x 2 /√(1-x 2 ) Let
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If depreciation/amortisation is done properly, impairment adjustments will not arise. Required: Do you agree with the above statement? Critically and fully explain your
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