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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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SOLVE AND GRAPH THE PARABOLA NOTE: WRITE YOUR SOLUTIONS AND COMPLETE EQUATION OF GRAPH SPOINTS EACH 1. V(0,0) (0.2) P-2 2. V(0,0) E-5,0) P=-5 3. V(4-3) F(4,-2) P=1 4. V-1,5)
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if 1/x+2, 1/x+3, 1/x+5 are in AP find x Ans 1/x+2,1/x+3, 1/x+5 are in AP find x. 1/x+3 - 1/x+2 = 1/x+5-1/x+3 => 1/x 2 +5x+6 = 2/ x 2 +8x +15 => On solving we get x
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