Particular solution for 2t
[DOC File]A Level Mathematics Questionbanks
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( particular solution is: – ln = 2t – 1 B1 [8] 13. a) 2xy = y2 – 1 = M1. ln = ln + C M1 A1 A1. Let C = ln A: ln = ln + ln A M1. y > 1; x > 0 modulus signs not needed: ln (y2 – 1) = ln x + ln A. ln (y2 – 1) = ln Ax A1. y2 – 1 = Ax. y2 = Ax + 1 A1 [7] b) y = 2, x = 1 4 = A + 1 ( A = 3 M1 A1 ...
[DOCX File]HW2
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Its general solution is the sum of a particular solution and the general solution of the corresponding homogeneous differential equation, dV dt =-2V , namely V(t)=C e -2t . To find a particular solution of the inhomogeneous equation, we use the method of undetermined coefficients, a.k.a. judicious guessing.
[DOC File]Section 1
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Example 4: Find the particular solution for the initial condition y(0) = 2 for the equation ex² yy’ + x = 0. Example 5: Solve , x > 0, with the initial condition x0 = 1, y0 = ⅓ Example 6: A point is moving along a line in such a way that its velocity is 4- 2t.
[DOC File]Paper Reference(s)
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2 + 5 + 2x = 2t + 9. (6) (b) Find the particular solution of this differential equation for which x = 3 and = –1 when t = 0. (4) The particular solution in part (b) is used to model the motion of a particle P on the x-axis. At time t seconds (t ( 0), P is x metres from the origin O.
[DOC File]COT 5405: Design and Analysis of Algorithms
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If T(n)= (n is a solution, then T(n)= C(n is also a solution (n = 3(n-1 – 2(n-2 . C(n = C(3(n-1 – 2(n-2 ) C(n = 3(C(n-1 )– 2(C(n-2 ) (II) Use (I) & (II) to find the solver for C0=-1 and C1=3. Example 2; T(n)=2T(n-1) +1; T(1)=1. Step 1. Assume: T(n) = C based on the last element in the above equation. C= 2C +1. C= …
[DOC File]1.1-diff-eqns?
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Consequently 2 independent solutions of the homogeneous equation are y1=e-0.2t and y2=e-0.5t. Trying a particular solution of the form yp = Acos(t) + B sin(t) upon differentiation and substituion yields. p(D)[ yp] = (0.7 B - 0.9 A) cos(t) - (0.9 B +0.7 A) sin(t), which has to equal cos(t).
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