Pure Mathematics

1)
x = 1/t
...
x^5 d^2z/dx^2 + (2x^4 - 5x^3) dz/dx +4xz = 6x +3 can be reduced to....

d^2z/dt^2 + 5 dz/dt + 4z = 3t+6


2)
use the substitution y=x^2 to show that...
x d^2x/dt^2 + (dx/dt)^2 + 5x(dx/dt) + 3x^2 = sin2t + 3cos2t

can be converted to

d^2y/dt^2 + 5dx/dt + 6y = 2sin2t + 6cos2t

hence solce for x in terms of t

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  1. assume a solution in the form of

    y=AsinBt+CcosDt then
    y'= BAcosBt -CDsinDt
    y"=-B^2 A sinBT + CD^2 cosDt

    put those in the equation..
    -B^2 A sinBT + CD^2 cosDt +5BAcosBt -6CDsinDt +6AsinBt+6CcosDt =2sin2t + 6cos2t

    work with the sin terms first:

    -B^2 A sinBT -6CDsinDt +6AsinBt =2sin2t
    It is clear B=D=2
    -4A-12C+12=2 **1
    then the cosine terms...
    CD^2 cosDt +5BAcosBt +6CcosDt = + 6cos2t
    again, D=B=2
    4C+10A+6C=6***2
    now, you have two equations, two unknowns, solve for A and C. ...and you have the solution

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  2. i worked that part out and got my terms but then noticed that there's a x'term mixed in with the y terms...the ones that are the real problem though are 1 and part a of number 2

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