If d^(2)x/dt^(2)+g/b(xa)=0,(a,b,g being positive constants) and x=a' and dx/dt=0 when t=0,show that x=a+(a'a)cos{√(g)/(b)t}?
2 Answers
# x(t) = (a'a)cos(sqrt(g/b)t) + a \ \ \ \ #
Explanation:
We have:
# (d^2x)/(dt^2)+g/b(xa)=0# with#x=a', dx/dt=0# when#t=0#
We can write the equation as:
# (d^2x)/(dt^2) + g/bx  (ag)/b = 0 iff (d^2x)/(dt^2) + g/bx = (ag)/b ..... [A]#
This is a second order nonHomogeneous Differentiation Equation. The standard approach is to find a solution,
Complementary Function
The homogeneous equation associated with [A] is
# x''  0x' +g/b = 0 #
And it's associated Auxiliary equation is:
# m^2 +g/b = 0 => m^2 = g/b#
And as we are given that
#m = +sqrt(g/b)#
The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.
 Real distinct roots
#m=alpha,beta, ...# will yield linearly independent solutions of the form#x_1=Ae^(alphat)# ,#x_2=Be^(betat)# , ...  Real repeated roots
#m=alpha# , will yield a solution of the form#x=(At+B)e^(alphat)# where the polynomial has the same degree as the repeat.  Complex roots (which must occur as conjugate pairs)
#m=p+qi# will yield a pairs linearly independent solutions of the form# x=e^(pt)(Acos(qx)+Bsin(qx))#
Thus the solution of the homogeneous equation [A] is:
# x = e^(0x)(Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t)) #
# \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t) #
Particular Solution
In order to find a particular solution of the nonhomogeneous equation:
# (d^2x)/(dt^2) + g/bx = f(t) \ \ # with#f(t) = (ag)/b #
So, we should probably look for a solution of the form:
# x = C # ..... [B]
Where the constant
Differentiating [B] wrt
# x' \ \= 0 #
# x'' = 0 #
Substituting these results into the DE [A] we get:
# (0) + g/bC = (ag)/b => C=a#
And so we form the Particular solution:
# x_p = a #
General Solution
Which then leads to the GS of [A}
# x(t) = x_c + x_p #
# \ \ \ \ \ \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t) + a#
Next we apply the initial conditions:
# x=a', dx/dt=0# when#t=0#
So that:
# a' = Acos0 + Bsin0 + a => A = a'a#
And differentiating the above result wrt
# x'(t) = Asqrt(g/b)sin(sqrt(g/b)t) + Bsqrt(g/b)cos(sqrt(g/b)t) #
And again applying the initial conditions:
# 0 = Asqrt(g/b)sin0 + Bsqrt(g/b)cos0 => B = 0 #
And so the complete solution is:
# x(t) = (a'a)cos(sqrt(g/b)t) + a \ \ \ \ # QED
The Problem:

#x'' +g/b(xa)=0 qquad qquad a,b,g gt 0# 
#x_o=a' qquad x'_o =0#
Show that:
Substitute :

#bb(xi = x  a) qquad xi' = x' qquad xi'' = x''# 
#xi_o=a'  a qquad xi'_o =0#
This makes it homogeneous , which means we can race to a solution.
#xi'' + omega^2 xi=0 qquad qquad omega^2 = g/b#
This is the Harmonic Oscillator, which can be solved in any number of ways, but always with wellknown solution:
# {(xi =A cos omegat + B sin omega t), (xi' =  omega A sin omegat + omega B cos omega t):}#
Applying the IV's at
# {(a'  a =A ), (0 = omega B ):}#
So:
Therefore: