(a) Newton’s Law F=mx"=20m(x-1000)/1000-kx'
The force is the gravitational force acting on the mass of the particle and the resistance force proportional to the particle’s velocity (independent of mass)
Note that x-1000 is a negative quantity, because the gravitational force opposes vertical motion. Mass × overall acceleration equals the resultant force of gravity and resistance.
m=2 kilograms, k is the constant of proportionality.
2nd order DE is 2x"+kx'+40(x/1000-1)=0.
So 2x"+kx'+x/25=40, or
x"+kx'/2+x/50=20.
(b) Split this into the homogeneous equation:
x₁"+kx₁'/2+x₁/50=0 and particular solution (guess) x₂=a, (a is a constant to be found), such that x=x₁+x₂, implying that x' and x'' are the individual sums of the derivatives.
To find a in the particular solution, we solve x₂"+kx₂'/2+x₂/50=20, knowing that when x₂=a, x₂'=x₂"=0 so a/50=20, making a=1000, so x₂=1000.
To solve the homogeneous equation we use the characteristic equation:
r²+kr/2+1/50=0. Completing the square:
r²+kr/2+k²/16-k²/16+1/50=0,
(r+k/4)²-k²/16+1/50=0, r+k/4=±√(k²/16-1/50)
There are three cases depending on k:
(1) k²/16>1/50: this leads to two real roots for r: r₁ and r₂;
(2) k²/16=1/50: this leads to one real root for r=-k/4;
(3) k²/16<1/50: this leads to two complex roots for r: z₁ and z₂.
Corresponding solutions for x₁:
(1) x₁=Ae^(r₁t)+Be^(r₂t);
(2) x₁=Ae^(-kt/4)+Bxe^(-kt/4)
(3) z₁=a+ib, z₂=a-ib⇒Ae^((a+ib)t)+Be^((a-ib)t)⇒
A(e^(at))(e^(ibt))+B(e^(at))(e^(-ibt))⇒
e^(at)[A(cos(bt)+isin(bt))+B(cos(bt)-isin(bt))]⇒
e^(at)[(A+B)cos(bt)+i(A-B)sin(bt)]⇒
e^(at)(Pcos(bt)+Qsin(bt)). P and Q are constants, irrespective of whether they’re complex or not. In other examples, A and B, a and b are constants.
(c) All three cases involve exponential terms. Case 3 combined exponential with trigonometric. Each case involves two constants which are found by using initial conditions such as when t=0, x=0 because the particle is launched from the ground. Also by differentiating x wrt t, we get velocity and we know the initial velocity is 15 m/s at t=0. These two initial conditions are all that’s required to find the constants (designated A, B, P, Q in the examples).
(d) When t=0, x=0. Let’s see what happens when we use these conditions for each of the three cases:
(1) x=Ae^(r₁t)+Be^(r₂t)+1000 (we need to add x₂ to x₁ to get the full solution for x). So A+B+1000=0 initially. x'=Ar₁e^(r₁t)-Br₂e^(r₂t), 15=Ar₁-Br₂.
B=-(1000+A), 15=Ar₁+r₂(1000+A), 15=Ar₁+Ar₂+1000r₂, A=(15-1000r₂)/(r₁+r₂), B=-(1000+A)=-(1000+(15-1000r₂)/(r₁+r₂)),
B=-(1000r₁+1000r₂+15-1000r₂)/(r₁+r₂)=-(1000r₁+15)/(r₁+r₂).
Note that if r₁+r₂=0, there would be no x' term in the differential equation, implying that k=0, which is only possible if there is no atmospheric drag.
x=[(15-1000r₂)/(r₁+r₂)]e^(r₁t)-[(1000r₁+15)/(r₁+r₂)]e^(r₂t)+1000.
When x=1000, the particle will escape the planet’s atmosphere, so:
[(15-1000r₂)/(r₁+r₂)]e^(r₁t)=[(1000r₁+15)/(r₁+r₂)]e^(r₂t).
That is e^(t(r₁-r₂))=(15+1000r₁)/(15-1000r₂).
If the term on the right is negative or zero the particle cannot leave the planet’s atmosphere, because there would be no solution for t. The reason is that, to solve for t, we need to take logs of each side of the equation, and we can’t do this if a quantity≤0. Case (1) is k²/16>1/50, so k>√(8/25), k>(2√2)/5, k>0.5657 approx. 1000r₁=-250k+1000√(k²/16-1/50), 1000r₂=-250k-1000√(k²/16-1/50). (15-250k+1000√(k²/16-1/50))/(15+250k+1000√(k²/16-1/50))>0 to leave the planet’s atmosphere.
I don’t think I have enough space left to go through the other cases, but I hope you have enough to know how to examine them yourself.
