Differential Equations
Linear
Separable
y' = -2xy
y'/y = -2x
ln y = -x2 + C
y = e-x^2+c =C1e-x^2
Exact Equations
M(x, y) + N(x, y) y' = 0 is exact if F exists such that
Fx = M and Fy = N.
Example: 3x2 – 2y2 + (1 – 4xy)y' = 0
M = 3x2 – 2y2 ; N = (1 – 4xy)
My = -4y = Nx => exact
Fx = 3x2 – 2y2 and Fy = 1 – 4xy
F = x3 – 2xy2 + f(y) => Fy = -4xy + f'(y) = 1 – 4xy
Conclude: F = x3 – 2xy2 + y
y = cx2
y = cx2 ; c = y/x2
y' = 2cx = 2y/x is the equation of the slope.
Suppose a second family given by
y' = -x/2y; negative reciprocals
2ydy = -x dx or y2 = -x2/2 + k
x2/2k + y2/k = 1 for k > 0; family of ellipses
Decay and Mixing Problems
dx/dt = -kx
ln x = -kt + ln C
ln (x/C) = -kt
x = Ce-kt
Simple Harmonic Motion
x" = -k2x
dv/dt = (dv/dx)(dx/dt) = v dv/dx
x" = vdv/dx = -k2x
vdv = -k2xdx
v2/2 = -k2x2/2 + C; Assume (x0, v0)
C = (1/2)(k2x02)
v2 = k2(x02 – x2)
v = [k2(x02 – x2)]1/2
Falling Bodies
x' = dx/dt
x" = d2x/dt2
m d2x/dt2 = Force; Weight = mg
m = w/g
mx" = -mg
x' = -gt + v0; tc = v0/g
x = -gt2/2 + v0t + s0
x(tc) = (-g/2) (v0/g)2 + v02 /g + s0
= v02/2g + s0 is the max height
Solve for x = 0 for time when object hits earth.
Damping Force
F = -cx' + mg (falling body) assuming damping force is
proportional to the velocity
mx" = -cx' + mg with initial conditions x(0) = 0; x'(0) = 0
mv' + cv = mg; Solve for v to get
mdv + cvdt = mgdt
mdv = (mg – cv)dt
mdv = dt Integrate to get ln (mg – cv) = -ct/m
(mg – cv)
mg – cv = e-ct/m and v = mg[1 – e-ct/m]/c
vterminal = mg/c at t   Let v = dx/dt and solve for x
Linear Differential Equations
y" + 5y' + 6y = 0
r2 +5r + 6 = 0
(r + 2)(r + 3) = 0
r = -2, -3
y = C1e-2t + C2e-3t
Linear Equations
Solve 3x1 – 2x2 = 1
-2x1 + 2x2 = 5
Add to get x1 = 6; x2 = 17/2
Matrix Solution
2 2 1 = 6
2 3 5 17/2
2
Angle of Intersection
Find the angle of intersection of curves 2x2 + y2 = 20 and
4y2 – x2 = 8.
4(20 - 2x2) – x2 = 8 or 9x2 = 72; x = 81/2 ; y =  2
(1) y' = -2x/y
(2) y' = x/4y
at (81/2, 2)
y'= -2* 81/2/2 = -81/2
y' = 81/2/8 = 8-1/2
negative reciprocals => 90 and symmetry