Differential Equations Linear. Separable y' = -2xy y'/y = -2x ln y = -x 2 + C y = e -x^2+c =C 1 e -x^2.

Presentation on theme: "Differential Equations Linear. Separable y' = -2xy y'/y = -2x ln y = -x 2 + C y = e -x^2+c =C 1 e -x^2."— Presentation transcript:

1 Differential Equations Linear

2 Separable y' = -2xy y'/y = -2x ln y = -x 2 + C y = e -x^2+c =C 1 e -x^2

3 Exact Equations M(x, y) + N(x, y) y' = 0 is exact if F exists such that F x = M and F y = N. Example: 3x 2 – 2y 2 + (1 – 4xy)y' = 0 M = 3x 2 – 2y 2 ; N = (1 – 4xy) M y = -4y = N x => exact F x = 3x 2 – 2y 2 and F y = 1 – 4xy F = x 3 – 2xy 2 + f(y) => F y = -4xy + f'(y) = 1 – 4xy Conclude: F = x 3 – 2xy 2 + y

4 y = cx 2 y = cx 2 ; c = y/x 2 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 y 2 = -x 2 /2 + k x 2 /2k + y 2 /k = 1 for k > 0; family of ellipses

5 Decay and Mixing Problems dx/dt = -kx ln x = -kt + ln C ln (x/C) = -kt x = Ce -kt

6 Simple Harmonic Motion x" = -k 2 x dv/dt = (dv/dx)(dx/dt) = v dv/dx x" = vdv/dx = -k 2 x vdv = -k 2 xdx v 2 /2 = -k 2 x 2 /2 + C; Assume (x 0, v 0 ) C = (1/2)(k 2 x 0 2 ) v 2 = k 2 (x 0 2 – x 2 ) v = [k 2 (x 0 2 – x 2 )] 1/2

7 Falling Bodies x' = dx/dtx" = d 2 x/dt 2 m d 2 x/dt 2 = Force; Weight = mg m = w/g mx" = -mg x' = -gt + v 0; t c = v 0 /g x = -gt 2 /2 + v 0 t + s 0 x(t c ) = (-g/2) (v 0 /g) 2 + v 0 2 /g + s 0 = v 0 2 /2g + s 0 is the max height Solve for x = 0 for time when object hits earth.

8 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 v terminal = mg/c at t   Let v = dx/dt and solve for x

9 Linear Differential Equations y" + 5y' + 6y = 0 r 2 +5r + 6 = 0 (r + 2)(r + 3) = 0 r = -2, -3 y = C 1 e -2t + C 2 e -3t

10 Linear Equations Solve 3x 1 – 2x 2 = 1 -2x 1 + 2x 2 = 5 Add to get x 1 = 6; x 2 = 17/2 Matrix Solution 2 2 1 = 6 2 3 5 17/2 2

11 Angle of Intersection Find the angle of intersection of curves 2x 2 + y 2 = 20 and 4y 2 – x 2 = 8. 4(20 - 2x 2 ) – x2 = 8 or 9x 2 = 72; x =  8 1/2 ; y =  2 (1) y' = -2x/y (2) y' = x/4y at (8 1/2, 2) y'= -2* 8 1/2 /2 = -8 1/2 y' = 8 1/2 /8 = 8 -1/2 negative reciprocals => 90  and symmetry