diff eq final

take half of the coefficient of s, square it, then add and subtract the result to the polynomial

A/Ax+b

A/Ax+B + B/(Ax+B)² + … + K/(Ax+B)^k

Ax+b/ax²+bx+c

Ax+B/ax²+bx+c + … + Kx+L/(ax²+bx+c)^k

(f*g)(t)=∫f(t-Γ)g(Γ)dΓ evaluated from 0 to t

L[f(t-c)H(t-c)] = e^(-cs)F(s)

write the function outside of the heaviside, in brackets write the H function as [H(t-x₀)-H(t-x₁)] from the interval the function occurs

f(t-c)H(t-c)

multiply the convolution by the delta function

the point where the x and y nullclines intersect

when both eigenvalues are positive real numbers

when both eigenvalues are negative real numbers

when there is one positive eigenvalue and one negative

when the two complex eigenvalues have a negative real value (α<0)

when the two complex eigenvalues have a positive real value (α>0)

when the two complex eigenvalues have no real value (α=0)

dy/dt + p(t)y = g(t)

f(x) = A₀/2 + ∑Aₙcos(nπx/L) + ∑Bₙsin(nπx/L)

1/L ∫f(x)cos(nπx/L)dx evaluated from -L to L

1/L ∫f(x)sin(nπx/L)dx evaluated from -L to L

1/L ∫f(x)dx evaluated from -L to L

sin is odd, cos is even

(-1)ⁿ

0

0

2/L ∫even function evaluated from 0 to L

split the integrals into 2 with limits from -L→0 and 0→L

solve for equilibrium points, solve the Jacobian, solve Jacobian at equilibrium points, find eigenvalues from that matrix and categorize those points

matrix form of the partial derivatives of x and y for both equations

when the rod is insulated so the partial of u/x is equal to 0

the partial of u/t = k*second partial of u/x + u(x,t)/cp

u(x,t) = k*second partial of u/x

u(0,t)=T₀ and u(L,t)=Tₙ

u(x,0)=f(x)

u(x,t)= fourier series of f(x) *e^kn²t

C₁e^(r₁t) + C₂e^(r₂t)

C₁e^(r₁t) + tC₂e^(r₂t)

e^(αt) [C₁cos(βt)+C₂sin(βt)]

k=(mg)/x

y’’+(µ/m)y’+(k/m)y=0 or F(x)