1 First Order

Definition: A differential equation is an equation with an unknown function and it’s derivative(s)

Example 1.0.1: D_xy = 2x + y
Solution: y = ∫ 2x + 7dx = x² + 7x + C

Example 1.0.2: D_xy + y = 7
Solution here is a little more complicated

1.1 Linear First Order

Definition: A linear first order differential equation is one such that it can be written + P(x)y = Q(x)

Example 1.1.1: D_xy = y(1 - y) = y - y² (the logistic model)
Solution: D_xy = = y(1 -y) ⇔ = dx ⇔∫ = ∫ dx ⇔ u = 1 - ⇒ -∫ du = x + C ⇔-ln|u| = -ln|1 -| = x + C ⇔ ln|1 -| = -x + C ⇔ Assuming 1 - ≥ 0, 1 - = e^-x+C ⇔-1 = e^-x+Cy - y = y(e^-x+C - 1) ⇔ y =

Solving first order linear D.E.s The simplest general method for solving first order linear D.E.s ( + P(x)y = Q(x)) is to add an additional function μ(x):

And then μ(x) can be plugged back in to find y.

Theorem: If P,Q are continuous on an open interval, I, containing x₀, then the initial value problem (IVP) + P(x)y = Q(x), y(x₀) = y₀ has a unique solution y(x) on I given by y(x) = e^{-∫ P(x)dx} for an appropriate C

1.2 Substitution

If you have = f(x,y), substitute part of the e.q. with v = α(x,y), and then plug back into the e.q. at the end.

Usually you want to get a linear D.E. relative to v and D_xv (for example D_xv + P(x)v = Q(x) ) which is easier to solve.

Note 1: sometimes a second order D.E. can be reduced into a first order by substituting v = q(D_xy) ⇒ D_xv = w(D_x²y). (e.g. xD_x²y + D_xy = Q(x) sub v = D_xy ⇒ D_xv = D_x²y ⇒ xD_xv + v = Q(x) is 1st order)

Node 2: you can substitute implicitly (e.g. yD_xy + (D_x²y)² = 0 sub v = yD_xy ⇒ D_xv = (D_xy)²+yD_x²y ⇒ D_xv = 0 ⇒ yD_xy = y = v = c ⇒∫ ydy = ∫ cdx+C)

1.2.1 Special cases of substitution

homogenious equation: An equation of the type: D_xy = F() ⇒ substitute v = ⇒ y = vx ⇒ D_xy = D_x(v)x+v which can be solved by v‘x + v = F(v) ⇒ =

In general f(x,y)dy = g(x,y)dx substitute y = ux ⇒ = h(u)du ⇒ integrate.

Bernoulli euquation: An equation of the type: D_xy + P(x)y = Q(x)yⁿ ⇒ substitute v = y^1-n ⇒ D_xv = (1 - n)y^-nD_xy ⇒ D_xv + (1 - n)P(x)v = (1 - n)Q(x)

Remember that n can be negative (for example D_xy + P(x)y = Q(x))

1.3 Exact equation

An equation I(x,y) + J(x,y)D_xy = 0 ⇔ I(x,y)dx + J(x,y)dy = 0 is exact iff I_y = J_x. Then ∃ψ(x,y) s.t.

and ψ(x,y) = c is a solution.

For an IVP, given f(a) = b, plugin x = a,y = b into ψ, and solve for c.

1.3.1 Autonomous equation

An autonomous DE is one such that the independent variable (e.g. x, t) is not in the eq. (e.g. D_xy = P(y)).

An autonomous equation is always seperable

1.4 Seperable equation

A seperable equation is one of the form D_xy = f(x)g(y). A seperable equation can be solved as follows: D_xy = = f(x)g(y) ⇒∫ = ∫ f(x)dx + C.

2 Second Order

A second order differential equation is a differential equation that includes the second derivative.

2.1 Second Order Constant Coefficient Homogeneous

A second order constant coefficient homogeneous D.E. is one with the form aD_x²y + bD_xy + cy = 0.

Theorem (Super Position): If a second order homogeneous D.E. has 2 solutions a,b, then a + b is also a solution.

To solve a second order linear constant coefficient homogeneous differential equation aD_x²y + bD_xy + cy = 0 let y = e^rx. Then by plugging in we get:

which can be solved for r = r₁,r₂ ⇒ y = Ae^rx + Be^rx ∀A,B by super position. If there is a repeated root (r₁ = r₂), than let y = Ae^rx + Bxe^rx, plug in, and solve.

2.2 Second Order Non-Homogeneous

Let D_x²y + p(x)D_xy + q(x)y = g(x) be the 2nd order non-homogeneous D.E. For simplicity, let L[y] = D_x²y + p(x)D_xy + q(x)y (so the D.E. is L[y] = g(x) ). Then the solution is of the form y = (c₁y₁ + c₂y₂ = y_h) + y_p where y_h is the solution to L[y] = 0. To find y_p there is really two methods:

2.2.1 Undetermined Coefficients

Refer to the following table to find the general equation for y_p based on g(x):

Where s is the smallest integer ≥ 0 s.t. y_p is not a solution to L[y] = 0 and P_n(x),Q_n(x),R_n(x) are polynomials of degree n.

Then plug into L[y_p] = g(t) and solve for the coefficients.

2.2.2 Variation of Parameters

Given y_h = c₁y₁ + c₂y₂ we wantto find u₁,u₂ s.t. D_x(u₁)y₁ + D_x(u₂)y₂ = 0 and D_x(u₁)D_x(y₁) + D_x(u₂)D_x(y₂) = g(x). We can find precise values using the Wronskian.

Definition: The Wronskian of two functions w(f,g) is defined as w(f,g) :=

Lemma: If w(f,g) ≡ 0 (w(f,g)(x) = 0,∀x), then f,g are linearally dependent. If w(f,g) ⁄≡ 0 (∃x s.t. w(f,g)(x)≠0), then f,g are linearally independent.

Then D_xu₁ = and D_xu₂ = . This tells us that y_p = u₁y₁ + u₂y₂ (notice that u₁,u₂ are not derivatives).

Equivelently, this is:

Where x₀ is a convenient point int the interval I in which y₁,y₂ are defined.

3 Nth Order

A nth order differential equation is (as the name implies), a differential equation that includes derrivatives of n orders.

3.1 Nth Order Constant Coefficient Homogeneuous

A nth order constant coefficient homogeneous D.E. is one with the form a₀D_xⁿy + a₁D_x^n-1y + + a_n-1D_xy + a_ny = 0.

To solve, by following the same procedure as for the 2nd order parallel, let y = e^rx ⇒ a₀rⁿ + a₁r^n-1 + + a_n-1r + a_n = 0, and solve.

4 Systems of Differential Equations

4.1 First Order

Definition: A system of first order differential equations is defined as such: D_tx_i = ∑ _j=1ⁿ(P_ij(t)x_j) + g_i(t) for 1 ≤ i ≤ n ⇔ D_t = _ij =

A system of first order D.E.s is homogeneous if g_i(t) = 0,∀t.

Theorem: If {} is a standard basis for ℜⁿ, is a solution to the homogeneous system with the initial condidtion (t₀) = then {} is the fundimental solution set.

4.1.1 Eigenvalue Method for Homogeneous Systems

For a homogeneous system, you have D_x = A. If you can find the eigenvalues λ₁,,λ_n and eigenvectors ,,, then = ∑ _i=1ⁿc_ie^λ_i

If there are repeated eigenvalues, solve for the known one like normal: (In this example I’m just showing for a system of two, but it extends) = + , = c₁e^λ₁t, let = te^t ⇒ D_t = (e^t + te^t). Then plug back into initial D.E., and solve for .

4.1.2 Converting to and from systems

Given a second order, often you can convert to a system of first orders, or vice versa. This is done by assigning the variables in the system to be different level derivatives.

Example 4.1.2.1: aD_x²x + bx = 0 ⇒ let x₁ = x,x₂ = D_xx ⇒ D_xx₂ = ,D_xx₁ = x₂

You can also go from multiple of a higher order to lower orders:

Example 4.1.2.2: For two second orders we define y₁ = x₁, y₂ = D_xx₁, y₃ = x₂, y₄ = D_xx₂

4.1.3 Matrix Exponents

Consider D_tX = AX ∀X ∈ M_n×n(S). If = c₁ + c₂ + + c_n (means = e^λ_it most likely) is a solution to D_t = A, then there is a solution X = Φ(t) = where are the columns of the matrix. If there is an initial condition (0) = then Φ(t) = ⇔ = Φ^-1(t)

Then to solve D_tX = AX, (0) = , we have (t) = Φ(t)Φ^-1(0) To solve D_tX = AX, we really want X = e^At. We know e^x = ∑ _n=0^∞ ⇒ e^A = ∑ _n=0^∞ ⇒ e^At = ∑ _n=0^∞

Note 4.1.3.1: If AB = BA, e^A+B = e^Ae^B;(e^A)^-1 = e^-A;e^{0∈M_n×n(S)} = I If A = diag(a₁,a₂,,a_n) then e^A = diag(e^a₁,e^a₂,,e^a_n) If A = SDS^-1 for a diagonal D then e^A = Se^DS^-1. if A is non-diagonalizable, then cehck if there is n s.t. Aⁿ = 0. If so, than a polynomial can be constructed from e^A = ∑ _i=0ⁿ

Example 4.1.3.2 For A = , A²≠0, A³ = 0 ⇒ e^A = ∑ _i=0² = (A⁰ = I) + A + A²

Note 4.1.3.3: If AB = BA, C = A + B then e^Ct = e^Ate^Bt. Note that if A = nI, AB = nIB = nB = Bn = BnI = BA.

Thus the solution to D_t = a, (0) = is (t) = e^At (= Φ(t)Φ^-1(0) ⇒ e^At = Φ(t)Φ^-1(0))