文章目录

1. Separable Differentiable Equations
2. Linear Equations
- First-order linear equation
- - Solution of the homogeneous equation
  - Solution of the inhomogeneous equation
  - - An alternate solution -- Variation of parameters
  - Theorem -- Structure of the Solution
3. Exact Differential Equations
- Definition
- Theorem
- Solving Exact Differential Equation
- - Integrating Factors
  - Separable Equations
- Homogenous Equations
- - Solve Homogenous Equations

1. Separable Differentiable Equations

dydt=g(t)h(y)ordydt=g(t)f(y)\frac{dy}{dt} = \frac{g(t)}{h(y)} \\ or \\ \frac{dy}{dt} = g(t)f(y) dtdy=h(y)g(t)ordtdy=g(t)f(y)

Separate the variables
h(y)dy=g(t)dtordyf(y)=g(t)dth(y)dy = g(t)dt \\ or \\ \frac{dy}{f(y)} = g(t)dt h(y)dy=g(t)dtorf(y)dy=g(t)dt
Integrate both sides
∫h(y)dy=∫g(t)dtor∫dyf(y)=∫g(t)dt\int h(y)dy = \int g(t)dt \\ or \\ \int \frac{dy}{f(y)} = \int g(t)dt ∫h(y)dy=∫g(t)dtor∫f(y)dy=∫g(t)dt
Solve for the solution y(t), if possible

PS: For dyf(y)\frac{dy}{f(y)}f(y)dy, if f(y)=0f(y)=0f(y)=0, then if f(y0)=0f(y_0)=0f(y0)=0, y(t)=y0y(t)=y_0y(t)=y0 is a solution.

2. Linear Equations

First-order linear equation

Form:
x′=a(t)x+f(t)x' = a(t)x + f(t) x′=a(t)x+f(t)
Homogeneous if f(t)=0f(t)=0f(t)=0, form:
x′=a(t)xx' = a(t)x x′=a(t)x
a(t),f(t)a(t), f(t)a(t),f(t) are called coefficients of the equation.

Solution of the homogeneous equation

dxx=a(t)dt⟹ln∣x∣=∫a(t)dt+C∣x∣=e∫a(t)dt+C=eCe∫a(t)dt\frac{dx}{x} = a(t)dt \qquad \Longrightarrow \qquad ln|x| = \int a(t)dt + C \\ |x| = e^{\int a(t)dt + C} = e^C e^{\int a(t)dt} xdx=a(t)dt⟹ln∣x∣=∫a(t)dt+C∣x∣=e∫a(t)dt+C=eCe∫a(t)dt

Let AAA be a constant, which can be either >0, <0 or =0
x(t)=Ae∫a(t)dtx(t) = Ae^{\int a(t)dt} x(t)=Ae∫a(t)dt

Solution of the inhomogeneous equation

The equation
x′=ax+fx' = ax + f x′=ax+f

Rewrite the equation as
x′−ax=fx' - ax = f x′−ax=f
Multiply by the integrating factor
u(t)=e−∫a(t)dtu(t) = e^{-\int a(t)dt} u(t)=e−∫a(t)dt
So that the equation becomes
(ux)′=u(x′−ax)=uf(ux)' = u(x' - ax) =uf (ux)′=u(x′−ax)=uf
Integrate this equation
u(t)x(t)=∫u(t)f(t)dt+Cu(t)x(t) = \int u(t)f(t)dt + C u(t)x(t)=∫u(t)f(t)dt+C
Solve for x(t)

An alternate solution – Variation of parameters

The equation
y′=ay+fy' = ay + f y′=ay+f

Get a partialicular solution to the homogeneous equation yh′=ayhy_h' = ay_hyh′=ayh
yh(t)=e∫a(t)dty_h(t) = e^{\int a(t)dt} yh(t)=e∫a(t)dt
Substitute y=vyhy = vy_hy=vyh into the inhomogeneous equation to find v, or remember that
v′=fyhv' = \frac{f}{y_h} v′=yhf
And solve vvv
Write the v into the general solution y=vyhy = vy_hy=vyh

Theorem – Structure of the Solution

Every solution to the inhomogeneous equation is of the form
y(t)=yp(t)+Ayh(t)y(t) = y_p(t) + Ay_h(t) y(t)=yp(t)+Ayh(t)
Where AAA is an arbitrary constant,

ypy_pyp is a partialicular solution to the inhomogeneous equation y′=a(t)y+f(t)y' = a(t)y + f(t)y′=a(t)y+f(t),

and yhy_hyh is a partialicular solution to the associated homogeneous equation y′=a(t)yy' = a(t)yy′=a(t)y.

3. Exact Differential Equations

Definition

The differential of a continuously differentiable function F is the differential form
dF=∂F∂xdx+∂F∂ydydF= \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy dF=∂x∂Fdx+∂y∂Fdy
A differential form is said to be exact if it is the differential of a continuously differentiable function.

Theorem

Let ω=P(x,y)dx+Q(x,y)dy\omega = P(x,y)dx + Q(x,y)dyω=P(x,y)dx+Q(x,y)dy be a differential form where both P and Q are continuous and differentiable

(a) if ω\omegaω is exact, then
∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ∂y∂P=∂x∂Q
(b) if (a) is true in a rectangle R, then ω\omegaω is exact in R

Solving Exact Differential Equation

If the equation P(x,y)dx+Q(x,y)dy=0P(x, y)dx + Q(x, y)dy = 0P(x,y)dx+Q(x,y)dy=0 is exact, the solution is given by F(x,y)=CF(x, y) = CF(x,y)=C, where F is found by

Solve ∂F∂x=P\frac{\partial F}{\partial x} = P∂x∂F=P by integration:
F(x,y)=∫P(x,y)dx+ϕ(y)F(x, y) = \int P(x, y)dx + \phi (y) F(x,y)=∫P(x,y)dx+ϕ(y)
Solve ∂F∂y=Q\frac{\partial F}{\partial y} = Q∂y∂F=Q
∂F∂y=∂∂y∫P(x,y)dx+ϕ′(y)=Q(x,y)\frac{\partial F}{\partial y} = \frac{\partial}{\partial y} \int P(x, y)dx + \phi'(y) = Q(x, y) ∂y∂F=∂y∂∫P(x,y)dx+ϕ′(y)=Q(x,y)

Integrating Factors

The form Pdx+QdyPdx + QdyPdx+Qdy has an integrating factor depending on one of the variables under the following conditions.

If
h=1Q(∂P∂y−∂Q∂x)h = \frac{1}{Q} \left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) h=Q1(∂y∂P−∂x∂Q)
is a function of x only, then μ(x)=e∫h(x)dx\mu (x) = e^{\int h(x)dx}μ(x)=e∫h(x)dx is an integrating factor.
If
g=1P(∂P∂y−∂Q∂x)g = \frac{1}{P} (\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}) g=P1(∂y∂P−∂x∂Q)
is a function of y only, then μ(y)=e−∫g(y)dy\mu(y) = e^{-\int g(y)dy}μ(y)=e−∫g(y)dy is an integrating factor.

Separable Equations

If the equation has the form
P(x)dx+Q(y)dy=0P(x)dx + Q(y)dy = 0 P(x)dx+Q(y)dy=0
The solution is given by
F(x,y)=∫P(X)dx+∫Q(y)dy=CF(x, y) = \int P(X)dx + \int Q(y)dy = C F(x,y)=∫P(X)dx+∫Q(y)dy=C

Homogenous Equations

A function G(x, y) is homogenous of degree n if
G(tx,ty)=tnG(x,y)G(tx, ty) = t^nG(x, y) G(tx,ty)=tnG(x,y)
A differential equation Pdx+Qdy=0Pdx + Qdy = 0Pdx+Qdy=0 is said to be homogenous if both of the coefficients P and Q are homogeneous of the same degree

Solve Homogenous Equations

Substitute y = xv

Then P(x,y)dx+Q(x,y)dy=0P(x, y)dx + Q(x, y)dy = 0P(x,y)dx+Q(x,y)dy=0 turns into P(x,xv)dx+Q(x,xv)(vdx+xdv)=0P(x, xv)dx + Q(x, xv)(vdx + xdv) = 0P(x,xv)dx+Q(x,xv)(vdx+xdv)=0
P(x,xv)=xnP(1,v)Q(x,xv)=xnQ(1,v)P(x, xv) = x^nP(1, v)\\ Q(x, xv) = x^nQ(1, v) P(x,xv)=xnP(1,v)Q(x,xv)=xnQ(1,v)
Dividing xnx^nxn, and collecting terms
(P(1,v)+vQ(1,v))dx+xQ(1,v)dv=0(P(1, v) + vQ(1, v))dx + xQ(1, v)dv = 0 (P(1,v)+vQ(1,v))dx+xQ(1,v)dv=0
The integrating factor is
1x(P(1,v)+vQ(1,v))\frac{1}{x(P(1, v) + vQ(1, v))} x(P(1,v)+vQ(1,v))1
Separate variables, and solve the equation
dxx+Q(1,v)dvP(1,v)+vQ(1,v)=0\frac{dx}{x} + \frac{Q(1, v)dv}{P(1, v) + vQ(1, v)} = 0 xdx+P(1,v)+vQ(1,v)Q(1,v)dv=0

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