Linear Analasys Week 1 Definitions

ODE

Ordinary Differential Equation

Equations involving an unknown function of one variable and its derivatives

EX: xy''' -x^2y'+y^2=5

Order

Highest appearing unknown derivative. y''' = 3rd Order

Linear vs Unlinear

Unknown is non-linear,

x^2 is fine

y^2 is not fine

DEF: Linear ODE

Explicit Solution

Explicit vs. implicit asks: Is y solved by itself?
When defined solution is in form: y=f(x) Direct
A function y=f(x) that satisfies differential equation on an interval I, is called a solution to a differnetial equation on I.
EX: Show y =e^xcosx is a solution to y’’ - 2y’ +3xy = e^x(3x-2)cosx on (-inf, inf)

ANS:
y’=e^(cosx-sinx), y’’ = e^x(cosx-sinx - sinx - cosx) = -2e^xsinx
Plug in: (-2e^xsinx) -2(e^x(cosx-sinx)) + 3x(e^xcosx)

SIMPLIFY: -2e^xcosx + 3xe^xcosx

Implicit Solution

Defines solution with F(x,y) = 0. Relationship, and solutions do not need to be unique
EX:

Show that xy-ln(y) = x² is an implicit solution to (x- 1/y)y’’ + (2+y’/(y²))y’ = 2

ANS:

Implicit Differentiaion

d/dx(xy-ln(y) = d/dx(x²)

y+xy’ - 1/y * y’ =2x
and

d/dx(y+xy’=y’/y) = d/dx(2x)
y’+y’+xy’’ - (-1/y²(y’)²+1/y * y’’) = 2
(x-1/y)y’’ + (2+y’/y²)y’ = 2

Particular Solution

A solution to an ODE that has no unknown constants

General solution

general vs. particular asks: How many constants are still free?
A solution to an nth order ODE is written to include n arbitrary constants
Literally meaning it doesnt matter as long as it fits.
c₁, c₂, …, c_n, such that every particular solution can be obtained by picking suitable values for c_1, c_2,.. , c_n.
EX:

Find the general solution to y’’ = x-10sin2x

Sol:

y’=∫(x-10sin2x)dx → =x²/2 + 5cos2x + c₁
y= ∫(x²/2 + 5cos2x + c₁)dx → y=x³/6+ 5/2sin2x + c₁x + c₂
(Theorems form claculs guarentee all solutions have this form)

How do we get a particular solution from a general solution

We need initial Conditions.

EX": Find the particular solution to y’’ = x-10sin2x
if y(0) = 1 and y’(0) = 3

ANS:
solve the general solution, we did this is general solution card.

y=x³/6+ 5/2sin2x + c₁x + c₂

when y(0) = 1, we get c₂ = 1

when y’(0) = 3, we get c₁ = -2
Particular solution is:

y = x³/6 + 5/2sin2x - 2x + 1

Initial value problem

An nth order ODE together with n conditons of the form y(x₀)=y₀,…y^(n-1)(x₀)=y_n-1 for some constants x₀, y₀, y₁, … , y_n-1 is called an initial value problem(IVP).

Theorem: Let a₀, …, a_n-1, F be continuous functions on an interval I. Then, for any x₀ for I, the IVP
y⁽ⁿ⁾ + a_n-1(x)y^(n-1) + … +a₁(x)y’+a₀(x)y = F(x)
with y(x₀) = y₀, … < y^(n-1)(x₀) = y_n-1 has a unique solution on I

First Order ODE is form:

dy/dx = f(x,y)
reminder dy/dx is slope

When solving for y yeild all curves y = y(x), called SOLUTION CURVES, whose sloope at (x,y(x)) is given by f(x,y(x))

EX: Sketch some solution curves to y’ = 3x²
we have y = ∫3x²dx = x³+C, so solution curves are shiftied cubics

*specifying an initial condition y(x₀) = y₀ specifies which curve.

Theorem: Existance and Uniqueness for First order ODE

Let f(x,y) be a continuous on the rectangle R [a,b] x [c,d] = {(x,y) | a <= x <= b, c <= y <= d}

and that f_{y(partial deriv)} is continuous on R Then for any interior point (x₀,y₀) \(\in \) R, there exists an interval I containing x₀, such that the IVP. dy/dx = f(x,y), y(x₀) = y₀ Has a unique solution on I.
Basically: Solution curves dont overlap and cover R² <sub>(Range).

EX:</sub>

show that the IVP dy/dx = root(xy) , y(x₀) = y₀ has a unique solution for all x₀, y₀ > 0. Let (x₀, y₀) in the first quadrant (x₀ y₀ > 0).

ANS:

let (x₀,y₀) in the first quadrant (x₀, y₀ > 0) Let R be a rectangle in the first quadrant containing (x₀,y₀) as an interior point.
Since f(x,y) = root(xy) is continuous on R(first quad) and f_y,(x,y) = x/(2root(xy)) which is also continious on R(the rectangle in Q1).
By the theorem, there is an interval I containing x₀ such that the IVP has a unique solution on I. Since x₀, y₀ > 0 were arbitrary, the IVP has aunique solution everywhere as long as x₀ y₀ > 0.
In fact, y = 0 and y = 1/9x³ - 2/9 * x^3/2 + 1/9 are TWO SOLUTIONS TO THE IVP.

Slope Fields

WE can use slope fields to get an idea of what solution curves to dy/dx = f(x,y) look like.
At a selection of points (x,y), we plot a small line segment with slope f(x,y).

EX:

Sketch the slope field for dy/dx = x - y and draw some plausible solution curves.
Choose when the slope is constant.

When your professor says “choose when the slope is constant,” they mean choose values of the slope, like:

k=−2,−1,0,1,2

Then set:

x−y=k

These are the places where the slope is constant.

Solve for y:

x−y=k

−y=k−x

y=x−k

So the isoclines are diagonal lines:

y=x−k

Along each of those lines, every small slope mark has the same slope k.