First order Differential Equations

1st order: 1 constant - 1 Initial condition to fully define a solution

Linear Differential Equations

Initial form of 1st order linear D.E. :

\frac{d y}{d t} + p (t) \cdot y = g (t)

where: $p (t), g (t)$ continuous functions

General Solution

Integrating factor: $μ (t)$ where $μ (t) \cdot p (t) = μ^{'} (t)$

μ (t) \cdot \frac{d y}{d t} + μ^{'} (t) \cdot y = μ (t) \cdot g (t)

(μ (t) \cdot y)^{'} = μ (t) \cdot g (t)

μ (t) \cdot y + c = \int μ (t) \cdot g (t) d t

y (t) = \frac{\int μ (t) \cdot g (t) d t + c}{μ (t)}

where the integrating factor μ(t) can be found:

μ (t) \cdot p (t) = μ^{'} (t)

p (t) = \frac{μ^{'} (t)}{μ (t)}

p (t) = l n^{'} (μ (t))

\int p (t) d t = l n (μ (t)) + k

μ (t) = e^{\int p (t) d t + k} = k \cdot e^{\int p (t) d t}

Final form of general solution y=y(t):

y (t) = \frac{\int e^{\int p (t) d t} \cdot g (t) d t + c}{e^{\int p (t) d t}}

Solution Process

Put the differential equation in the correct initial form.
Find the integrating factor, μ (t).
Multiply everything in the differential equation by μ (t) and verify that the left side becomes the product rule (μ (t) y (t))′ and write it as such.
Integrate both sides, make sure you properly deal with the constant of integration.
Solve for the solution y(t).
Solve for constant using the initial condition.

Separable Differential Equations

Non-Linear first order Differential Equations
Form :

N (y) \frac{d y}{d x} = M (x)

Solution :

\int N (y) \frac{d y}{d t} = \int M (x)

u = y (x), d u = \frac{d y}{d x} d x

\int N (u) d u = \int M (x) d x = \int N (y) d y

Do not forget the interval of validity of the solution:

It mus be a continuous interval
It must contain the initial condition

Exact Differential Equations

If there is a function Ψ so that the differential can be written as :

\frac{\partial Ψ}{\partial x} + \frac{\partial Ψ}{\partial y} \frac{d y}{d x} = 0

\frac{\partial Ψ}{\partial x} d x + \frac{\partial Ψ}{\partial y} d y = 0

X d x + Y d y = 0

then we call D.E. exact.

Solution :

Finding Ψ(x,y) :
Provided that it is continuous and its first order derivatives are also continuous:

\frac{\partial^{2} Ψ}{\partial x \partial y} = \frac{\partial^{2} Ψ}{\partial y \partial x}

We use this to check if the equation is exact.
Then start from either part:

Ψ = \int X d x = [i n t] + C (y)

The constant of integration is now a function of the other parameter. To find C(y) we integrate Ψ with respect to y and equate it Y:

\frac{\partial Ψ}{\partial y} = Y

Solution of initial equation :

\frac{d}{d x} Ψ (x, y (x)) = 0

Ψ (x, y) = c

and we soooolve for explicit solution within the interval of validity.

Bernoulli Differential Equations

Form :

\frac{d y}{d t} + p (x) y = q (x) y^{n}

where p(x) and q(x) are continuous functions and n is a real number.

Solution

If n=0 or n=1 then this is a #Linear Differential Equations
if n≠0,1 :

divide by $y^{n}$

y^{- n} y^{'} + p (x) y^{1 - n} = q (x)

substitute with $u = y^{1 - n}$ , $u^{'} = (1 - n) y^{- n} y^{'}$

\frac{1}{1 - n} u^{'} + p (x) u = q (x)

Voilà! You have once again a #Linear Differential Equations

Existence of solutions & Intervals of Validity

Theorem 1

$y^{'} + p (t) y = g (t), y (t_{0}) = y_{0}$

If p(t) and g(t) are continuous functions on an open interval $a < t < b$ and the interval contains $t_{0}$ then there is a unique solution to the IVP on that interval

Through this theorem we can know if a solution exists, as well as the interval of validity without actually solving the D.E.

Theorem 2

$y^{'} = f (t, y), y (t_{0}) = y_{0}$

If $f (t, y)$ and $\frac{\partial f}{\partial y}$ are continuous functions in some rectangle $α < t < β$ , $γ < y < δ$ containing the point $(t_{0}, y_{0})$ then there is a unique solution to the IVP in some interval $t_{0} - h < t < t_{0} + h$ that is contained in $α < t < β$ .

This indicates that there is a unique solution but we cannot find the interval of validity without solving

Euler's Method

Numerical approximation of differential equations

βλέπε pauls notes