Calculus II: Introduction to differential equations
1. Differential equation
1.1. Definition
A differential equation is an equation that contains
a derivative.
The general solution of a differential equation is all the
functions that satisfy the related equation.
The particular solution of the differential equation
dy/dx = f(x) is given by its related initial conditions, that
is for x = xo, we have y = y = f(xo) = yo.
1.2. Example
dy/dx + x - 1 = 0
dy/dx = - x + 1
dy = (- x + 1)dx
∫ dy = ∫ (- x + 1) dx
y + cst1 = - (1/2) x2 + x + cst2
or
y = - (1/2) x2 + x + cst
if for x = 0 , yo = 3, then
cst = 3, and the particular solution is :
y = - (1/2) x2 + x + 3
2. Separable differential equations
2.1. Definition
If the differential equation contains variables that we can
separate, the equation is called separable differential
equations.
It has the form:
g(y) dy = f(x) dx
For this equation, to obtain the general solution, we separate
first the variable and calculate the related primitives of the
two sides of the equation.
2.2. Example
Let's solve the following differential equation with
y = 0 if x = 1.
dy/dx = 3x2y3 .
We separate the variables as follows:
dy/y3 = 3x2 dx
The primitives of the two memmbers are:
(-1/2)/y2 = x3 + cst.
That is
y2 = (-1/2)/(x3 + cst)
Or
y2 = - 1/2x3 + cst.
According to the initial conditions y = 0 if
x = 1, we have:
0 = - 1/2 + cst, so cst = + 1/2
The general solution of the equation is:
y2 = - 1/2x3 + 1/2.
3. Linear first order differential equation
3.1. Definition
A Linear first order differential equation is of
the form:
dy/dx + P(x)y = Q(x)
That we cannot separate as before.
- Linear stands for the power of the terms is 1.
- First order stands for the derivative in the equation
is a first derivative.
To find the general solution of this equation, we
multiply its two sides by a factor called
integrating factor I(x) such that the
left side of the equation dy/dx + P(x)y becomes
[dy/dx + P(x)y] I(x) = d(I(x) y)/dx
.
So
I(x)[dy/dx + P(x)y] = I(x) Q(x)
The first side must be: d(I(x) y)/dx. So
d(I(x) y)/dx = I(x)[dy/dx + P(x)y] , or
y dI(x)/dx + dy/dx . I(x) = I(x). dy/dx + I(x) . P(x) y I(x)
y dI(x)/dx = I(x) . P(x)y
dI(x)/I(x) = P(x) dx
ln(I(x)) = ∫ P(x) dx + cst
We are searching for a particular integrating factor,
so we consider I(x) > 0 and no constant .
I(x) = exp {∫ P(x) dx}
I(x) = exp {∫ P(x) dx}
The equation
d(I(x) y)/dx = I(x) Q(x) gives
I(x) y = ∫ I(x) Q(x) dx
Therefore
y = [∫ I(x) Q(x) dx]/I(x)
Differential equation:
dy/dx + P(x)y = Q(x)
Integrating factor:
I(x) = exp {∫ P(x) dx}
General solotion:
y = (1/I(x)) ∫ I(x) Q(x) dx
3.2. Example
Solve the differential equation:
dy/dx + 2 xy = 3 x
P(x) = 2 x
Q(x) = 3 x
∫ P(x) dx = x2
I(x) = exp {∫ P(x) dx} = exp {x2}
∫ I(x) Q(x) dx = ∫ exp {x2} 3 x dx =
3 ∫ exp {x2} x dx = (3/2) exp {x2} + const
Therefore
y = (1/I(x)) ∫ I(x) Q(x) dx =
(3/2) exp {- x2} [exp {x2} + const] =
(3/2) [1 + const . exp {- x2}] =
(3/2) + const . exp {- x2}
4. Exercises
Solve the differential equation
dy/dx - 1/x = 0
with yo = 1 when x = e
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