(x^(2)((dy)/(dx))^(2)+2x((dy)/(dx))-12y)=0
Answers
Answer:
Ordinary Differential Equations
1 Introduction
A differential equation is an equation relating an independent variable, e.g. t, a dependent variable, y,
and one or more derivatives of y with respect to t:
dx
dt
= 3x y2 dy
dt
= e
t d
2y
dx
2
+ 3x
2
y
2 dy
dx
= 0.
In this section we will look at some specific types of differential equation and how to solve them.
2 Classifying equations
We can classify our differential equation by four properties:
• Is it an ordinary differential equation?
• Is it linear?
• Does it have constant coefficients?
• What is the order?
Ordinary
An Ordinary Differential Equation or ODE has only one independent variable (for example, x, or t).
The alternative (with more than one) is called a partial differential equation and will not be covered in
this course.
Linearity
A differential equation is linear if every term in the equation contains none or exactly one of either the
dependent variable or its derivatives. There are no products of the dependent variable with itself or its
derivatives. Each term has at most one power of the equivalent of x or ˙x or ¨x or . . . ; or f(x) and its
derivatives.
Examples:
f(x)
df
dx
= −ω
2x is not linear df
dx
= f
3
(x) is not linear d
2f
dx
2
= −x
2
f(x) + e
x
is linear.
Constant coefficients
A differential equation has constant coefficients if the dependent variable and all the derivatives are only
multiplied by constants.
Examples: which have constant coefficients?
3
df
dx
= −ω
2x: yes
d
2f
dx
2
= −x
2
f(x) + e
x
: no
d
2f
dx
2
+ 3
df
dx
+ 2f(x) = sin xex
Finally, a “trick” one:
3e
x df
dx
+ e
x
f(x) = x
3 does have constant coefficients: divide the whole equation by e
x
.
Order
The order of a differential equation is the largest number of derivatives (of the dependent variable) ever
taken.
Examples:
f(x)
df
dx
= −ω
2x is 1st order d
2f
dx
2
= −x
2
f(x)+e
x
is 2nd order d
2f
dx
2
+3
d
2f
dx
2
df
dx
= 0 is 2nd order.
3 First order linear equations
First the general theory. A first order linear differential equation for y(x) must be of the form
dy
dx
+ p(x)y = q(x).
If there is something multiplying the dy/dx term, then divide the whole equation by this first.
Now suppose we calculate an integrating factor
I(x) = exp µZ
p(x) dx
¶
.
Just this once, we won’t bother about a constant of integration.
We multiply our equation by the integrating factor:
I(x)
dy
dx
+ I(x)p(x)y = I(x)q(x).
and then observe that
d
dx
(yI(x)) = dy
dx
I(x) + y
dI
dx
=
dy
dx
I(x) + yp(x)I(x)
which is our left-hand-side. So we have the equation
d
dx
(yI(x)) = I(x)q(x)
which we can integrate (we hope):
yI(x) = Z
I(x)q(x) dx + C
y =
1
I(x)
Z
I(x)q(x) dx +
C
I(x)
.
We sort out the constant C from the initial conditions at the end.
: no.
Step-by-step explanation:
By this , you can do like this statement.
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