Filetype PDF | Posted on 28 Jan 2023 | 2 years ago
Authentication
digital resources
164x Filetype PDF File size 0.48 MB Source: jmahaffy.sdsu.edu
Calculus 2
Lia Vas
Differential Equations of First Order. Separable Differential
Equations. Euler’s Method
A differential equation is an equation in an unknown function that contains one or more
derivatives of the unknown function.
Afunction y = y(x) is a solution of a differential equation if the equation is satisfied for every
value of the variable x when y and all its derivatives are substituted into the equation.
Forexample,thefunctiony = e2x isasolutionoftheequationy′′+2y′−8y = 0sincethederivatives
′ 2x ′′ 2x 2x 2x 2x 2x
y =2e and y =4e yield an identity 4e +4e −8e =0⇒(4+4−8)e =0⇒0=0 when
plugged in the equation. Note that this is identity does not depend on a specific value of x.
A differential equation of the first order is the equation in function y featuring only the
first derivative y′ of y. Implicit form of such equation is F(y′,y,x) = 0 and, in cases when can solve
for y′, the form is
y′ = f(x,y).
The general solution of the first order differential equation is a family of all functions that
satisfy the equation. For example, the function y = 2x + c is the general solution of the differential
equation y′ = 2.
In many applications however, a solution passing a certain point or satisfying a certain condition
may be more relevant than the general solution. The condition y(x ) = y is called an initial
0 0
condition of the equation y′ = f(x,y) and the differential equation
y′ = f(x,y) together with the initial condition y(x ) = y
0 0
is called an initial value problem. The solution that satisfies the equation and the condition
y(x ) = y is called the particular solution.
0 0
For example, if the condition y(0) = 5 is also considered with the equation y′ = 2,, then the
general solution y = 2x + c does not satisfy it for every, but only for one value of the constant c.
Plugging the initial condition values in the general solution, we obtain that 5 = 2(0)+c and so c = 5.
Thus, y = 2x+5 is the particular solution of this initial value problem.
Separable Differential Equations. A first order differential equation in unknown function
y(x) is separable if we can separate the variables x and y.
Separable Inseparable
1
To solve a separable differential equation,
1. Write y′ as dy.
dx
2. Rewrite the equation so that the left side has just one, and the right side just the other variable.
3. Integrate both sides.
4. If possible, solve for the dependent variable.
Wehave already encountered the simplest form of a separable differential equations, those of the
form
y′ = f(x)
for a given function f(x). For example, the first exam material problem of the form “find a function
such that the given function is its derivative” is of this form. Let us illustrate this with the following
example.
Example 1. Find the general solution of the differential equation y′ = 2x + 3. Then find the
solution with y(0) = 5.
Solution. Integrating the rate y′ = 2x + 3 produces the antiderivative y. Framing this in the
context of separable differential equations, separating the variables in dy = 2x + 3 produces
dx
dy = (2x+3)dx.
Integrating both sides, we have that Z dy = Z (2x + 3)dx ⇒ y = x2 +3x+c.
Thus, the general solution is a family of parabolas of the form y = x2 + 3x + c.
2
Considering the initial condition y(0) = 5, we have that 5 = 0 + 3(0) + c ⇒ 5 = c. Hence, the
parabola y = x2 +3x+5 is the particular solution.
The next example has a bit more of “separating” present.
Example 2. Find the general solution of the differential equation y′ = 2y. Discuss the nature of
the general solution.
Solution. Writing y′ as dy produces dy = 2y. Divide by y and multiply by dx to obtain the
dx dx
variables separated and have that dy = 2dx.
y
Integrate both sides to have
ln|y| = 2x +c.
Solving for |y| produces |y| = e2x+c so that y =
±e2x+c. Note that e2x+c is equal to e2xec. Thus,
c
denoting ±e by C eliminates the ± (as well as
the absolute value) so that we can write the gen-
eral solution as
y = Ce2x.
2
This is a family of exponential functions, increasing and positive if C > 0 and decreasing and
negative if C < 0. If C = 0, the solution is y = 0.
The previous examples leads us to one of the most frequently occurring differential equations
because it models the situation when the rate of change is proportional to the size of the quantity
considered. If y denotes the size of the quantity considered at time x and k denotes the proportionality
constant, the highlighted sentence translates to the following equation.
y′ = ky
Example 3. Find the general solution of y′ = ky where k is a parameter.
Solution. Follow the steps of the previous problem to have that
dy = ky ⇒ dy =kdx ⇒ ln|y|=kx+c ⇒ |y|=ekx+c =ekxec ⇒ y =±ecekx ⇒ y =Cekx
dx y
whereC againstandsfor±ec.Thus, thesolutionsareexponentialfunctions. Note that C corresponds
to the initial size of y since y(0) = Ce0 = C. If we denote it by y , we have the familiar format
0
kx
y = y e
0
of the exponential growth for k > 0 or exponential decay for k < 0. If y0 is positive, the solutions
are increasing exponential functions for k > 0 and decreasing exponential functions for k < 0.
Euler’s Method. In some cases, it is impossible to find analytical solution of an equation (a
formula for a solution even in the implicit form). In those cases, one looks for a numerical solution, a
list of (x,y) points that represents an approximation of the analytical solution. Thus, the difference
between an analytical and numerical solution is that the first is given by an exact formula y = y(x)
of the solution, while the second is a list of points that approximate the points on the exact solution.
Numerical methods of solving differential equations are important because many relevant differential
2 sinx
equations cannot be solved exactly. For example, y′ = ex and y′ = x .
One of the simplest numerical methods for solving a first order differential equation y′ = f(x,y)
with the initial condition y(x ) = y , is Euler’s method.
0 0
Euler’s method approximates the values of the solution at equally spaced x-values x , x =
0 1
x0 +h, x2 = x1 +h,... where h is the step size.
3
Wetreat x as the x-initial value and y as the y-initial value. At the point (x ,y ), the slope
0 0 0 0
of the solution is given by y′ = f(x ,y ) so the tangent line to the solution curve at the initial point
0 0
is y −y
0 = f(x ,y )
x−x 0 0
0
or, in point-slope form,
y −y =f(x ,y )(x−x ) or y = y +f(x ,y )(x−x ).
0 0 0 0 0 0 0 0
For the next point x1 = x0 + h, we can compute the y-value of the approximate solution by
y =y +f(x ,y )(x −x )=y +f(x ,y )h.
1 0 0 0 1 0 0 0 0
Havingobtainedthepoint(x ,y ), we note that the slope of the solution is given by y′ = f(x ,y ).
1 1 1 1
So, the tangent line to the solution curve at (x ,y ) is
1 1
y −y =f(x ,y )(x−x )
1 1 1 1
For the point x2 = x1 + h, the y-value computed using the tangent line is
y =y +f(x ,y )(x −x )=y +f(x ,y )h.
2 1 1 1 2 1 1 1 1
Continuing in this way, we obtain a sequence of (x,y) values
x =x +h
n+1 n
y =y +f(x ,y )h
n+1 n n n
The accuracy of Euler’s method can be increased by decreasing the step size h.
Example 4. Find the first three approximations of y′ = y +1, y(0) = 1 with step size 0.1.
Solutions. We have that
y =y +(y +1)0.1=1+(1+1)0.1=1.2
1 0 0
y =y +(y +1)0.1=1.2+(1.2+1)0.1=1.42
2 1 1
y =y +(y +1)0.1=1.42+(1.42+1)0.1=1.662
3 2 2
Continuing in this way, we can approximate the value of solution at x = 1 to be y =4.187.
10
The calculator program below calculates the values of the numerical solution of the differential
equation y′ = f(x,y) using Euler’s method. The program asks for f(x,y), initial x, final x, step size
h and initial y.
After pressing ENTERforthefirsttime, valuesx1 andy1 aredisplayed. WhenpressingENTER
one more time, values x and y are displayed. So, keep pressing ENTER until you display x and
2 2 n
y . All the used (x,y) values are also stored in lists L and L so that you can display them if
n 1 2
necessary.
4
no reviews yet
Please Login to review.
