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Euler’s Method
Section 6.6
Suppose we are given a differential equation and initial condition:
,dyf x y
dx 0 0y x y
Then we can approximate the solution to the differential equationby its linearization (which is “close enough” in a short intervalabout x ).0 0 0 0L x y x y x x x
0 0 0 0,y x f x y x x
y y x
0x
(solution curve)
L x
0y 0 0,x y
The basis of Euler’smethod is to “string
together” manylinearizations to
approximate a curve.
y y x
0x
0y 0 0,x y
1 0x x dx
Now, let’s specify a new valuefor the independent variable:
1 1 0 0 0,y L x y f x y dx If dx is small, then we have a new linearization:
1 1,x y x
1 1,x L x
1 0x x dx dx
0 0,x yFrom the point , which lies exactly on the solution curve,we have obtained the point , which lies very close to thepoint on the solution curve.
1 1,x y 1 1,x y x
1 1,f x ySecond Step: We use the point and the slopeof the solution curve through .
1 1,x y 1 1,x y
2 1 1 1,y y f x y dx
2 1x x dx 1 1,x y
Setting , we use the linearization of the solutioncurve through to calculate
2 2,x yThis gives us the next approximation to values along thesolution curve . y y x
Continue the pattern to find the third approximation:
3 2 2 2,y y f x y dx
Let’s see this process graphically…
Three steps in the Euler approximation to the solution of theinitial value problem ,y f x y 0 0y x y,
True solution curve
y y x
Euler approximation
3 3,x y 2 2,x y
1 1,x y
0 0,x y
Error
dx dx dx0x 1x 2x 3x
Using Euler’s MethodFind the first three approximations using Euler’smethod for the initial value problem
1 2 3, ,y y y
0 1y 1y y starting at with 0.1dx 0 0x
We have:
1 0 0.1x x dx 0 1y 0 0x
2 0 2 0.2x x dx
3 0 3 0.3x x dx
Using Euler’s MethodFind the first three approximations using Euler’smethod for the initial value problem
1 2 3, ,y y y
0 1y 1y y starting at with 0.1dx 0 0x
First Approximation
1 0 0 0,y y f x y dx
0 01y y dx
1 1 1 0.1 1.2
Using Euler’s MethodFind the first three approximations using Euler’smethod for the initial value problem
1 2 3, ,y y y
0 1y 1y y starting at with 0.1dx 0 0x
Second Approximation
2 1 1 1,y y f x y dx
1 11y y dx
1.2 1 1.2 0.1 1.42
Using Euler’s MethodFind the first three approximations using Euler’smethod for the initial value problem
1 2 3, ,y y y
0 1y 1y y starting at with 0.1dx 0 0x
Third Approximation
3 2 2 2,y y f x y dx
2 21y y dx
1.42 1 1.42 0.1 1.662
Using Euler’s MethodUse Euler’s method to solve the given initial value problem onthe interval starting at with .Compare the approximations to the values of the exactsolution.
0 2y y x y
0 1x 0 0x 0.1dx
Let me show you a new calculator program!!!
x y (Euler) y (exact) Error
0 2y y x y 1 xf x x e
0 –2 –2 0
0.1 –1.8000 –1.8048 0.0048
0.2 –1.6100 –1.6187 0.0087
0.3 –1.4290 –1.4408 0.0118
0.4 –1.2561 –1.2703 0.0142
0.5 –1.0905 –1.1065 0.0160
0.6 –0.9314 –0.9488 0.0174
0.7 –0.7783 –0.7966 0.0183
0.8 –0.6305 –0.6493 0.0189
0.9 –0.4874 –0.5066 0.0191
1.0 –0.3487 –0.3679 0.0192
Other Practice ProblemsUse differentiation and substitution to show that the givenfunction is the exact solution of the given initial value problem.
0 2y y x y 1 xf x x e
y x y x f x 1 xx x e
1 xe
1 xd dy f x x e
dx dx 1 xe
Initial Condition: 00 0 1f e 2
The same!
The same!
Other Practice ProblemsUse analytic methods to find the exact solution of the given initialvalue problem.
1 ,dy
x ydx
2 0y
1
dyxdx
y
21
ln 12
y x C
2 21
x Cy e
2 21
x Cy e
2 2 1C xy e e 2 2 1xy Ae
Other Practice ProblemsUse analytic methods to find the exact solution of the given initialvalue problem.
1 ,dy
x ydx
2 0y
2 2 1xy Ae Initial Condition:
22 20 1Ae 20 1Ae
2A e
Solution:22 2 1xy e e
or 2 2 21
xy e
Other Practice ProblemsUse analytic methods to find the exact solution of the given initialvalue problem.
2 1 2 ,dy
y xdx
1 1y
2 1 2y dy x dx 1 2y x x C
2
1y
x x C
Initial Condition: Solution:
2
11
1 1 C
1C
2
1
1y
x x
