Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Differential Equations
Understanding Euler Method: Finding Initial Condition of y(0)=1
Reply to thread
Message
[QUOTE="chwala, post: 6854827, member: 287397"] Consider the first -order differential equation that is given in example 3.1.1; ##\dfrac{dy}{dx}+2y=x^3e^{-2x}## The equation of the curve will be given by as follows; ... integrating factor= ##e^{2x}## therefore ... ##ye^{2x} = \int x^3 dx## ##ye^{2x} = \dfrac{x^4}{4} +k## ##y=\dfrac{x^4}{4e^{2x}}+ \dfrac{k}{e^{2x}}## Using ##y(0)=1## we shall get, ##1=0+k, ⇒k=1## thus, ##y=\dfrac{x^4e^{-2x}}{4}+ e^{-2x}## I can see from the graph that the point ##(0,1)## lies on the graph... For the exact values i have; ##y_1=0.8188, y_2=0.6706, y_3=0.5499## and so on... these values are quite close to the ones that were found by Euler's method. Cheers guys! [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Differential Equations
Understanding Euler Method: Finding Initial Condition of y(0)=1
Back
Top