Difference between revisions of "Solutions of Differential Equations"

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* <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
 
* <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
 
* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} = C</math>, so the particular solution is <math> y = 3e^{x} </math>.
 
* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} = C</math>, so the particular solution is <math> y = 3e^{x} </math>.
* <math> y'' + y' - 2y = 0 </math>. The G.S. is <math> y = Ce^{x} + De^{-2x} </math>. <math> y' = Ce^{x} - 2De^{-2x} </math> and <math> y'' = Ce^{x} + 4De^{-2x} </math>, so <math> 0 = y'' + y' - 2y </math> becomes <math> 0 = Ce^{x} + 4De^{-2x} + Ce^{x} - 2De^{-2x} - 2(Ce^{x} + De^{-2x}) = Ce^{x} + Ce^{x} - 2Ce^{x} + 4De^{-2x} - 2De^{-2x} - 2De^{-2x}) = 0 </math>.
+
* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.
 
 
  
 
==Resources==
 
==Resources==
 
* [http://www.maths.gla.ac.uk/~cc/2x/2005_2xnotes/2x_chap5.pdf Differential Equations], University of Glascow
 
* [http://www.maths.gla.ac.uk/~cc/2x/2005_2xnotes/2x_chap5.pdf Differential Equations], University of Glascow
 
* [https://www.api.simply.science/index.php/math/calculus/differential-equations/types-of-differential-equations/9869-general-and-particular-solutions-of-a-differential-equation General and Particular Solutions], Simply Math
 
* [https://www.api.simply.science/index.php/math/calculus/differential-equations/types-of-differential-equations/9869-general-and-particular-solutions-of-a-differential-equation General and Particular Solutions], Simply Math

Revision as of 19:53, 17 September 2021

A solution of a differential equation is an expression of the dependent variable that satisfies the relation established in the differential equation. For example, the solution of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' + y - 2 = 4x } will be some equation y = f(x) such that y and its first derivative, y', satisfy the relation . The general solution of a differential equation will have one or more arbitrary constants, depending on the order of the original differential equation (the solution of a first order diff. eq. will have one arbitrary constant, a second order one will have two, etc.).

Examples:

  • . Through simple integration, we can calculate the general solution of this equation to be , where C is an arbitrary constant.
  • . The G.S. is . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' = Ce^{x} } , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' - y = 0 \to Ce^{x} - Ce^{x} = 0} , so this solution satisfies the relationship for all arbitrary constants C.
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' + y' - 2y = 0 } . The G.S. is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = Ce^{x} + De^{-2x} } . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' = Ce^{x} - 2De^{-2x} } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' = Ce^{x} + 4De^{-2x} } , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 = y'' + y' - 2y } becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 = Ce^{x} + 4De^{-2x} + Ce^{x} - 2De^{-2x} - 2(Ce^{x} + De^{-2x}) = Ce^{x} + Ce^{x} - 2Ce^{x} + 4De^{-2x} - 2De^{-2x} - 2De^{-2x}) = 0 } .

The particular solution of a differential equation can be solved if we have enough points to solve for the arbitrary constants.

Examples:

  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' = 2x } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(2) = 0 } . With this point and the general solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2 + C} , we can calculate the constant C to be -4. Thus the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2 - 4} .
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' - y = 0 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 3 } . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3 = Ce^{0} = C} , so the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = 3e^{x} } .
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' + y' - 2y = 0 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 2 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(0) = -1 } . So, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 = Ce^{0} + De^{0} = C + D } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1 = Ce^{0} - 2De^{0} = C - 2D} . Thus C = 1 and D = 1, and the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = e^{x} + e^{-2x} } .

Resources