community project mathcentre community project First Order Ordinary All mccp resourcesdifferential are released under a Creative Equation Commons licence Introduction mccp-richard- Prerequisites: You will need to know about trigonometry, dierentiation, integration, complex numbers in order to make the most of this teach-yourself resource. We are looking at equations involving a function y(x) and its rst derivative: + P(x)y = Q(x) () We want to nd y(x), either explicitly if possible, or otherwise implicitly. Direct Integration These ODEs can be solved as follows: = f(x) = f(x) so: y = f(x) = 3x2 6x + 5 y = x 3 3x 2 + 5x + C with C constant of integration The constant of integration can be anything, unless you have boundary conditions. In the previous example, if you have y(0)=0 then: y(0) = C = 0
Separation of Variables = f(x)g(y) mathcentre community project These ODEs can be solved as follows: All mccp resources are released under a Creative Commons licence = f(x) g(y) = f(x) g(y) = 2x y + (y + ) = 2x (y + ) = 2x 2 y2 + y = x 2 + C with C constant of integration Homogeneous Equations A rst order homogeneous dierential equation is a dierential equation in the form: = f(y x ) These ODEs can be solved by making the substitution y(x) = v(x) x where v is a function of x. Then we have: Using the product rule: Inserting in the ODE: Re-arranging: = x + v x + v = f(v) f(v) v = x f(v) v = x
Example Rearranging, we get: = x + 3y 2x y mathcentre = 2 community + 3 2 x x project + v = 2 + 3 2 v All mccp resources x are released under a Creative Commons licence = 2 + 2 v + v = 2x Taking the log on both sides: v = e C x /2 with y = v x, ln( + v) = 2 ln(x) + C = ln(x/2 ) + C withc constant of integration y = kx 3/2 x, with k = e C Integrating Factor + P(x)y = Q(x) These ODEs can be solved as follows: multiply both sides of the equation by the integrating factor: e P(x). Then: e P(x) + P(x)e P(x) y = Q(x)e P(x) Now we see that, using the product rule and the chain rule: e P(x) + P(x)e P(x) y = d ( e ) P(x) y Therefore: d ( e ) P(x) y = Q(x)e P(x) y = e P(x) Q(x)e P(x) Example The integrating factor is e = e x and: y = x with P(x) = and Q(x) = x x e e x y = xe x d ( ) ye x = xe x Using integration by part: ye x = e x ( + x) + C with C constant of integration y = ( + x) + Ce x
Bernouilli Equations Bernouilli equations are of the form: + P(x)y = Q(x)yn mathcentre These equations are solved with the following community method: project All mccp resources are released under a Creative Commons licence Multiply both sides of the equation by y n, the equation becomes: Make the change of variable v = y n, then: n y + P(x)y n = Q(x) so + P(x)v = Q(x) n = ( n)y n The latest form of the equation can be solved for v using the method of the integrating factor. Finally, y can be found from the relationship y = v n. y 2 + x y = xy2 + x y = x v = y, v x = x = y 2 The integrating factor is IF = e x = e ln x = x x x v = 2 x v = v = x 2 + Cx y = x 2 + Cx
Exercises (a) x = 5x3 + 4 (b) x(y 3) (d) x y = x3 + 3x 2 2x (e) ( + x2 ) mathcentre = 4y + 3xy = 5x with y() = 2 (c) (2y x) community project = 2x + y with y(2) = 3 (f) 2 Answers encouraging academics to share + y = y3 (x ) maths support resources All mccp resources are released under a Creative Commons licence (a) y = 5 3 x3 + 4 ln x + C (d) y = 2 x3 + 3x 2 2x ln x + Cx (b) y 3 ln y = 4 ln x + C (e) y = 5 3 + 8 3 ( + x2 ) 3/2 (c) ( y x )2 y x + x 2 = 0 (f) y = (x + Ce x ) /2