# Solution 197 403 198 403 family of solutions x y c 9

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Solution: 197 / 403 198 / 403 Family of solutions x y c = 9 c = 4 c = 1 c = 0 c = 9 c = 4 c = 1 c = 0 y = arcsin x + c y =- arcsin x + c x =- 1 x = 1 * Solution 199 / 403 Linear First Order ODEs Example 5.7: Solve x dy dx + y = e x . Solution: 200 / 403
A linear first order ODE has the form: dy dx + P ( x ) y = Q ( x ) To solve: Multiply ODE by I ( x ) I ( x ) dy dx + P ( x ) I ( x ) y = Q ( x ) I ( x ) If the left side can be written as the derivative of y ( x ) I ( x ) , then d dx y ( x ) I ( x ) = Q ( x ) I ( x ) which can be solved by integrating with respect to x . 201 / 403 Aim: Find an integrating factor I so the left side will be the derivative of y I . Then d dx ( y I ) ≡ I dy dx + PI y dy dx I + y d I dx = I dy dx + PI y y d I dx = PI y To solve for all y d I dx = PI (separable) 202 / 403 1 I d I dx = P Z 1 I d I = Z P dx log |I| = Z P dx + c ⇒ |I| = e R P dx + c = e R P dx · e c ⇒ I = ± e c |{z} constant · e R P dx 203 / 403 So one integrating factor is I ( x ) = e R P dx Note: Since we only need one integrating factor I , we can neglect the ‘ + c ’ and modulus signs when calculating I . 204 / 403
Example 5.8: Find the general solution of dy dx + y x = sin x ( x , 0) . Solution: 205 / 403 206 / 403 207 / 403 Family of solutions x y c = 0 c = 1 c =- 2 c =- 2 c = 1 - 2 2 - 2 2 y ( x ) = - cos x + 1 x sin x + c x 208 / 403
Example 5.9: Solve 1 2 dy dx - xy = x if y (0) = - 3 . Solution: 209 / 403 210 / 403 Note: 211 / 403 Other First Order ODEs Sometimes it is possible to make a substitution to reduce a general first order ODE to a separable or linear ODE. A homogeneous type ODE has the form dy dx = f y x Substituting u = y x reduces the ODE to a separable ODE. Bernoulli’s equation has the form dy dx + P ( x ) y = Q ( x ) y n Substituting u = y 1 - n reduces the ODE to a linear ODE. 212 / 403
Example 5.10: Solve the homogeneous type differential equation dy dx = y x + cos 2 y x - π 2 < y x < π 2 by substituting u = y x . Solution: 213 / 403 214 / 403 215 / 403 Example 5.11: Solve the Bernoulli equation dy dx + y = e 3 x y 4 ( y , 0) by substituting u = y - 3 . Solution: 216 / 403
217 / 403 218 / 403 219 / 403 Population Models Malthus (Doomsday) model Rate of growth is proportional to the population p at time t . dp dt p dp dt = kp (separable/linear) where k is a constant of proportionality representing net births per unit population per unit time. If the initial population is p (0) = p 0 , then the solution is p ( t ) = p 0 e kt You should check that you can derive this on your own! 220 / 403
Note: The Doomsday model is unrealistic since if k > 0 - unbounded exponential growth k < 0 - population dies out k = 0 - population stays constant 221 / 403 Equilibrium Solutions Definition An equilibrium is a constant solution of an ODE. Note: For the ODE dx dt = f ( x , t ) , this means I x ( t ) = C where C is a constant I dx dt = 0 Terminology The plural form of equilibrium is equilibria . 222 / 403 Example 5.12: Find all equilibrium solutions of the ODE dx dt = 3 x - 2 . Solution: 223 / 403 Phase plots A phase plot is a plot of dx dt as a function of x . A phase plot will give I the equilibria I the behaviour of solutions close to the equilibria Note: Phase plots are only useful for ODEs of the form dx dt = f ( x ) i.e., when the right-hand side has no explicit dependence on t .
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