Separable Differential Equations MATH 6 Calculus I J. Robert Buchanan Department of Mathematics Fall 207
Background We have previously solved differential equations of the forms: y (t) = k y(t) (exponential growth/decay) y (t) = k(y(t) T a ) (Newton s Law of Cooling) These are examples of first-order separable ordinary differential equations.
First-order Separable ODEs Every first-order ordinary differential equation can be written in the form: y = f (x, y) A first-order ordinary differential equation is separable if the equation above can be re-written in the form: g(y)y = h(x) Note: the expressions involving y and y are on one side of the equation and the expressions involving x are on the other.
Examples Determine which of the following first-order ordinary differential equations are separable.. y = 2xy(cos y ) 2. y = x 2 (x y) 3. y = 2x sin y x 3 y 4. y = y 2 (x 2 3x + 2)
Solution. y = 2xy(cos y ) 2. y = x 2 (x y) y(cos y ) y = 2x (separable) x y y = x 2 (not separable) 3. y = 2x sin y x 3 y 2 sin y x 2 y y = x (not separable) 4. y = y 2 (x 2 3x + 2) y 2 y = x 2 3x + 2 (separable)
Solving a Separable First-order ODE We may solving a separable first-order ordinary differential equation by. separating the variables x and y, 2. integrating both sides of the equation, and 3. solving for y (if possible). g(y)y = h(x) g(y) dy dx = h(x) g(y) dy = h(x) dx g(y) dy = h(x) dx
Example Solve the following separable first-order ordinary differential equation: y = (y ) cos x Assume y >.
Solution y = (y ) cos x dy y dx = cos x dy = cos x dx y y dy = cos x dx ln(y ) = sin x + K (since y >, K is constant) e ln(y ) = sin x+k e y = e sin x e K y(x) = + C e sin x (where C = e K )
Family of Solutions y(x) = + C e sin x y x c=0. c=0.2 c=0.3 c=0.4 c=0.5 c=0.6 c=0.7 c=0.8 c=0.9 c=.
Example Solve the following separable first-order ordinary differential equation: y y 2 = x ln x Assume y.
Solution y = dy dx = y 2 x ln x y 2 x ln x dy = y 2 x ln x dx dy = y 2 y 2 dy = u du x ln x dx (substitute u = ln x, du = x dx) sin y = ln u + C = ln ln x + C y(x) = sin(c + ln ln x )
Family of Solutions y(x) = sin(c + ln ln x ) y.0 0.5 0.0-0.5 -.0 2 3 4 5 x c=-0.5 c=-0.4 c=-0.3 c=-0.2 c=-0. c=0. c=0. c=0.2 c=0.3 c=0.4 c=0.5
Solving an Initial Value Problem If we have an initial condition, then we may solve for the arbitrary constant C in the solution to the ordinary differential equation.
Solving an Initial Value Problem If we have an initial condition, then we may solve for the arbitrary constant C in the solution to the ordinary differential equation. Example Solve the following initial value problem. y = x y 2 y(0) =
Solution ( of 2) y = x y 2 dy dx = x y 2 y 2 dy = (x ) dx y 2 dy = (x ) dx 3 y 3 = 2 x 2 x + C y 3 = 3 2 x 2 3x + 3C y(x) = 3 3 2 x 2 3x + 3C (general solution) If y(0) = then = 3 3 2 (0)2 3(0) + 3C which implies C = /3.
Solution (2 of 2) y(x) = 3 3 2 x 2 3x + 5 4 3 2 y 0 - -2-4 -2 0 2 4 x
Logistic Equation Our previous model for exponential growth has a significant flaw. y (t) = k y(t) y(0) = A lim y(t) = lim A t t ek t = when k > 0.
Logistic Equation Our previous model for exponential growth has a significant flaw. y (t) = k y(t) y(0) = A lim y(t) = lim A t t ek t = when k > 0. We can introduce a slightly more complicated model called the logistic equation to bound the growth of the population. y (t) = k y(t)(m y(t)) y(0) = A The constant M is called the carrying capacity.
Equilibrium Solutions Consider the logistic equation with the special initial condition: y (t) = k y(t)(m y(t)) y(0) = 0. We can verify that y(t) = 0 is a solution to this initial value problem.
Equilibrium Solutions Consider the logistic equation with the special initial condition: y (t) = k y(t)(m y(t)) y(0) = 0. We can verify that y(t) = 0 is a solution to this initial value problem. Likewise the constant function y(t) = M satisfies the initial value problem: y (t) = k y(t)(m y(t)) y(0) = M.
Equilibrium Solutions Consider the logistic equation with the special initial condition: y (t) = k y(t)(m y(t)) y(0) = 0. We can verify that y(t) = 0 is a solution to this initial value problem. Likewise the constant function y(t) = M satisfies the initial value problem: y (t) = k y(t)(m y(t)) y(0) = M. These are called equilibrium solutions.
Non-equilibrium Solution ( of 3) Consider the logistic equation with initial condition: y = k y(m y) y(0) = A. dy = k dt y(m y) ( /M + /M ) dy = k dt y M y ( M y + ) dy = k dt M y ( M y + ) dy = k dt M y (ln y ln M y ) = k t + C M
Non-equilibrium Solution (2 of 3) So far we have M (ln y ln M y ) = k t + C ln y M y = km t + CM y M y = e km t+cm = e km t e CM
Non-equilibrium Solution (2 of 3) So far we have M (ln y ln M y ) = k t + C ln y M y = km t + CM y M y = e km t+cm = e km t e CM When t = 0, y = A and thus we have A M A = ecm. Now we may solve for y(t).
Non-equilibrium Solution (3 of 3) Case: 0 < A < M Case: M < A y M y = A M A y M y ekm t = y = ( y + A ) M A ekm t = y M y y(t) = y(t) = A M A ekm t A M A ekm t (M y) AM M A ekm t AMe km t M A + Ae km t = A M A ekm t AMe km t M A + Ae km t
Family of Solutions M = 500, k = /0, y(t) = 500Ae 50t 500 A + Ae 50t. 600 y 400 200 0 0.00 0.02 0.04 0.06 0.08 0.0 t
Example Solve the following case of the logistic equation. y = y(5 y) y(0) = 2 y(t) = AMe km t M A + Ae km t
Example Solve the following case of the logistic equation. y = y(5 y) y(0) = 2 y(t) = y(t) = = AMe km t M A + Ae km t (2)(5)e ()(5) t 5 2 + 2e ()(5) t 0e 5t 3 + 2e 5t
Homework Read Section 7.2 Exercises: 33 odd