1st Order Linear D.E.

Transcription

1 1st Order Linear D.E. y + p(t)y = g(t) y(t) = c 1 y 1 where y 1 is obtained from using the integrating factor method. Derivation: Define the integrating factor as: µ = e p(t) Observe that taking the derivative with respect to t yields: µ = d ( p(t))e p(t) = p(t)e p(t) = p(t)µ Distributing the integrating factor through the differential equation reveals an internal product rule on the left hand side of the equation. µ(y + p(t)y) = µg(t) µy + (µp(t))y = µy + µ y = µg(t) (yµ) = µg(t) Integrating the sides with respect to t and dividing by the integrating factor solves the differential equation (yµ) = µg(t) Solution Steps: yµ = y = 1 µ µg(t) µg(t) y + p(t)y = g(t) µ = e p(t) y = 1 µg(t) µ 1

2 Exact Equations M(x, y) + N(x, y) dy dx = 0 the method of exact equations can be used to solve for y(x) Testing Exactness: If the function is exact in its current state, then M y = N x and dµ dx = 0 otherwise an integrating factor µ 0 is required. This integrating factor can take two forms depending on assumption that it is purely a function of x or y: which yields Solution Method: dµ dx = M y N x µ = M y N x N M µ µm y = µn x µ(m(x, y) + N(x, y) dy ) = µm(x, y)dx + µn(x, y)dy = 0 dx Ψ(x, y) = (µm(x, y)dx + µn(x, y)dy) = 0dx = c Ψ(x, y) = µm(x, y)dx + µn(x, y)dy = c Removing Redundant Terms: The resulting solution equation Ψ must be such that, omitting constants C, Ψ(x, y) = µm(x, y)dx = µn(x, y)dy which allows for the reduction of redundant terms.

3 Tank Mixture Problems Given a tank problem drawn below where x i (t) is the amount of substance mixed with water, V i is the volume of the tank, the solution is as follows: Flow Diagram.png General Differential Equation: For the tank system, a differential equation for each tank is written as the rate of mixture in minus the rate of mixture out. Note the concentrations (γ) and flow rate ( V ) of each stream leaving and entering the tank. d(x i ) = Σ(γ V ) in Σ(γ V ) out Note that fluid leaving a well stirred tank has concentration γ well stirred = x i(t) V i Solution Method: Summing all the concentrations and flow rates in and out of tank 1 and two yields the following system of equations: d(x 1 ) = γ in Vin + γ 1 V1 γ out,1 Vout,1 γ 1 V1 d(x ) = γ 1 V1 γ 1 V1 γ out, Vout, Replacing the concentration terms with their well-stirred terms: d(x 1 ) = γ in Vin + x V V1 x 1 Vout,1 x 1 V1 = γ in Vin + x V1 V x 1 ( V out,1 + ) d(x ) = x 1 V1 x V V1 x V Vout, = x 1 V1 x ( V out, + V ) Note: If the tank system has only one tank, only the first equation is required with = = 0 3

4 Tank Mixture Problems (Cont.) Elimination Method: Similar to how linear systems of equations were solved in College Algebra, adding and subtracting the differential equations here will be used to reduce the equation to a function of only one variable. First each eqation is set equal to 0: d(x 1 ) V1 = γ in Vin + x x 1 ( V out,1 + ) d(x 1) V1 γ in Vin x + x 1 ( V out,1 + ) = 0 V V d(x ) V1 = x 1 x ( V out, + ) d(x ) V1 x 1 + x ( V out, + ) = 0 V V Let D = d such that D[x i] = x i and factor out the x 1, x terms as follows: (D + ( V out,1 + V1 ))[x 1 ] γ in Vin x = 0 V (D + ( V out, + V1 ))[x ] x 1 = 0 V Multiplying equation one by and equation two by (D + ( V out,1 + )) allows us to cancel the x 1 terms through addition of the two equation and results in the differential equation: (D + ( V out,1 + ))(D + ( V out, + V ))[x ] (γ in Vin + x V1 V ) = 0 (D +( V out,1 + + V out, + V )D+ ( V out,1 + )( V out, + ) V )[x ] (γ in Vin +x V1 V ) = 0 d (x ) +( V out,1 + + V out, + ) d(x ) V +( ( V out,1 + )( V out, + ) V1 V )x = γ in Vin x can be solved by use of the following methods of solving second order non-homogeneous differential equations. Further, x 1 can be solved for by another elimination method or by substituting x, x back into the above differential equation 4

5 Homogeneous Equations d y + p(t)dy + q(t)y = 0 y(t) = c 1 y 1 + c y where y 1 and y are obtained by solving the characteristic equation of the differential equation. Characteristic Equation: If p(t) and q(t) are constants, assume a solution of y = e λt with first and second derivatives λe λt and λ e λt respectively. Plugging this into the differential equation yields: e λt (λ + p(t)λ + q(t)) = 0 which is solved through the quadratic formula to yield homogeneous solutions y 1 and y. λ 1, = p(t) ± p(t) 4q(t) y 1 = e λ 1t, y = e λ t Repeated Roots: If λ 1 = λ then the repeated homogeneous solution is multiplied by t such that y 1 = e λt, y = te λt Complex Roots: If p(t) 4q(t) < 0 such that the roots of the characteristic equation are complex, they take the form: λ 1, = p(t) 4q(t) p(t) ± i = η ± ωi y 1 = e (η+ωi)t, y = e (η ωi)t by the Euler identity e iα = cos(α) + isin(α) and rearranging of the variables, the solution becomes y 1 = e ηt cos(ωt), y = e ηt sin(ωt) 5

6 Nonhomogeneous Equations - Method of Undetermined Coefficients d y + p(t)dy + q(t)y = g(t) y(t) = c 1 y 1 + c y + y p where y 1 and y are obtained by solving the differential equation as a homogeneous equation and method of undetermined coefficients is utilized to obtain y p. For the method of undetermined coefficients, y p is defined based off of g(t) and the solutions to the homogeneous case. Choosing y p s Form from g(t): -If g(t) has a term t n, then y p should be assumed to contain a polynomial of degree n such that y p = (A n t n + A n 1 t n A 0 t 0 ) h(t) where h(t) is any other functions from this section. -If g(t) has a term sin(ωt) or cos(ωt), then y p should be assumed as y p = (A 0 cos(ωt) + A 1 sin(ωt) h(t) where h(t) is any other functions from this section. -If g(t) has a term e αt, then y p should be assumed as y p = A 0 e αt h(t) where h(t) is any other functions from this section. Choosing y p s Form from Homogeneous Solutions: -If the proposed solution created from the above g(t) rules has a term that matches the form of any solutions to the homogeneous equation, the anticipated solution should be multiplied by t. Solution of Coefficients: Take all required derivatives of the proposed solution and plug in to the differential equation: and solve for all constants in y p. y p + p(t)y p + q(t)y p = g(t) 6

7 Nonhomogeneous Equations - Variation of Parameters d y + p(t)dy + q(t)y = g(t) y(t) = c 1 y 1 + c y + y p where y 1 and y are obtained by solving the differential equation as a homogeneous equation and variation of parameters is utilized to obtain y p. Particular Solution: For variation of parameters, y p is defined: g(t)y y p = ν 1 y 1 + ν y, ν 1 = W, and ν g(t)y1 = W with W in this case being the Wronskian. Note that ν 1 includes a negative sign but ν does not. The Wronskian can be calculated by the following formula: [ ] y1 y W = det( y 1 y = y 1 y y y 1 Note that the Wronskian cannot be zero as used in the integration above. Reduction of Order: If y 1 (t) is not identically 0, then the homogeneous solutions to the differential equation are related by e p(t) y = y 1 y 1 7

8 Laplace Transforms Given the differential equation and initial conditions, d y + p(t)dy + q(t)y = g(t), y(0) = k 1, y (0) = k y(t) = c 1 y 1 + c y + y p where y 1, y, and y p are obtained by transforming the differential into the Laplace domain and returning the solution to the time domain. Properties of the Laplace Transform: -The Laplace Transform is a linear operator which means it can be distributed over terms being added or subtracted. I.e. L[A(t) + B(t)] = L[A(t)] + L[B(t)] -Scalar multiples of functions undergoing Laplace Tranforms are as follows: L[cA(t)] = cl[a(t)] Laplace Transform of y (n) (t): The Laplace Transform of any derivative of the solution function yields, L[y (n) (t)] = s n Y (s) s n 1 y(0) s n y (0)... y n 1 (0) which, for second order differential equations, means L[y(t)] = Y (s), L[y (t)] = sy (s) y(0), L[y (t)] = s Y (s) sy(0) y (0) Second Order D.E. Solution: (Assuming p(t) and q(t) are constants where p(t)=p and q(t)=q) Y (s) = L[g(t)] + (s + p)y(0) + y (0) s + ps + q y(t) = L 1 [Y (s)] = L 1 [ L[g(t)] + (s + p)y(0) + y (0) ] s + ps + q Laplace Transform Table: A table of common Laplace Transforms and their inverses is located in the textbook. 8

9 Laplace Transforms (Cont.) Unit Step Functions and Impulse Functions: Unit step functions and impulse functions are piece-wise inputs to a system modelled in a differential equation. They are defined: { 0 t < c u(t c) = u c = 1 c t δ(t c) = δ c = { 0 t c 1 t = c Properties of Laplace Transform Revisited: Given a Laplace Transform of a function multiplied by a unit step function or impulse function, the transform has the property: L[u c f(t c)] = e cs L[f(t)] = e cs [F (s)] L[kδ c ] = ke cs Note that an impulse does not have a function multiplied by it, but can have a magnitude Inverse Laplace Transform of Unit and Impulse Functions: L 1 [e cs f(s)] = u c L 1 [F (s)] = u c f(t c) 9