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Outline Week 12: Application of vector differential equations to electrical networks; vector differential equations Course Notes: 6.3, 6.4 Goals: be able to solve a linear system of differential equations; find characteristics of electiral networks involving inductors and capacitors using methods learned this term. Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), Solutions: y (t) = Ce λt , constant C λ constant Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Solutions: y (t) = Ce λt , constant C Example: a population’s growth is 0.3 times the number of individuals in that population per year. Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Solutions: y (t) = Ce λt , constant C Example: a population’s growth is 0.3 times the number of individuals in that population per year. y 0 (t) = 0.3y (t) Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Solutions: y (t) = Ce λt , constant C Example: a population’s growth is 0.3 times the number of individuals in that population per year. y 0 (t) = 0.3y (t) y (t) = Ce 0.3t for some constant C Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Solutions: y (t) = Ce λt , constant C Example: a population’s growth is 0.3 times the number of individuals in that population per year. y 0 (t) = 0.3y (t) y (t) = Ce 0.3t for some constant C At year t = 0, there are 100 individuals. Differential Equations We’re going to doing this in a linear-systems context soon. y 0 (t) = λy (t), λ constant Solutions: y (t) = Ce λt , constant C Example: a population’s growth is 0.3 times the number of individuals in that population per year. y 0 (t) = 0.3y (t) y (t) = Ce 0.3t for some constant C At year t = 0, there are 100 individuals. y (t) = 100e 0.3t Differential Equations Example: a radioactive substance decays at a rate of 2% of its mass every year. Differential Equations Example: a radioactive substance decays at a rate of 2% of its mass every year. y 0 (t) = −0.02y (t) Differential Equations Example: a radioactive substance decays at a rate of 2% of its mass every year. y 0 (t) = −0.02y (t) y (t) = Ce −0.02t where C is the amount at t = 0 Systems of Linear Differential Equations y10 (t) = ay1 (t) + by2 (t) y20 (t) = cy1 (t) + dy2 (t) Systems of Linear Differential Equations y10 (t) = ay1 (t) + by2 (t) y20 (t) = cy1 (t) + dy2 (t) 0 y (t) y := 10 y2 (t) 0 a b A= c d y0 = Ay y (t) y := 1 y2 (t) Guessing Solutions: Eigenvectors Differential Equation: y0 = Ay Let’s take a guess from our previous examples: what if y = e λt x for some constant λ and some vector x? Guessing Solutions: Eigenvectors Differential Equation: y0 = Ay Let’s take a guess from our previous examples: what if y = e λt x for some constant λ and some vector x? Then: So, if y0 = Ay: Hence: y0 = λe λt x λe λt x = A(e λt x) λx = Ax so λ and x are an eigenvalue/eigenvector pair of A. Guessing Solutions: Eigenvectors Differential Equation: y0 = Ay Let’s take a guess from our previous examples: what if y = e λt x for some constant λ and some vector x? y0 = λe λt x Then: λe λt x = A(e λt x) So, if y0 = Ay: Hence: λx = Ax so λ and x are an eigenvalue/eigenvector pair of A. We’ve successfully guessed a solution! y = e λt x where λ, x are an eigenvalue/eigenvector pair Systems of Linear Differential Equations: Adding Solutions Adding Solutions Suppose y1 and y2 are both solutions to the system of differential equations Ay = y0 . Systems of Linear Differential Equations: Adding Solutions Adding Solutions Suppose y1 and y2 are both solutions to the system of differential equations Ay = y0 . Then (y1 + y2 ) is also a solution. Systems of Linear Differential Equations: Adding Solutions Adding Solutions Suppose y1 and y2 are both solutions to the system of differential equations Ay = y0 . Then (y1 + y2 ) is also a solution. Further, (c1 y1 + c2 y2 ) is also a solution, for any constants c1 and c2 . (Home exercise: prove this is true!) Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. Example: y0 = Iy, y ∈ R2 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. 2 Example: y0 = Iy, y∈R 1 λ1 = 1, x1 = ; 0 0 λ2 = 1, x2 = ; 1 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. 2 Example: y0 = Iy, y∈R 1 λ1 = 1, x1 = ; 0 0 λ2 = 1, x2 = ; 1 1 0 t t y(t) = c1 e + c2 e . 0 1 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. General Question: Is there a solution to y0 = Ay that also has y(0) = y0 , for some constant vector y0 ? Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. General Question: Is there a solution to y0 = Ay that also has y(0) = y0 , for some constant vector y0 ? Suppose it has the form above: y(0) = c1 x1 + c2 x2 + · · · + ck xk ? =? y0 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. General Question: Is there a solution to y0 = Ay that also has y(0) = y0 , for some constant vector y0 ? Suppose it has the form above: y(0) = c1 x1 + c2 x2 + · · · + ck xk ? =? y0 If the eigenvalues of A form a basis then there is always exactly one solution to y0 = Ay with any desired initial condition y(0) = y0 , and it has the form given above. Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. 2 Example: y0 = Iy, y∈R 1 λ1 = 1, x1 = ; 0 0 λ2 = 1, x2 = ; 1 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. 2 Example: y0 = Iy, y∈R 1 λ1 = 1, x1 = ; 0 0 λ2 = 1, x2 = ; 1 7 Suppose we have initial conditions y(0) = . −3 Solutions to Systems of Linear Differential Equations Theorem Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck , y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk is a solution to the equation y0 = Ay. 2 Example: y0 = Iy, y∈R 1 λ1 = 1, x1 = ; 0 0 λ2 = 1, x2 = ; 1 7 Suppose we have initial conditions y(0) = . −3 t 1 0 7e y = 7e t − 3e t = . 0 1 −3e t Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t) y20 (t) = 2y2 (t) + 6y3 (t) 0 y3 (t) = 3y3 (t) with initial condition 0 y(0) = 11 2 Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t) y20 (t) = 2y2 (t) + 6y3 (t) 0 y3 (t) = 3y3 (t) with initial condition 1 4 5 A = 0 2 6 0 0 3 0 y(0) = 11 2 solving y0 = Ay Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t) y20 (t) = 2y2 (t) + 6y3 (t) 0 y3 (t) = 3y3 (t) with initial condition 1 4 5 A = 0 2 6 0 0 3 1 λ1 = 1, x1 = 0 0 0 y(0) = 11 2 solving y0 = Ay 4 λ2 = 2, x2 = 1 0 29 λ3 = 3, x3 = 12 2 The form of the solution will be: y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk That is: 1 4 29 y(t) = c1 e t 0 + c2 e 2t 1 + c3 e 3t 12 0 0 2 To find the constants c1 , c2 , c3 we solve: 0 1 4 29 11 = c1 0 + c2 1 + c3 12 2 0 0 2 So c1 = −25, c2 = −1, and c3 = 1. Our solution is: 1 4 29 −25e t − 4e 2t + 29e 3t −e 2t + 12e 3t y(t) = −25e t 0 − 1e 2t 1 + e 3t 12 = 3t 0 0 2 2e Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) y20 (t) = 3y1 (t) − y2 (t) with initial condition y(0) = 4 1 Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) y20 (t) = 3y1 (t) − y2 (t) with initial condition y(0) = 1 0 A= 3 −1 4 1 Example Find the solution to the system of linear differential equations y10 (t) = y1 (t) y20 (t) = 3y1 (t) − y2 (t) with initial condition y(0) = 1 0 A= 3 −1 2 λ1 = 1, x1 = 3 4 1 0 λ2 = −1, x2 = 1 The form of the solution will be: y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk That is: 2 −t 0 y(t) = c1 e + c2 e 3 1 t To find the constants c1 , c2 we solve: 2 0 4 = c1 + c2 8 3 1 So c1 = 2, c2 = −5. Our solution is: 2 4e t −t 0 y(t) = 2e − 5e = 3 1 6e t − 5e −t t Complex Eigenvalues 0 −2 y (t) = y(t) 8 0 0 −4 y(0) = 12 Complex Eigenvalues 0 −2 y (t) = y(t) 8 0 0 Eigenvalues: λ1 = 4i, λ2 = −4i −4 y(0) = 12 Complex Eigenvalues 0 −2 y (t) = y(t) 8 0 0 Eigenvalues: λ1 = 4i, λ2 = −4i i −i Eigenvectors: x1 = , x2 = 2 2 −4 y(0) = 12 Complex Eigenvalues 0 −2 y (t) = y(t) 8 0 0 −4 y(0) = 12 Eigenvalues: λ1 = 4i, λ2 = −4i i −i Eigenvectors: x1 = , x2 = 2 2 General solution: y(t) = c1 e 4it x1 + c2 e −4it x2 for some constants c1 and c2 . Complex Eigenvalues 0 −2 y (t) = y(t) 8 0 0 −4 y(0) = 12 Eigenvalues: λ1 = 4i, λ2 = −4i i −i Eigenvectors: x1 = , x2 = 2 2 General solution: y(t) = c1 e 4it x1 + c2 e −4it x2 for some constants c1 and c2 . Particular solution: y(t) = (3 + 2i)e 4it x1 + (3 − 2i)e −4it x2 Complex Eigenvalues: Closer Look Particular solution: y(t) = (3 + 2i)e 4it x1 + (3 − 2i)e −4it x2 = (3 + 2i)[cos(4t) + i sin(4t)]x1 + (3 − 2i)[cos(−4t) + i sin(−4t)]x2 = (3 + 2i)[cos(4t) + i sin(4t)]x1 + (3 − 2i)[cos(4t) − i sin(4t)]x2 i i = (3 + 2i)[cos(4t) + i sin(4t)] + (3 − 2i)[cos(4t) − i sin(4t)] 2 2 = ··· −4 cos(4t) − 6 sin(4t) = 12 cos(4t) − 8 sin(4t) Complex Eigenvalues: Closer Look Suppose c1 = a + bi = c2 . Then the general solution is: y(t) = c1 e 4it x1 + c2 e −4it x2 = c1 e 4it x1 + c1 e −4it x2 = c1 e 4it x1 + c1 e 4it x1 = c1 e 4it x1 + c1 e 4it x1 −2 sin(4t) −2 cos(4t) =a +b 4 cos(4t) −4 sin(4t) which is real. Complex Eigenvalues: Closer Look Suppose c1 = a + bi = c2 . Then the general solution is: y(t) = c1 e 4it x1 + c2 e −4it x2 = c1 e 4it x1 + c1 e −4it x2 = c1 e 4it x1 + c1 e 4it x1 = c1 e 4it x1 + c1 e 4it x1 −2 sin(4t) −2 cos(4t) =a +b 4 cos(4t) −4 sin(4t) which is real. Whence c1 and c2 ? Are they always conjugates when y(0) is real? Complex Eigenvalues: Closer Look Suppose c1 = a + bi = c2 . Then the general solution is: y(t) = c1 e 4it x1 + c2 e −4it x2 = c1 e 4it x1 + c1 e −4it x2 = c1 e 4it x1 + c1 e 4it x1 = c1 e 4it x1 + c1 e 4it x1 −2 sin(4t) −2 cos(4t) =a +b 4 cos(4t) −4 sin(4t) which is real. Whence c1 and c2 ? Are they always conjugates when y(0) is real? c1 x1 + c2 x2 = y(0) c1 x1 + c2 x1 = y(0) If y(0) has real only entries and x1 does not, then c1 = c2 . Complex Eigenvalues 1 1 A= −1 1 Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1 + i, λ2 = 1 − i Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1+ i, λ2 = 1 − i −i Eigenvectors: x1 = , 1 Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1+ i, λ2 = 1−i −i i Eigenvectors: x1 = , x2 = 1 1 Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1+ i, λ2 = 1−i −i i Eigenvectors: x1 = , x2 = 1 1 One solution: Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1+ i, λ2 = 1−i −i i Eigenvectors: x1 = , x2 = 1 1 One solution: (1+i)t −i e 1 −i =e t e it 1 −i =e (cos(t) + i sin(t)) 1 t t e sin t −e cos t = t +i e cos t e t sin t t Complex Eigenvalues 1 1 A= −1 1 Eigenvalues: λ1 = 1+ i, λ2 = 1−i −i i Eigenvectors: x1 = , x2 = 1 1 One solution: (1+i)t −i e 1 t t e sin t −e cos t = t +i e cos t e t sin t General Solution: =c1 t e t sin t −e cos t + c2 e t cos t e t sin t Complex Eigenvalues 1 0 4 A= −5 −2 Complex Eigenvalues 1 0 4 A= −5 −2 Eigenvalues: λ1 = −1 + 12 i, λ2 = −1 − 12 i Complex Eigenvalues 1 0 4 A= −5 −2 Eigenvalues: λ1 = −1 + 12 i, 2+i Eigenvectors: x1 = , −10 λ2 = −1 − 12 i 2−i x2 = −10 Complex Eigenvalues 1 0 4 A= −5 −2 Eigenvalues: λ1 = −1 + 12 i, 2+i Eigenvectors: x1 = , −10 Choosing one: 1 λ2 = −1 − 12 i 2−i x2 = −10 e λ1 t x1 = e (−1+ 2 i)t x1 = e −t e it/2 x1 = e −t (cos(t/2) + i sin(t/2))x1 2+i −t = e (cos(t/2) + i sin(t/2)) −10 −t 2 cos(t/2) − sin(t/2) −t cos(t/2) + sin(t/2) =e + ie −10 cos(t/2) −10 sin(t/2) Complex Eigenvalues 1 0 4 A= −5 −2 Eigenvalues: λ1 = −1 + 12 i, λ2 = −1 − 12 i 2+i 2−i Eigenvectors: x1 = , x2 = −10 −10 Choosing one: λ1 t −t 2 cos(t/2) − sin(t/2) −t cos(t/2) + sin(t/2) e x1 = e + ie −10 cos(t/2) −10 sin(t/2) General solution: c1 2 cos(t/2) − sin(t/2) c2 cos(t/2) + sin(t/2) + t −10 cos(t/2) −10 sin(t/2) et e