Homework 8 Solution - Introduction to Dierential Equations Math 286 X1 Fall 2009 Homework 8 Solutions 1 Solve 1 1 4 1 x = 1 1 x(0 = x Solution We rst

Homework 8 Solution - Introduction to Dierential Equations...

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Unformatted text preview: Introduction to Differential Equations – Math 286 X1 Fall 2009 Homework 8 Solutions 1. Solve −1 1 −4 −1 x′ = 1 1 x(0) = x, . Solution: We first compute the eigenvalues and eigenvectors. The characteristic polynomial is λ2 + 2λ + 5 = 0, which has roots λ= −2 ± √ 4 − 20 = −1 ± 2i. 2 Choosing λ = −1 + 2i, we have −2i 1 −4 −2i A − (−1 + 2i)I = and solving −2i 1 −4 −2i gives x y 0 0 = −2ix + y = 0, −4x − 2iy = 0. One solution to this is x = 1, y = 2i, so our eigenvector can be chosen as 1 2i v= . Our two solutions will be the real and imaginary parts of eλt v, which we compute as e(−1+2i)t 1 2i 1 2i = e−t (cos(2t) + i sin(2t)) = e−t cos(2t) + i sin(2t) 2i cos(2t) − 2 sin(2t) Taking real and imaginary parts gives two solutions: x1 = e−t , x2 = e−t sin(2t) 2 cos(2t) . cos(2t) −2 sin(2t) + C2 e−t sin(2t) 2 cos(2t) . cos(2t) −2 sin(2t) Thus the general solution is x(t) = C1 e−t Plugging in t = 0 gives x(0) = C1 2C2 , so C1 = 1, C2 = 1/2, so the solution is x(t) = e−t cos(2t) −2 sin(2t) 1 1 + e−t 2 sin(2t) 2 cos(2t) . . 2. Solve 3 0 x′ = 1 3 1 1 x(0) = x, . Solution: The characteristic polynomial is λ2 − 6λ + 9 = 0, which is factored as (λ − 3)2 = 0, so it has a double root of 3. We have that 0 1 0 0 , = A − 3I = 0 0 and to get the eigenvector we get 0 0 1 0 x y 1 0 or the equation y = 0. A solution to this is v1 = , . This gives us one solution to the ODE, namely 1 0 e3t . We need to now find the generalized eigenvector, which we find by solving the system (A − 3I)v2 = v1 , or 0 0 1 0 x y 0 1 or the equation y = 1. So we can choose v2 = is , . Then we know another solution to this problem 1 0 x2 = e3t (tv1 + v2 ) = e3t t 1 0 = 0 1 + t 1 = e3t Thus our general solution is x(t) = C1 e3t 1 0 + C2 e3t t 1 . Plugging in t = 0 gives x(0) = C1 C2 , so C1 = C2 = 1, and our solution is x(t) = e3t 1 0 + e3t 2 t 1 = e3t t+1 1 . . 3. Solve 3 0 x′ = 1 2 1 1 x(0) = x, . Solution: The characteristic polynomial is λ2 − 5λ + 6 = 0, which has roots λ = 2, 3. We have A − 2I = 1 −1 so the eigenvector for λ = 2 is 1 0 , , and 0 0 A − 3I = so the eigenvector for λ = 3 is 1 1 0 0 1 −1 , . Thus the general solution is x(t) = C1 e2t 1 −1 x(0) = C1 1 −1 1 0 + C2 e3t + C2 . Plugging in t = 0 gives 1 0 , which gives the system C1 + C2 = 1, −C1 = 1, which has solution C1 = −1, C2 = 2. Thus our solution is 1 −1 x(t) = −e2t 1 0 + 2e3t . 4. For each of these matrices A, compute etA : A= 2 0 0 2 , 2 0 A= 0 −1 , A= 7 −4 8 −5 . Solution: In each case, we need to compute the eigenvalues and eigenvectors of the matrix first. So we proceed: (a) The characteristic polynomial is λ2 − 4λ + 4 = 0, which has a double root of 2. We have A − 2I = 3 0 0 0 0 , so that all vectors are eigenvectors and we can choose what we like. Choose 1 0 v1 = , 0 1 v2 = , and this gives 1 0 0 1 S= . Since S is the identity matrix, S −1 = S, so we have e2t 0 etA = S 0 e2t e2t 0 S −1 = I 0 e2t I= (b) the characteristic polynomial is λ2 − λ − 2 = 0, which has roots λ = 2, −1. For λ = 2, we have 0 0 and 0 0 0 1 x y 0 1 , = A − 2I = 0 0 0 0 , gives y = 0, so we have v1 = 1 0 . For λ = −1, we have A+I = and we have v1 = 0 1 3 0 . Thus we have S= 1 0 0 1 . Similarly to above, we have S e2t 0 0 e −t S= e2t 0 0 e−t since S is the identity. (c) The characteristic polynomial is λ2 − 2λ − 3 = 0, which has roots λ = 3, −1. We have A − 3I = and our eigenvector can be 1 1 4 8 −4 −8 , and A+I = 4 8 8 −4 −4 , e2t 0 0 e2t . so our eigenvector can be 1 2 . Thus S= and S −1 = 1 det S 1 1 1 2 , 2 −1 −1 1 = 2 −1 −1 1 2 −1 −1 1 = −e−t + 2e3t −2e−t + 2e3t . Thus etA = 1 1 1 2 e3t 0 0 e −t 5 e−t − e3t 2e−t − e3t . ...
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