Math : Solutions to Assignment 10

Transcription

1 Math -3: Solutions to Assignment. There are two tanks. The first tank initially has gallons of pure water. The second tank initially has 8 gallons of a water/salt solution with oz of salt. Both tanks drain into the other at a rate of gallons per minute. Find formulas to express the amount of salt in each tank. Solution: Let x = ( x y ) t where x,y represent the amount of salt in the first and second tanks respectively. x in = y out = y 8 = y 4 and x out = y in = x = x 5. So we reach the IVP system x = 5 4 x, for x() = 5 4 The characteristic equation is λ + 9 λ = λ(λ + 9 ) = For λ =, one such eigenvector is v = ( 5 4 ) t. ( x (t) = e t 5 5 = 4 4) For λ = 9, we have an eigenvector v = ( ) t x (t) = e 9 t e 9 = t e 9 t So x(t) = C ( 5 4 ) e 9 + C t e 9 t Solving x() = ( ) t yields C = 9 and C = e 9 t x(t) = e 9 t

2 . Solve the following IVP: x = 4 x for x() = Solution: The characteristic equation is λ 9 = (λ 3)(λ + 3) =. For λ = 3, one such eigenvector is v = ( ) t. So we get x (t) = e 3t e 3t = e 3t For λ = 3, one such eigenvector is v = ( x (t) = e 3t = ) t. So we get ( e 3t e 3t ) So x(t) = C ( e 3t e 3t ) + C ( e 3t e 3t ) Solving x() = ( ) t yields C = 3 and C = 3 x(t) = 4e 3t e 3t 3 e 3t + e 3t

3 3. Find a general solution to the following system x = x Solution: The characteristic equation is λ + =. We have the complex root λ = i = + i with eigenvector ( v = v + i v = + i ) So x (t) = e αt cos t + sin t [cos βt v sin βt v ] = cos t x (t) = e αt sin t cos t [cos βt v + sin βt v ] = sin t Our general solution is x(t) = C ( cos t + sin t cos t ) sin t cos t + C sin t

4 4. Solve the following IVP: x = x for x() = 3 Solution: The characteristic equation is λ 4λ + 4 = (λ ) =. So we only will be able to find one independent eigenvector for λ =. One such vector is v = ( ) t. So x (t) = e t = ( e t e t ) Now we need to find w such that A w = w + v. So we try any vector that s not a multiple of v ( ( ( ( ( u = A u = = = = u v ) 3) ) ) ) So we want w = u. It can be verified that A w = w + v. So ( ( x (t) = e λt ( w + t v) = e t e + t = )) t (t ) te t So e t e x(t) = C e t + C t (t ) te t Solving for x() = ( ) t yields C = C = x(t) = ( te t (t + )e t )

5 5. Find a general solution to the following system x 3 = x + Solution: The characteristic equation is λ 5λ + 4 = (λ )(λ 4) =. For λ =, one such eigenvector is v = ( ) t. So we get x (t) = e t = For λ = 4, one such eigenvector is v = ( x (t) = e 4t = ( e t e t ) ) t. So we get ( e 4t e 4t ) So now we need to find x p = X X bdt ( e t e X = [ x, x ] = 4t e t e 4t, and b = ) X = e 4t e 4t 3e 5t e t e t = e t e t 3 e 4t e 4t X e t e t e t bdt = 3 e 4t e 4t dt = e 4t dt = ( e t 3 e 4t ) So x p = X ( X e t e 4t e t bdt = e t e 4t 3 e 4t ) = 3 = So the general solution is e t e 4t x(t) = C e t + C e 4t

6 6. Given an n n matrix A, show that the set E λ := { v A v = λ v} is a vector subspace whenever E λ { }. Easy way: Since A v = λ v (A λi) v =, E λ = null(a λi). As we covered in class, a nullspace is a vector subspace, so E λ is one as well. Brute force way: We need to verify the two properties of a vector space: So And x, y E λ so A x = λ x and A y = λ y A( x + y) = A x + A y = λ x + λ y = λ( x + y) ( x + y) E λ A(α x) = αa x = αλ x = λ(α x) α x E λ So E λ is a vector subspace of R n

7 7. Find the values of α such that the system α α x = b α is guaranteed to have a solution for any choice of b. Solution: Guaranteed solution A is non-singular deta. α α α = α α α α = α3 α = α(α )(α + ) So our solution set is (, ) (, ) (, ) (, )

8 8. Determine which of the following vectors are in V, where V = span,, v =, v =, v 3 =, v 4 = Solution: v = v V v R 4, so it can not be in V R 3 v 3 will yield no solution v 3 is not in V v 4 = = v 4 V In fact is always in any span (assuming it is of the same dimension).

9 9. Determine which of the following vectors are in V, where V = span,, Solution: Consider v =, v = 3 e π, v 3 = 5, v 4 = 4 cos A =,, The question now can be stated as finding a solution to A c = v i for i =,, 3 or 4. But det(a) = A is non-singular. So a solution must exist for any choice of vector (yes even for v ). In fact v = π + e cos + cos + e π π + cos e + 7

10 . Find a basis for V = span 3,, 4, Solution: We need to see if there is a solution to A c =, where A = v = 3, v =, v 3 = 4, v 4 = Doing so would show that the vectors are not independent, allowing us to remove vectors to get a basis. Consider the augmented system By performing the following operations R R R, R 3 3R R 3, and R 4 4R R 4, we get the system Again by performing the operations R 3 R R 3 and R 4 3R R 4, we get 4 We get the solution set (t s) v +(s 4t) v +s v 3 +t v 4 =. By setting s = and t =, we see that v 4 v + v 4 =, so v 4 may be removed, as it can be expressed in terms of v and v. Likewise, using s = and t =, we see that v + v + v 3 = so v 3 may be removed as well. So our basis will be following two vectors 3, 4

11 . A = a) What is the dimension of null(a). b) Find a basis for null(a). c) Find the general solution for A x = Solution: a) There are two free columns two free variables dim(nulla) =. b) Assigning x = s and x 4 = t,we have the following two equations: x 3 t = and x + x 3 + t =. null(a) = s + t = span, These two vectors form a basis for null(a). c) Any solution to the equation must be of the form x(t) = p + v, where v = s + t and p is any vector satisfying A p = One such vector is p = ( ) t. So we get x(t) = + s + t

12 . Solve the following IVP: x = x cos t for x() = 5 Solution: This is a separable st order IVP. So dx dt = x cos t dx x = cos t dt assuming x By integrating we get ln x = sin t + C x(t) = Ae sin t x() = Ae sin = A = 5 So we get the solution x(t) = 5e sin t

13 3. Find a general solution for x 3x + x = sin t Solution: We can either solve this ODE using the undetermined coefficients method or by the Laplace transform (using arbitrary initial conditions). We will do the former. Consider the characteristic equation λ 3λ + = (λ )(λ ) =. So we have two roots λ =, λ =. This gives us our two independent homogeneous solutions, x (t) = e t and x (t) = e t Now we find x p (t). Our trial solution will be x p (t) = A cos t + B sin t. x p = A cos t+b sin t, x p = A sin t+b cos t, and x p = A cos t+ B sin t Plugging these functions into our ODE yields (A 3B) cos t + (3A + B) sin t = sin t A = 3, B = So our general solution is x(t) = C e t + C e t + 3 cos t + sin t

14 4. Suppose (λ, v) is an eigenpair for an n n matrix A. Suppose also that (µ, v) is an eigenpair for n n matrix B. Show that (e λ+µ, v) is an eigenpair for e A+B Solution: To find this solution, we first show that (λ+µ, v) is an eigenpair for A + B. So we calculate (A + B) v = A v + B v = λ v + µ v = (λ + µ) v Now we use the example from class to conclude that, because (λ + µ, v) is an eigenpair for A + B, (e λ+µ, v) is an eigenpair for e A+B.