Solutions to Homework 5, Introduction to Differential Equations, 3450: , Dr. Montero, Spring y 4y = 48t 3.

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1 Solutions to Homework 5, Introduction to Differential Equations, 3450: , Dr. Montero, Spring 2009 Problem 1. Find a particular solution to the differential equation 4y = 48t 3. Solution: First we look at the polynomial associated to the right h side: λ 2 4 = (λ 2)(λ + 2). The roots of this polynomial are ±2. Next we observe that annihilator for the right h side is D 4, that has the associated polynomial λ 4. This polynomial has only one root, 0, with multiplicity 4, this is, of course, different from ±2. Hence we seek a particular solution of the form y(t) = α 1 + α 2 t + α 3 t 3 + α 4 t 3. We find next y (t) (t) to obtain y (t) = α 2 + 2α 3 t + 3α 4 t 2 (t) = 2α 3 + 6α 4 t. We plug this into the equation, group together the terms with t 0, t 1, t 2 t 3, to obtain 2α 3 + 6α 4 t 4(α 1 + α 2 t + α 3 t 2 + α 4 t 3 ) = (2α 3 4α 1 ) + (6α 4 4α 2 )t 4α 3 t 2 4α 4 t 3 = 48t 3. We solve from here to obtain 4α 4 = 48 so α 4 = 12, α 3 = 0, 6α 4 4α 2 = 0, so α 2 = 18, 2α 3 4α 1 = 0 so α 1 = 0.

2 Problem 2. Find a particular solution of + 6y + 8y = 13te 5t. Solution: The annihilator for the right h side is (D 5) 2, that has associated polynomial of if one prefers, This also says that the function satisfies the equation (λ 5) 2, λ 2 10λ g(t) = 13te 5t 10y + 25y = 0. We will make use of this fact later to simplify our computations. We also note that the roots of λ 2 10λ are 5 5 (or 5 with multiplicity 2). On the other h, the polynomial for the left h side is λ 2 + 6λ + 8 = (λ + 4)(λ + 2), that has roots 2 4. Neither of these is 5, so we seek a particular solution of the form y p (t) = α 1 e 5t + α 2 te 5t = (α 1 + α 2 t)e 5t. Notice now that, due to the way this function was built, y p satisfies We must plug y p into the equation To do this we take advantage of the fact that p 10y p + 25y p = y + 8y = 13te 5t. p 10y p + 25y p = 0. as follows: + 6y + 8y = ( 10y + 25y) + 16y + 17y,

3 so p + 6y p + 8y p = ( p 10y p + 25y p) + 16y p + 17y p = 16y p + 17y p. Because of this, when we plug y p into + 6y + 8y = 13te 5t the equation reduces to 16y p + 17y p = 13te 5t. Now we compute y p (t) = α 2e 5t + 5(α 1 + α 2 t)e 5t = (5α 1 + α 2 )e 5t + 5α 2 te 5t. Plugging this into 16y p + 17y p = 13te 5t. gives us 16(5α 1 + α 2 )e 5t + 80α 2 te 5t + 17(α 1 + α 2 t)e 5t = (97α α 2 )e 5t + 97α 2 te 5t = 13 t e 5t. We deduce from here that 97α 2 = 13, so α 2 = 13 97, 97α α 2 = 0, so α 1 = Problem 3. Find the solution of + 2y = 72 sin(2t) + 16 cos(2t), that also satisfies y(0) = 8 y (0) = 3. Solution: First, the annihilator for the right h side is (D 2 + 4). This means that the function g(t) = 72 sin(2t) + 16 cos(2t)

4 satisfies + 4y = 0. The roots of the polynomial associated to this equation are ±2i. On the other h, the polynomial associated to the left h side of the equation is λ 2 + 2λ = λ(λ + 2). The roots of this polynomial are 0 2. We first seek a particular solution of the form y p (t) = α 1 sin(2t) + α 2 cos(2t). Plugging this into the equation, re-arranging a bit, we obtain 4α 1 sin(2t) 4α 2 cos(2t) + 2(2α 1 cos(2t) 2α 2 sin(2t)) = ( 4α 1 4α 2 ) sin(2t) + ( 4α 2 + 4α 1 ) cos(2t) = 72 sin(2t) + 16 cos(2t). From here we get the system α 1 + α 2 = 18 α 1 α 2 = 4. The solutions are α 1 = 7 α 2 = 11. The particular solution we seek is y p (t) = 7 sin(2t) 11 cos(2t). This means that the general solution to the equation + 2y = 72 sin(2t) + 16 cos(2t) is y(t) = α 1 + α 2 e 2t 7 sin(2t) 11 cos(2t). We use now the initial data to determine α 1 α 2. We obtain y(0) = 8 = α 1 + α 2 11 y (0) = 3 = 2α The second equation says that α 2 = 17 2.

5 Using this in the first equation gives us The solution we seek is α 1 = y(t) = e 2t 7 sin(2t) 11 cos(2t). Problem 4. Find a particular solution to the equation 14y + 62y = 96e 7t cos(3t) + 16e 7t sin(3t) + 9 by the method of undetermined coefficients (annihilators). Solution: First, the polynomial associated with the left h side is λ 2 14λ + 62 = (λ 7) On the other h, we need to annihilate the right h side. For the function g 1 (t) = 96e 7t cos(3t) + 16e 7t sin(3t), the annihilator is (D 7) = D 2 14D This means that g 1 (t) solves the equation 14y + 58y = 0. On the other h, the annihilator for the function g 2 (t) = 9 is just This means that g 2 satisfies D. y = 0. Thus, the annihilator for the whole right h side is D(D 2 14D + 58). We seek a particular solution of the form y p (t) = α 1 e 7t cos(3t) + α 2 e 7t sin(3t) + α 3.

6 We will split this function in two as follows: y p,1 = α 1 e 7t cos(3t) + α 2 e 7t sin(3t), y p2 = α 3. Now we need to plug y p = y p,2 + y p,2 into the equation 14y + 62y = 96e 7t cos(3t) + 16e 7t sin(3t) + 9. Now we observe that the function y p,1 solves the equation 14y + 58y = 0. Hence, the result of plugging y p,2 into the left h side of is 14y + 62y = 96e 7t cos(3t) + 16e 7t sin(3t) + 9 p,1 14y p,1 + 62y p,1 = p,1 14 p, p,1 + 4y p,1 = 4y p,1. On the other h, the result of plugging y p,2 into the left h side of our equation is As a result, we plug y p in the equation to obtain p,2 14y p,2 + 62y p,2 = 62y p,2. 14y + 62y = 96e 7t cos(3t) + 16e 7t sin(3t) + 9 4y p,1 + 62y p,2 = 4α 1 e 7t cos(3t) + 4α 2 e 7t sin(3t) + 62α 3 = 96e 7t cos(3t) + 16e 7t sin(3t) + 9. We obtain 4α 1 = 96, so α 1 = 24, 4α 2 = 16, so α 2 = 4, 62α 3 = 9 so α 3 = Problem 5. Use the method of undetermined coefficients to find a particular solution of + 2y + 2y = (7 + 10t)e t cos(t) + ( t)e t sin(t).

7 Solution: First, the polynomial associated to the left h side of the equation is λ 2 + 2λ + 2 = (λ + 1) On the other h, the annihilator of the right h side is ((D + 1) 2 + 1) 2. This means that the roots of the polynomial associated to this annihilator are λ = 1 ± i, both with multiplicity 2. These are the same roots of the polynomial associated to the left h side of the equation, so we must seek a particular solution of the form y p (t) = (α 1 t + α 2 t 2 )e t cos(t) + (α 3 t + α 4 t 2 )e t sin(t). I will use the notation y 1 (t) = e t cos(t) y 2 (t) = e t sin(t). The particular solution then looks like this: y p (t) = (α 1 t + α 2 t 2 )y 1 (t) + (α 3 t + α 4 t 2 )y 2 (t). Note that both y 1 y 2 solve the equation + 2y + 2y = 0. We now need to compute y p y p. What to do. y p (t) = (α 1 + 2α 2 t)y 1 + (α 1 t + α 2 t 2 )y 1 + (α 3 + 2α 4 t)y 2 + (α 3 t + α 4 t 2 )y 2. Next y p (t) = 2α 2y 1 + 2(α 1 + 2α 2 t)y 1 + (α 1t + α 2 t 2 )y 1 + 2α 4 y 2 + 2(α 3 + 2α 4 t)y 2 + (α 3 t + α 4 t 2 ) 2. Next, we plug these computations in the equation p + 2y p + 2y p. The result of this is y p + 2y p + 2y p = 2α 2 y 1 + 2(α 1 + 2α 2 t)y 1 + (α 1 t + α 2 t 2 )y 1 + 2α 4 y 2 + 2(α 3 + 2α 4 t)y 2 + (α 3t + α 4 t 2 ) 2 + 2(α 1 + 2α 2 t)y 1 + 2(α 1 t + α 2 t 2 )y 1 + 2(α 3 + 2α 4 t)y 2 + 2(α 3 t + α 4 t 2 )y 2 + 2(α 1 t + α 2 t 2 )y 1 (t) + 2(α 3 t + α 4 t 2 )y 2 (t).

8 We now re-arrange this as follows: p + 2y p + 2y p = (α 1 t + α 2 t 2 ) 1 + 2(α 1t + α 2 t 2 )y 1 + 2(α 1t + α 2 t 2 )y 1 + (α 3 t + α 4 t 2 ) 2 + 2(α 3t + α 4 t 2 )y 2 + 2(α 3t + α 4 t 2 )y 2 + 2α 2 y 1 + 2(α 1 + 2α 2 t)y 1 + 2(α 1 + 2α 2 t)y 1 + 2α 4 y 2 + 2(α 3 + 2α 4 t)y 2 + 2(α 3 + 2α 4 t)y 2. Notice that we can re-arrange this a bit further to obtain p + 2y p + 2y p = (α 1 t + α 2 t 2 )( 1 + 2y 1 + 2y 1) Now we recall that y 1 y 2 both satisfy Hence, we obtain Next we note that + (α 3 t + α 4 t 2 )( 2 + 2y 2 + 2y 2 ) + 2α 2 y 1 + 2(α 1 + 2α 2 t)y 1 + 2(α 1 + 2α 2 t)y 1 + 2α 4 y 2 + 2(α 3 + 2α 4 t)y 2 + 2(α 3 + 2α 4 t)y y + 2y = 0. p + 2y p + 2y p =2α 2 y 1 + 2(α 1 + 2α 2 t)y 1 + 2(α 1 + 2α 2 t)y 1 + 2α 4 y 2 + 2(α 3 + 2α 4 t)y 2 + 2(α 3 + 2α 4 t)y 2. y 1 = y 1 y 2 y 2 = y 2 + y 1. We use this in the above equation to obtain p + 2y p + 2y p = (2α 1 + 2α 2 2α 1 + 2α 3 )e t cos(t) + (4α 2 4α 2 + 4α 4 )te t cos(t) + (2α 3 2α 1 + 2α 4 2α 3 )e t sin(t) + (4α 4 4α 2 4α 4 )te t sin(t) = 7e t cos(t) + 10te t cos(t) + 25e t sin(t) + 11te t sin(t).

9 We solve this to obtain 4α 4 = 10, so α 4 = 5 2, 4α 2 = 11 so α 2 = 11 4 then we use the other two equations to obtain α 1 = α = 10 α 3 = Problem 6. Find a particular solution of 3 + 2y y = 2t 2 2t + 2e 3t. Solution: First note that the polynomial associated to the left h side is 3λ 2 + 2λ 1 = (3λ 1)(λ + 1). The roots are λ = 1 λ = 1. On the other h, the annihilator for the right h side 3 is D 3 (D 3). The roots of the polynomial associated to this operator are 0, with multiplicity 3, 3, which do not coincide with the roots of the polynomial associated to the left h side. We seek a particular solution of the form We compute y p y p (t) = α 1 + α 2 t + α 3 t 2 + α 4 e 3t. y p : y p(t) = α 2 + 2α 3 t + 3α 4 e 3t, p (t) = 2α 3 + 9α 4 e 3t. Plug this in the equation, group the terms with t 0, t 1, t 2 e 3t, to obtain (27α 4 + 6α 4 α 4 )e 3t + (6α 3 + 2α 2 α 1 ) + (4α 3 α 2 )t α 3 t 2 = 2t 2 2t + 2e 3t. This gives us finally α 3 = 2, so α 3 = 2, 4α 3 α 2 = 0 so α 2 = 10, 6α 3 + 2α 2 α 1 = 0 so α 1 = 32, 32α 4 = 2 so α 4 = 1 16.