MTH 3311 Test #1. order 3, linear. The highest order of derivative of y is 2. Furthermore, y and its derivatives are all raised to the

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1 MTH 3311 Test #1 F 018 Pat Rossi Name Show CLEARLY how you arrive at your answers. 1. Classify the following according to order and linearity. If an equation is not linear, eplain why. (a) y + y y = 4 order 3, linear. The highest order of derivative of y is 3. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear. (b) y (5) + yy = sin () order 5, non-linear. The highest order of derivative of y is 5. (y (5) is the fifth derivative of y it is NOT y 5.) Since y is a co-factor of y, the equation is non-linear. (c) y (4) + y + y = 6 6 order 4, non-linear. The highest order of derivative of y is 4. (y (4) is the fourth derivative of y it is NOT y 4.) Since y is raised to a power other than 1, the equation is non-linear. (d) sin () y 3y + 3 y = e + sin () order 3, linear. The highest order of derivative of y is 3. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear. (e) y y + 4y = +1 order, linear. The highest order of derivative of y is. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear.. Show that the function y = c 1 e 3 + c e is a solution of the differential equation y 9y = Observe: y = c 1 e 3 + c e y = 3c 1 e 3 + 3c e y = 9c 1 e 3 + 9c e Plugging into the left side of the equation, we have: y 9y = ( 9c 1 e 3 + 9c e ) 9 ( c 1 e 3 + c e ) = (9 9) c 1 e 3 + (9 9) c e 3 + ( ) = i.e., y 9y = Hence, y = c 1 e 3 + c e is a solution of the differential equation: y 9y =

2 3. Solve: dy = y + y + + ; subject to the initial condition y (0) = 1 (Assume that, y 0) Solve the equation by separating the variables, and then solve the equation using the integrating factor method. The variables can be separated. = ( + ) y + + = ( + ) y + ( + ) = ( + ) (y + 1) 1 y+1dy = ( + ) 1 y+1 dy = ( + ) ln (y + 1) = C e ln(y+1) = e 1 ++C = e 1 + e C = e 1 + C 1 = C 1 e 1 + i.e., y + 1 = C 1 e 1 + y = C 1 e Recall: y (0) = 1 1 = C 1 e 1 (0) +(0) 1 = C 1 1 = C 1 y = e Alternative Solution on the net page

3 Alternatively: We can solve this equation using the Integrating Factor Method i) Re-epress the equation in the form: y + p () y = Q () dy = y + y + + y = y + y + + y = ( + ) y + + y ( + ) y = ( + ) y + ( ) y = ( + ) p() ii) Compute the integrating factor: e p() = e ( ) = e 1 iii) Multiply both sides by the integrating factor e 1 y + ( ) y = ( + ) e 1 iv) Epress the left side as the derivative of a product [ ] e 1 y = ( + ) e 1 v) Integrate! [ ] d e 1 y = ( + ) e 1 i.e., [ ] d e 1 y = e 1 ( + ) = e u ( du) = e u du = e u + C = e 1 + C e u [ ] e 1 y = e 1 + C vi) Solve for y y = 1 + Ce 1 + i.e., y = Ce du Incorporating the initial condition y (0) = 1, we have: 1 = Ce 1 (0) +(0) 1 = C 1 i.e., 1 = C 1 C = y = e

4 4. Solve: y y = 1 +1 sin () ; using the integrating factor method. (Assume that, y > 0.) i) Epress the equation in the form: y + p () y = Q () y + 1 y = 1 sin () p() Q() ii) Compute the integrating factor: e p() = e ( +1) 1 = e } ln +1 = {{ e ln(+1) } = ( + 1) because > 0 iii) Multiply both sides by the integrating factor ( + 1) y 1 + ( + 1) +1 y = ( + 1) 1 +1 sin () i.e., ( + 1) y + y = sin () iv) Epress the left side as the derivative of a product d [( + 1) y] = sin () v) Integrate! d [( + 1) y] = sin () i.e., ( + 1) y = cos () + C vi) Solve for y y = cos() +1 + C +1 = cos()+c1 +1 4

5 5. Determine whether or not the equation is eact. If the equation is eact, solve it. ( 3 y + y cos () + y 3 + e ) + ( 3 + 6y + sin () + cos (y) e sin(y)) dy = 0 M N The equation will be eact if M y = N Check: M y = [ y 3 y + y cos () + y 3 + e ] = cos () y N = [ 3 + 6y + sin () + cos (y) e sin(y)] = cos () y i.e., M y = cos () y = N Hence, the original equation is eact. The solution to the Differential Equation is of the form: U (, y) = C, where U (, y) = M = Ndy. U (, y) = M = [ 3 y + y cos () + y 3 + e ] = 3 y + y sin () + y 3 + e + f (y) = C U (, y) = N = [ 3 + 6y + sin () + cos (y) e sin(y)] = 3 y + y 3 + y sin () + e sin(y) + g () = C Comparing terms, we see that f (y) = e sin(y) and g () = e Thus, U = 3 y + y 3 + sin () y + e + e sin(y) = C Our solution y is given implicitly by the equation: 3 y + y 3 + sin () y + e + e sin(y) = C 5

6 6. Solve: y dy = 3 + 4y using the substitution v = y. (Assume that, y > 0) dy i) Re-write in the form: = f ( ) y y dy = 3 + 4y y dy = ( y ) = 3 1 ( y ( y + ) ) f( ) y ii) Make the following substitutions: v = y ; dy = v + dv v + dv dv = 3 1 = 3 1 i.e. dv = +3 iii) Separate! +3 dv = 1 iv) Integrate: +3 dv = 1 v + v + v = 3 + v = 3 (Eq. 1) + = +3 ***************************************************************** Scratchwork: +3 dv = 1 dv u = 1 u du ( 1 du) = 1 1 u du = 1 ln u = 1 ln + 3 = 1 ln ( + 3 ) i.e., +3 dv = 1 ln ( + 3 ) ***************************************************************** Substituting this into Eq. 1, we have: 1 ln ( + 3 ) = ln () + C ln ( + 3 ) = ln () + C 1 ln ( + 3 ) = ln ( ) + C 1 e ln( +3) = e ln( )+C 1 = e ln( ) e C 1 = C e ln( ) = C i.e., + 3 = C ( ) y + 3 = C ( ) y + 3 = C 6

7 y + 3 = C 4 Our solution is given implicitly by the equation: y + 3 = C 4 7