#$&* course Mth 279 1/1/14 around 4pm Section 2.2*********************************************
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: 2. t^2 y ' - 9 y = 0, y(1) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t^2 y ' - 9 y = 0 t^2 dy/dt = 9 y dy/y = 9 dt/t^2 int (dy/y) = 9 * int (dt/t^2) ln(y) = -9/t + c y(t) = e^(c - 9/t) y(t) = c/e^(9/t) y(1) = 2 y(1) = c/e^(9/1) = 2 c = 2e^9 y(t) = 2e^9 / e^(9/t) confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (t^2 + t) y' + (2t + 1) y = 0 dy/dt = -(2t + 1) y /(t^2 + t) dy/y = -(2t + 1)/(t^2 + t) dt int (dy/y) = -int((2t + 1)/(t^2 + t) dt) ln(y) = -ln(t^2 + t) + c y(t) = c/(t^2 + t) y(0) = 1 y(0) = c/(0^2 + 0) = c/0 = 1 C is undefined due to not being able to divide by zero confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: OK &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 4. y ' + sin(3 t) y = 0, y(0) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y ' + sin(3 t) y = 0 dy/dt + sin(3 t) y = 0 dy/dt = -sin(3 t) dt int (dy/dt) = - int (sin(3 t) dt) ln(y) = cos(3t) /3 + c y(t) = e^(cos(3t) /3 + c) y(t) = c * e^(cos(3t) /3) y(0) = 2 y(0) = c * e^(cos(3* 0) /3) = 2 y(0) = c * e^(1/3) = 2 c = 2/e^(1/3) y(t) = 2/e^(1/3) * e^(cos(3t) /3) y(t) = 2e^(cos(3t) /3) /e^(1/3) confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did. y ' - t^2 y = 0 **** E, the only direction where slope changes to zero and goes back to where it began. #$&*
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I know how to find the average slope but I’m not sure how that will help since it yields an exponential equation ------------------------------------------------ Self-critique rating:2 ********************************************* Question: 7. The equation y ' - y = 2 is first-order linear, but is not homogeneous. If we let w(t) = y(t) + 2, then: What is w ' ? **** w' = (2+y) + 2 We get 2 + y from y' = 4 + y
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NONE ------------------------------------------------ Self-critique rating: ********************************************* Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?  YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y(0) = y_0 is equal to 1 which is the C value approximate with (-1, 0.5) y = e^(-b t) 0.5 = e^(-b (-1)) 0.5 = e^(b) —> natural log ln(0.5) = ln(e^(b)) b = - 0.69
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 "