#$&* course Mth 279 1/3/14 around 5 pm Solve each equation:*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 2. y ' + t y = 3 t YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = w + 3 y' = w' w ' + t (w + 3) = 3 t w ' + tw + 3t = 3 t w ' + tw = 0 y' + p(t)y = 0 -> y = c*e^(-int(p(t)dt)) w = c*e^(-int(t dt)) = c*e^(-t^2/2) y - 3 = c*e^(-t^2/2) y = c*e^(-t^2/2) + 3 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 3. y ' - 4 y = sin(2 t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u(t) = e^(-int (4 dt)) —> y’e^(-int (4 dt)) - 4 y e^(-int (4 dt)) = sin(2 t)e^(-int (4 dt)) —> (e^(-int (4 dt)) * y)' = sin(2 t)e^(-int (4 dt)) —> int ((e^(-int (4 dt)) * y)') = int (sin(2 t)e^(-int (4 dt))) —> e^(-int (4 dt)) * y = c - (2sin(2t) + cos(2t))/10e^(4t) —> y = c*e^(4t) - (1/5)sin(2t) - (1/10)cos(2t)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 4. y ' + y = e^t, y (0) = 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u(t) = e^(int (1 dt)) = e^t —>y’ e^t + ye^t = e^(2t) —> (ye^t)' = e^(2t) —> int ((ye^t)') = int (e^(2t)) —> ye^t = e^(2t)/2 + c —> y = e^t/2 + c/e^t —> y(0) = 2 —> y(0) = e^0/2 + c/e^0 = 2 —> y(0) = 1/2 + c = 2 c = 3/2 y(t) = (1/2)e^t + (3/2)e^(-t) confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u(t) = e^(int (3 dt)) = e^(3t) y ' + 3 y = 3 + 2 t + e^t —> y 'e^(3t) + 3 ye^(3t) = 3e^(3t) + 2 te^(3t) + e^(4t) —> (ye^(3t))' = 3e^(3t) + 2 te^(3t) + e^(4t) —> int ((ye^(3t))') = int (3e^(3t) + 2 te^(3t) + e^(4t)) —> ye^(3t) = e^(3t) + (2/9) e^(3t) (3t - 1) + (1/4)e^(4t) + c —> y(t) = c e^(-3t) + (2/3)t + (1/4)e^t + (7/9) y(1) = e^2 y(1) = c e^(-3) + (2/3) + (1/4)e + (7/9) = e^2 c = e^5 - (1/4)e^4 - (13/9)e^3 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: 6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = C e^(-t^2) + 1 —> ye^(t^2) = C + e^(t^2) —> (ye^(t^2))' = 2te^(t^2) I used the Power rule here y 'e^(t^2) + 2yte^(t^2) = 2te^(t^2) —> y ' + 2t y = 2t p(t) = 2t g(t) = 2t confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3"