#$&* course MTH 279 Solve each equation:*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: 2. y ' + t y = 3 t YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dy/dt = -ty + 3t dy/dt = t(-y+3) integrate(dy/(-y+3)) = integrate(tdt) -ln(y-3) = t^2/2 + C ln(y-3) = -t^2/2 + C y-3 = Ae^(-t^2/2) y = Ae^(-t^2/2) + 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 3. y ' - 4 y = sin(2 t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dy/dt - 4y = sin(2t) µ = e^(ʃ-4dt) = e^(-4t) e^(-4t)dy/dt - 4e^(-4t)y = sin(2t)e^(-4t) d/dt(e^(-4t)y) = sin(2t)e^(-4t) ʃd/dt(e^(-4t)y)dt = ʃsin(2t)e^(-4t)dt Integration by parts: u = e^(-4t) du = -4e^(-4t) dt dv = sin(2t) dt v = -cos(2t)/2 ʃsin(2t)e^(-4t)dt = -e^(-4t)cos(2t)/2 - 2ʃe^(-4t)cos(2t)dt u = e^(-4t) du = -4e^(-4t) dt dv = cos(2t) dt v = sin(2t)/2 ʃsin(2t)e^(-4t)dt = -e^(-t)cos(2t)/2 - sin(2t)e^(-4t) - 4 ʃsin(2t)e^(-4t)dt 5 ʃsin(2t)e^(-4t)dt = -e^(-4t)cos(2t)/2 - sin(2t)e^(-4t) ʃsin(2t)e^(-4t)dt = -e^(-4t)cos(2t)/10 - e^(-4t)sin(2t)/5 Substituting back into original: e^(-4t)y + c_1 = -e^(-4t)cos(2t)/10 - e^(-4t)sin(2t)/5 + c_2 e^(-4t)y = -e^(-4t)cos(2t)/10 - e^(-4t)sin(2t)/5 + C y = cos(2t)/10 - sin(2t)/5 + e^(4t)C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Self-critique (if necessary): I’m confident I did this correctly. It took me a few minutes to remember how to do this cyclical integration by parts, but once I realized the method I needed to use, it was pretty simple, just time consuming. Self-critique rating: 3 ********************************************* Question: 4. y ' + y = e^t, y (0) = 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Solving the first order homogeneous linear diff. eq.: y’ + y = 0 dy/dt = -y -dy/y = dt ʃ-dy/y = ʃdt -ln(y) = t + C 1/y = Ae^(t) y_1 = Ae^(-t) Solving the first order nonhomogeneous linear diff. eq. using undetermined coefficients: We can guess since the nonhomogeneous portion is e^(t) that the form our solution will take will be y_2 = Be^(t). Thus, y_2’ = Be^(t). We substitute into the original diff. eq. to get: Be^(t) + Be^(t) = e^(t) 2Be^(t) = e^(t) 2B = 1 B = ½ This means that y_2 = e^(t)/2. y_g = y_1 + y_2 = Ae^(-t) + e^(t)/2 Imposing the initial conditions, we have: 2 = Ae^(0) + e^(0)/2 = A + ½ A = 3/2 y_p = 3e^(-t)/2 + e^(t)/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: 5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Solving the homogeneous portion: y’+3y = 0 dy/dt = -3y ʃdy/y = ʃ-3dt ln(y) = -3t + C y_1 = Ae^(-3t) Solving the non-homogeneous portion: We can guess that y_2 will take a form of Be^(t) + Dt + E. y_2 = Be^(t) + Dt + E y_2’ = Be^(t) + D y_2’ + 3y_2 = 3 + 2t + e^(t) Be^(t) + Dt + E + Be^(t) + D = 3 + 2t + e^(t) B = ¼ D = 2/3 E = 7/9 y_2 = e^(t)/4 + 2t/3 + 7/9 y_g = y_1 + y_2 y_g = Ae^(-3t) + e^(t)/4 + 2t/3 + 7/9 Imposing initial conditions: e^2 = Ae^(-3) + e/4 + 2/3 + 7/9 Ae^(-3) = e^2 - e/4 - 13/9 A = e^(3)((36e^2 - 9e - 52)/36) y_p = e^(3)((36e^2 - 9e - 52)/36)(e^(-3t)) + e^(t)/4 + 2t/3 + 7/9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: µ(t) = e^(ʃp(t)dt) µ(t)dy/dt + µ(t)p(t)y = g(t)µ(t) ʃ(d/dt(µ(t)y)dt) = ʃg(t)µ(t)dt µ(t)y + c_1 = ʃg(t)µ(t)dt + c_2 y = (ʃg(t)µ(t)dt + C)/(µ(t)) y = e^(ʃ-p(t)dt)( ʃg(t)e^(ʃp(t)dt) + Ce^(ʃ-p(t)dt) = Ce^(-t^2) + 1 By looking at this and realizing that the derivative of -t^2 is -2t, we can determine that p(t) = 2t. We can also rationalize based on this that the value of g(t) is simply 1. y = Ae^(ʃ-2tdt) + e^(ʃ-2tdt)*ʃe^(ʃ2tdt)dt We can plug this back into the original equation and find that this is true. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I spent a good deal of time working on this problem. I tried the method of undetermined coefficients at first, but failed to get an answer. I’m not entirely sure why. It may have been a simple mistake. I worked through the long method and came up with the answer above, which works.