#$&* course Phy 2479 3/6 3 Query 07 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: change of dependent variable v = y^(2/3), v(0) = y(0)^(2/3) = -9^(2/3) v’ + (2/3)(-1)v = (2/3)t v’ - (2/3)v = (2/3)t, linear nonhomgenious mut thru by e^(Int(-2/3)) = e^((-2/3)t) e^((-2/3)t)v’ - (2/3)e^((-2/3)t)v = (2/3)t e^((-2/3)t) e^((-2/3)t)v = Int((2/3)t e^((-2/3)t) dt ), int by parts Int(((2/3)t e^((-2/3)t) dt) = (2/3)[ te^((-2/3)t) + (3/2)((-3/2)e^((-2/3)t))] = (2/3)te^((-2/3)t) - (3/2)e^((-2/3)t) + C e^((-2/3)t)v = (2/3)t e^((-2/3)t) - (3/2) e^((-2/3)t) + C v = [(2/3)t - (3/2)] + Ce^((2/3)t) For v(0) = y(0)^(2/3) = -9^(2/3) v = -(3/2) + C = -9^(2/3), C = (3/2) - 9^(2/3) ????This doesn’t seem right for init condition???? v(t) = [(2/3)t - (3/2)] + ((3/2) - 9^(2/3))e^((2/3)t) y = v^(1/3)= (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3) y(t) = (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: