Question 1 Not yet answered Marked out of 1.00 Flag question In order to solve with...

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Question 1 Not yet answered Marked out of 1.00 Flag question In order to solve with y(0) = 1 we first (2y-5)dy = (3x² e² )dx. After y²-5y = ³e +c of integration. Setting x= and where c is the have y²-5y-(x-e- 3) = 0. By using 13 y(x)= 5 2 +2²³-e². In the quadratic formula, negative square root is chosen because of value 0 disintegrate y=1 eliminate factor integration by parts c=-3 separate the variables. This gives both sides, we obtain y' = 3r² p² 2y-5 integrating the initial conditions c=3 the quadratic formula constant c=0 we have c=1 c=3 Thus, we x=1 y=0 Question 1 Not yet answered Marked out of 1.00 Flag question In order to solve with y(0) = 1 we first (2y-5)dy = (3x² e² )dx. After y²-5y = ³e +c of integration. Setting x= and where c is the have y²-5y-(x-e- 3) = 0. By using 13 y(x)= 5 2 +2²³-e². In the quadratic formula, negative square root is chosen because of value 0 disintegrate y=1 eliminate factor integration by parts c=-3 separate the variables. This gives both sides, we obtain y' = 3r² p² 2y-5 integrating the initial conditions c=3 the quadratic formula constant c=0 we have c=1 c=3 Thus, we x=1 y=0

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