dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1 I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
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Originally Posted by lord12 dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1 I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal? as i am feeling lazy today, i'll assume your solution is correct so far. as far as going from to use the quadratic formula ( , and )
Originally Posted by lord12 dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1 I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal? ... since
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