Inactive Tutor answered • 10/17/21
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New to Wyzant
dy(2y2 + 1)y = cosxdx; ∫(2y + 1/y)dy = ∫cosxdx; y2 + lny = sinx + C; 1 + ln1 = sin0 + C; C = 1.
Solution of IVP: y2 + lny = sinx + 1
Norman B.
asked • 10/05/21Solve initial value problem...
(dy/dx)=(y cosx)/(1+2y2) , y(0) = 1
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