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Math 2280 Quiz 2 1. Find the general solution of the equation dy = y sin x. dx It is clear that this equation is seperable, so we rewrite it as dy = sin x dx y and integrate to obtain Z dy = y Z sin x dx ⇒ ln y = − cos x + C ⇒ y = De− cos x for any choice of constant D. 2. Find the solution of the differential equation dy = 2xy 2 + 3x2 y 2 dx satisfying the initial condition y(1) = −1. Again we have a separable equation, which is more easily seen when we write it as dy = (2x + 3x2 )y 2 . dx Once we rearrange and integrate we have Z Z −1 dy = (2x + 3x2 ) dx ⇒ −y −1 = x2 + x3 + C ⇒ y = 2 . 2 y x + x3 + C The initial condition implies that −1 = y(1) = −1 ⇒ 2 + C = 1 ⇒ C = −1, 2+C so our particular solution is y(x) = −1 . x2 + x3 − 1 1