Q1ts - the standard procedure, in standard form this is dy...

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Quiz I 1. Find the solution to 2 y dy dx - x = xy 2 satisfying y (0) = - 1. Solution. This equation can be rewritten as dy dx = x ( y 2 + 1 2 y ) Note that this has no equilibrium solutions, and it is separable. So, following the standard procedure Z 2 y 1 + y 2 dy = Z xdx, and ln(1 + y 2 ) = x 2 / 2 + C. If y (0) = - 1, then C = ln2. So y 2 + 1 = e x 2 / 2+ln 2 = 2 e x 2 / 2 and y = - p 2 e x 2 / 2 - 1 . Note that you need the minus sign to make y (0) = - 1.
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2 2. Find the general solution to (1 + x ) dy dx + y = cos x . Solution . You may have noticed that this equation has already been multiplied by an integrating factor, and been able to skip some steps below. If not, following
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Unformatted text preview: the standard procedure, in standard form this is dy dx + 1 1 + x y = cos x 1 + x . So a ( x ) = 1 / (1 + x ), A ( x ) = R a ( x ) dx = ln | 1 + x | and the integrating factor is e A = | 1+ x | = (1+ x ). Either one will work. Using +(1+ x ), you get back to the original equation! So you have (1 + x ) dy dx + y = cos x, d dx (1 + x ) y = cos x, (1 + x ) y = sin x + C. So the general solution is y ( x ) = sin x 1 + x + C 1 + x ....
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Q1ts - the standard procedure, in standard form this is dy...

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