50_pdfsam_math 54 differential equation solutions odd

50_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 At the initial point, x = π/ 4, cos( π/ 4) > 0 and, therefore, we can take µ ( x )=(co s x ) 1 . Multiplying the standard form of the given equation by µ ( x )g ives 1 cos x dy dx + sin x cos 2 x y =2 x d dx ± 1 cos x y ² =2 x 1 cos x y = Z 2 xdx = x 2 + C y =cos x ( x 2 + C ) . From the initial condition, we ±nd C : 15 2 π 2 32 = y ³ π 4 ´ =cos π 4 µ ³ π 4 ´ 2 + C C = π 2 . Hence, the solution is given by y =cos x ( x 2 π 2 ). 23. We proceed similarly to Example 2 on page 52 and obtain an analog of the initial value problem (13), that is, dy dt +5 y =40 e 20 t ,y (0) = 10 . (2.10) Thus P ( t ) 5and µ ( t )=exp (R 5 dt ) = e 5 t . Multiplying the di²erential equation in (2.10) by µ ( t ) and integrating, we obtain e 5 t dy dt +5 e 5 t y =40 e 20 t
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