88_pdfsam_math 54 differential equation solutions odd

# 88_pdfsam_math 54 differential equation solutions odd - y =...

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Chapter 2 23. This is a Bernoulli equation with n = 2. Dividing it by y 2 and rewriting gives y 2 dy dx 2 x 1 y 1 = x 2 . Making the substitution v = y 1 and hence dv/dx = y 2 dy/dx , the above equation becomes dv dx +2 v x = x 2 . This is a linear equation in v and x . The integrating factor µ ( x )isg ivenby µ ( x )=exp ±Z 2 x dx ² =exp(2ln | x | )= x 2 . Multiplying the linear equation by this integrating factor and solving, we have x 2 dv dx +2 vx = x 4 D x ( x 2 v ) = x 4 x 2 v = Z x 4 dx = x 5 5 + C 1 v = x 3 5 + C 1 x 2 . Substituting y 1 for v in this solution gives a solution to the original equation. Therefore, we Fnd y 1 = x 3 5 + C 1 x 2 y = ± x 5 +5 C 1 5 x 2 ² 1 . Letting C =5 C 1
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Unformatted text preview: y = 5 x 2 x 5 + C . Note: y â‰¡ 0 is also a solution to the original equation. It was lost in the Frst step when we divided by y 2 . 25. In this Bernoulli equation, n = 3. Dividing the equation by x 3 , we obtain x âˆ’ 3 dx dt + 1 t x âˆ’ 2 = âˆ’ t. Now we make a substitution u = x âˆ’ 2 to obtain a linear equation. Since u = âˆ’ 2 x âˆ’ 3 x , the equation becomes âˆ’ 1 2 du dt + 1 t u = âˆ’ t â‡’ du dt âˆ’ 2 t u = 2 t 84...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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