46_pdfsam_math 54 differential equation solutions odd

46_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 7. In this equation, P ( x ) ≡ − 1 and Q ( x ) = e 3 x . Hence the integrating factor µ ( x ) = exp P ( x ) dx = exp ( 1) dx = e x . Multiplying both sides of the equation by µ ( x ) and integrating, we obtain e x dy dx e x y = e x e 3 x = e 2 x d ( e x y ) dx = e 2 x e x y = e 2 x dx = 1 2 e 2 x + C y = 1 2 e 2 x + C e x = e 3 x 2 + Ce x . 9. This is a linear equation with dependent variable r and independent variable θ . The method we will use to solve this equation is exactly the same as the method we use to solve an equation in the variables x and y since these variables are just dummy variables. Thus we have P ( θ ) = tan θ and Q ( θ ) = sec θ which are continuous on any interval not containing odd multiples of π/ 2. We proceed as usual to find the integrating factor µ ( θ ). We have µ ( θ ) = exp tan θ dθ = e ln | cos θ | + C = K · 1 | cos θ | = K | sec θ | , where K = e C . Thus we have
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