84_pdfsam_math 54 differential equation solutions odd

84_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 2 5. The given differential equation is not homogeneous due to the e−2x terms. The equation (ye−2x + y 3) dx − e−2x dy = 0 is a Bernoulli equation because it can be written in the form dy/dx + P (x)y = Q(x)y n as follows: dy − y = e2x y 3 . dx The differential equation does not have linear coefficients nor is it of the form y = G(ax + by ). 7. Here, the variables are y and θ. Writing dy y 3 − θy 2 (y/θ)3 − (y/θ)2 =− =− , dθ 2θ 2 y 2(y/θ) we see that the right-hand side is a function of y/θ alone. Hence, the equation is homogeneous. 9. First, we write the equation in the form −3x2 + y 2 y 3 − 3x2 y (y/x)3 − 3(y/x) dy = . = = dx xy − x3 y −1 xy 2 − x3 (y/x)2 − 1 Therefore, it is homogeneous, and we we make a substitution y/x = u or y = xu. Then y = u + xu , and the equation becomes u+x Separating variables and integrating yield u 3 − 3u 2u u2 − 1 2 du =2 −u= − 2 ⇒ du = − dx dx u −1 u −1 u x 2 1 u −1 2 dx u− ⇒ du = − dx ⇒ du = −2 u x u x 12 u − ln |u| = −2 ln |x| + C1 ⇒ u2 − ln u2 + ln(x4 ) = C. ⇒ 2 Substituting back y/x for u and simplifying, we finally get y x 2 u 3 − 3u du =2 . dx u −1 − ln y2 x2 + ln(x4 ) = C y2 x6 ⇒ y2 − 2 = K. x y2 + ln x2 x6 y2 = C, which can also be written as ln 80 ...
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