nagle_differential_equations_ISM_Part14

nagle_differential_equations_ISM_Part14 - Exercises 2.6...

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Unformatted text preview: Exercises 2.6 EXERCISES 2.6: Substitutions and Transformations 2. We can write the equation in the form dx dt = x 2- t 2 2 tx = 1 2 x t- t x , which shows that it is homogeneous. At the same time, it is a Bernoulli equation because it can be written as dx dt- 1 2 t x =- t 2 x- 1 , 4. This is a Bernoulli equation. 6. Dividing this equation by d , we obtain dy d- 1 y = 1 y 1 / 2 . Therefore, it is a Bernoulli equation. It can also be written in the form dy d = y + r y , and so it is homogeneous too. 8. We can rewrite the equation in the form dy dx = sin( x + y ) cos( x + y ) = tan( x + y ) . Thus, it is of the form dy/dx = G ( ax + by ) with G ( t ) = tan t . 10. Writing the equation in the form dy dx = xy + y 2 x 2 = y x + y x 2 and making the substitution v = y/x , we obtain v + x dv dx = v + v 2 dv v 2 = dx x Z dv v 2 = Z dx x - 1 v = ln | x | + C - x y = ln | x | + C y =- x ln | x | + C . In addition, separating variables, we lost a solution v 0, corresponding to y 0. 61 Chapter 2 12. From dy dx =- x 2 + y 2 2 xy =- 1 2 x y + y x , making the substitution v = y/x , we obtain v + x dv dx =- 1 2 1 v + v =- 1 + v 2 2 v x dv dx =- 1 + v 2 2...
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nagle_differential_equations_ISM_Part14 - Exercises 2.6...

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