de-t1-a

# Separating variables and applying the initial

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Separating variables and applying the initial conditions yields: , an implicit solution, or better , and explicit solution alive and well on the whole real line. (b) 4 e 2 x y 2 dx 3 y 2 2 xy dy 0 This varmint is plainly exact. A one-parameter family of solutions is given by , where C is an arbitrary constant. (c) dy dx 1 x y 2 x 2 1 ; y (1) ln(4). The ODE of problem (c) is linear as written with an obvious integrating factor of µ ( t ) e ln( x ) x for x >0. An explicit solution to the IVP is given by y ( x ) ln( x 2 1) ln(2) x for x >0.

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TEST1/MAP2302 Page 2 of 3 ______________________________________________________________________ (d) ( x sec y x y ) dx ( x ) dy 0; y (1) π 6 The ODE is homogeneous. The degree of homogeneity is 1. Look at the " y / x " as a hint. Then write the equation in the form of dy / dx = g ( y / x ) by doing suitable algebra carefully. After setting y = vx , substituting, and doing a bit more algebra, you will end up looking at the separable equation sec( v ) dx -( x ) dv =0 . Separating variables and integrating leads you to .
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Separating variables and applying the initial conditions...

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