de-t1-a

Separating variables and applying the initial

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

Separating variables and applying the initial conditions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online