69_pdfsam_math 54 differential equation solutions odd

69_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 2.4 Substituting the initial condition, y = π when x = 1, we find c. ln |1| + 12 π 2 − sin π = c ⇒ c = π2. Therefore, the answer is given implicitly by ln |x| + x2 y 2 − sin y = π 2 . (We also used the fact that at the initial point, (1, π ), x > 0 to skip the absolute value sign in the logarithmic term.) 23. Here M (t, y ) = et y + tet y and N (t, y ) = tet + 2. Thus My (t, y ) = et + tet = Nt (t, y ) and so the equation is exact. To find F (t, y ) we first integrate N (t, y ) with respect to y to obtain F (t, y ) = (tet + 2) dy = (tet + 2)y + h(t), where we have chosen to integrate N (t, y ) because this integration is more easily accomplished. Thus Ft (t, y ) = et y + tet y + h (t) = M (t, y ) = et y + tet y ⇒ h (t) = 0 ⇒ h(t) = C. We will incorporate this constant into the parameter of the solution. Combining these results gives F (t, y ) = tet y + 2y . Therefore, the solution is given by tet y + 2y = C . Solving for y yields y = C/(tet + 2). Now we use the initial condition y (0) = −1 to find the solution that passes through the point (0, −1). Thus y (0) = C = −1 0+2 ⇒ C = −1 2 2 . tet + 2 ⇒ C = −2. This gives us the solution y=− 25. One can check that the equation is not exact (∂M/∂y = ∂N/∂x), but it is separable because it can be written in the form y 2 sin x dx + ⇒ 1−y dy = 0 ⇒ x y−1 dy. x sin x dx = y2 y 2 sin x dx = y−1 dy x 65 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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