solution_1.2 - Solutions to selected problems in Section...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to selected problems in Section 1.2 1. This D.E. is of order two because the highest derivative in the equation is y 00 . 3. This D.E. is of order one because the highest derivative in the equation is y 0 . (Note: ( y 0 ) 3 6 = y 000 ). 4. This D.E. is of order three. 9(a). Since y ( t ) = Ce t 2 , we have y 0 ( t ) = 2 Cte t 2 . Putting this into the left hand side of the ODE, we have LHS = y 0 ( t ) - 2 ty ( t ) = 2 Cte t 2 - 2 tCe t 2 , which is equal to the RHS = 0. Hence y ( t ) = Ce t 2 is a solution for any C . 9(b). By the initial condition, we have Ce 1 2 = 2 , we have C = 2 /e . 10. From y 000 = 2, we have y 00 = Z y 000 dt = Z 2 dt = 2 t + C 1 , y 0 = Z y 00 dt = Z (2 t + C 1 ) dt = t 2 + C 1 t + C 2 , y = Z y 0 dt = Z ( t 2 + c 1 t + C 2 ) dt = 1 3 t 2 + 1 2 t 2 + C 2 t + C 3 . There are three arbitrary constants the solution process above. 11(a). Since y ( t ) = C 1 sin (2 t ) + C 2 cos (2 t ), we have y 0 ( t ) = 2 C 1 cos (2 t ) - 2 C 2 sin (2 t ) , y 00 ( t ) = - 4 C 1 sin (2 t
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/16/2009 for the course MATH 2214 taught by Professor Edesturler during the Fall '06 term at Virginia Tech.

Page1 / 3

solution_1.2 - Solutions to selected problems in Section...

This preview shows document pages 1 - 2. Sign up to view the full document.

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