solution_2.6 - Solutions 2.6 2. Since ODE can be reduced to...

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

View Full Document Right Arrow Icon
Solutions 2.6 2. Since ODE can be reduced to y 2 y 0 - 1 = 0 , this ODE is separable. We then and integrate this ODE in separable form: Z y 2 dt - Z 1 dt = C, y 3 3 - t = C. The initial condition yields y (0) 3 3 - 1 = C or 8 3 - 1 = C, hence C = 5 / 3. Hence the solution to IVP in implicit form is y 3 3 - t = 5 / 3 . In explicit form, we have the solution y = (3 t + 5) 1 / 3 . It exists for all t , but the slope becomes inFnite at t = - 5 / 3. 5. Since the ODE can be reduced to y 0 y 3 - t = 0 , this ODE is separable. Integration leads to - 1 2 y 2 - 1 2 t 2 = C, and the initial condition yields - 1 8 = C. 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
Hence the solution to IVP in implicit form is 1 2 y 2 + 1 2 t 2 = 1 8 . In explicit form, we have y = 2 1 - 4 t 2 (choose the plus sign in front of the square root because of the initial condi- tion). This solution exists for - 1 / 2 < t < 1 / 2. 9. Since this ODE can be reduced to y 0 1 - y 2 - t = , this ODE is separable. By integration 1 2 ln 1 + y 1 - y - t 2 2 = C. The initial condition yields
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.

Page1 / 4

solution_2.6 - Solutions 2.6 2. Since ODE can be reduced to...

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