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solution_2.6

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

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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 infinite 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

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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 1 2 ln 3 = C. Thus the solution to IVP in implicit form is ln 1 + y 1 - y - t 2 = ln 3 .
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