Question

Solve these initial value problems by using Separation of Variables. 23. dy/dt = (y^2 - 1)/(t^2 - 1), y(2) = 2 24. t^2 dy/dt = y - ty, y(-1) = -1 25. dy/dt = (2t + 1)/2y, y(-2) = 1 Extra Credit. Find a solution of t dy/dt = y^2 - y that passes through... 26. (t, y) = (0, 1) 27. (t, y) = (0, 0) 28. (t, y) = (1/2, 1/2)

          Solve these initial value problems by using Separation of Variables.
23. dy/dt = (y^2 - 1)/(t^2 - 1), y(2) = 2
24. t^2 dy/dt = y - ty, y(-1) = -1
25. dy/dt = (2t + 1)/2y, y(-2) = 1
Extra Credit. Find a solution of t dy/dt = y^2 - y that passes through...
26. (t, y) = (0, 1)
27. (t, y) = (0, 0)
28. (t, y) = (1/2, 1/2)

Solve these initial value problems by using Separation of Variables.
23. dy/dt = (y^2 - 1)/(t^2 - 1), y(2) = 2
24. t^2 dy/dt = y - ty, y(-1) = -1
25. dy/dt = (2t + 1)/2y, y(-2) = 1
Extra Credit. Find a solution of t dy/dt = y^2 - y that passes through...
26. (t, y) = (0, 1)
27. (t, y) = (0, 0)
28. (t, y) = (1/2, 1/2)

Added by Gregorio A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Drew Scalzo

10/15/2022

Step 1

Step 1:** Divide both sides by \(t^2\) and \(y\): \(\frac{dy}{dt} \cdot \frac{1}{y} = \frac{1}{t^2} - \frac{1}{t}\) ** Show more…

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Solve these initial value problems by using Separation of Variables. 23. dy/dt = (y^2 - 1)/(t^2 - 1), y(2) = 2 24. t^2 dy/dt = y - ty, y(-1) = -1 25. dy/dt = (2t + 1)/2y, y(-2) = 1 Extra Credit. Find a solution of t dy/dt = y^2 - y that passes through... 26. (t, y) = (0, 1) 27. (t, y) = (0, 0) 28. (t, y) = (1/2, 1/2)