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Wite a differential equation in which y^2= 4(t + 1) is a solution.

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Answer: dy/dt 2y = 4
Explanation:
The differential equation can be found by taking the derivative of y with respect to t of both sides of this equation.
Derivative of y^2 becomes 2y * dy/dt.
4(t+1) can be distributed to equal 4t + 4.
Derivative of 4t + 4 becomes 4.
Therefore, a possible differential equation is dy/dt * 2y = 4. Can check by separating and integrating.
User TehDorf
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