General question regarding solutions to differential equations.

[Wiz14]

My textbook defines a solution to a differential equation to be a function f(x) such that when substituted into the equation gives a true statement.

What I'm confused about are singular solutions.
For example the logistic equation: dP/dt = rP(1-P/K) where r and K are constants.
My textbook says that P = 0 and P = K are solutions(called equilibrium solutions) since plugging in 0 and K for P give you a true statement, but I thought solutions needed to be functions, not constants. For example why is P = 0 a solution but P = 1 isn't?

 

[Curious3141]

My textbook defines a solution to a differential equation to be a function f(x) such that when substituted into the equation gives a true statement.

What I am confused about is singular solutions.
For example the logistic equation: dP/dt = rP(1-P/K) where r and K are constants.
My textbook says that P = 0 and P = K are solutions(called equilibrium solutions) since plugging in 0 and K for P give you a true statement, but I thought solutions needed to be functions, not constants.

Constants can be functions too - they're simply called constant functions, e.g. $f(x) = 2$. They're denoted on the Cartesian plane by a flat (horizontal) line. In this case $P(t) = 0$ and $P(t) = K$ are valid solutions - the meaning is simply that P holds a constant value no matter what the value of t.

For example why is P = 0 a solution but P = 1 isn't?

P = 1 is a solution if and only if K = 1. For any other value of K, the statement is untrue. The LHS is zero (derivative of a constant function is zero), while the RHS is nonzero.

 

[Wiz14]

Constants can be functions too - they're simply called constant functions, e.g. $f(x) = 2$. They're denoted on the Cartesian plane by a flat (horizontal) line. In this case $P(t) = 0$ and $P(t) = K$ are valid solutions - the meaning is simply that P holds a constant value no matter what the value of t.



P = 1 is a solution if and only if K = 1. For any other value of K, the statement is untrue. The LHS is zero (derivative of a constant function is zero), while the RHS is nonzero.

Thank you for your response, it makes sense to me now. I also have another question. After you find the general solution, how do you find these singular solutions? How can you know that you are missing solutions or if you have all of them?

 

[Curious3141]

Thank you for your response, it makes sense to me now. I also have another question. After you find the general solution, how do you find these singular solutions? How can you know that you are missing solutions or if you have all of them?

For simple diff. eq.s you can find them by simple inspection, but the formal theory of finding them in difficult cases, and proving rigorously that they are singular solutions is quite involved, see: http://www.math24.net/singular-solutions-of-differential-equations.html