Determing If Two Functions Satisfy A Differential Equation

[Bashyboy]

Homework Statement

t^2y" + 5ty' + 4y = 0

Possible solutions:

y_1(t) = t^{-2} and y_2(t) = t^{-2} lnt

Homework Equations

The Attempt at a Solution

I was able to verify that y_1 was a solution, by the substituting the function, and its derivatives, into the differential equation, which resulted in the identity 0 = 0, this being true for every t.

However, for the second function, my work yielded the expression 16t^2 lnt +t^3 + 7t^2 = 0. My question is, does this function satisfy the differential equation only if, say, t_o is the value that results in the left side of equation being zero?

 

[Jufro]

Check your derivatives for y2. It comes out to be a solution 0=0. And in the future, a solution is only a solution if it satisfies the equation everywhere (that is not exactly true but I am not worrying about technicalities), not just at a particular point t0.

 

[Bashyboy]

So, you are saying that I need an interval, rather than a collection of points that don't make up an interval, to constitute a solution?

 

[STEMucator]

So, you are saying that I need an interval, rather than a collection of points that don't make up an interval, to constitute a solution?

When you're given a specific solution to a differential equation, that solution is potentially one of many solutions to the equation.

The only time you need to be concerned about the validity of the solution is if there are points of discontinuity. Notice at $t=0$ there's an issue. In fact, $y_1$ and $y_2$ are only solutions for $t>0$ on the real line.

The solution actually represents a particular integral curve from a large collection of integral curves dependent on the constant of integration (when you actually solve the differential equation).

 

[Jufro]

I am not saying that you need any particular interval for the problem at hand.

There are a lot more subtleties at hand that Zondrina points to but the problem itself just wants you to check the derivatives to make sure that the "guessed" solution satisfies the problem.

You need not worry about any interval. For now just try and see if a solution satisfies the equation independent of t. There should be no need for trying to find a t0 to satisfy the equation.

And if you check the derivatives and algebra of the problem you should find that y2(t)= t^(-2) Ln(t) does satisfy the equation i.e. 0=0.