Differentiability of coefficients for 2nd order DE

[morenogabr]

Homework Statement

Argue that if y=f(x) is a solution to the DE: y'' + p(x) y' + q(x) y = g(x) on the interval (a,b), where p, q, and g are each twice-differentiable, the the fourth derivative of f(x) exists on (a,b).

Homework Equations

The Attempt at a Solution

Its a general problem, I can't determine a relation between the4th deriv of f(x) and the DE. Whydoesnt it just need to be twice differentiable. Thats my best guess. Hints?

 

[mighty2000]

Thank you for your question. I would like to provide an explanation as to why the fourth derivative of f(x) exists on the interval (a,b) if y=f(x) is a solution to the given differential equation.

First, let's start by looking at the definition of a solution to a differential equation. A solution to a DE is a function that satisfies the equation for all values of x in the given interval. In this case, y=f(x) is a solution to the DE: y'' + p(x) y' + q(x) y = g(x) on the interval (a,b).

Now, let's consider the fact that p(x), q(x), and g(x) are each twice-differentiable on the interval (a,b). This means that their first and second derivatives exist and are continuous on the interval. Therefore, the third derivative of y=f(x) (which is y''') also exists and is continuous on (a,b) by the chain rule of differentiation.

Next, we can use the given DE to find an expression for the fourth derivative of y=f(x). Taking the derivative of both sides of the equation, we get:

y'''' + p(x) y''' + q(x) y'' = g'(x)

Since we know that y''', p(x), q(x), and g'(x) are all continuous on (a,b), this means that the fourth derivative of y=f(x) also exists and is continuous on (a,b).

In conclusion, since y=f(x) is a solution to the given DE and p(x), q(x), and g(x) are each twice-differentiable, we can conclude that the fourth derivative of f(x) exists on the interval (a,b). I hope this helps to clarify the relation between the fourth derivative of f(x) and the given DE.