Painguy said:Homework Statement
Show that for any linear equation of the form
[itex]\frac{dy}{dx}[/itex] + P(x)y = 0
if y(x) is a soltuion, then for any constant C the function Cy(x) is also a solution
Homework Equations
The Attempt at a Solution
(dy/dx) + P(x)y=0
∫P(x)y=∫0
y=-C/P(x)
C(dy/dx) + CP(x)y=0
C((dy/dx) + P(x)y)=0
∫P(x)y=∫0
y=-C/P(x)
This all seems a little to simple. Is what I did correct?
No, it isn't correct at all. That isn't how you solve this DE and you don't need to solve it anyway. To show that Cy satisfies the DE, plug it into the DE and verify it works. You will need to use the fact that y solves it.Simon Bridge said:Yes. In fact what you did was more complicated than it needs to be: you don't need the integration.
... the 2nd line does not follow from the 1st, and the 3rd does not follow from the 2nd.(dy/dx) + P(x)y=0
∫P(x)y=∫0
y=-C/P(x)
A linear differential equation is an equation that involves a function and its derivatives, with the highest derivative having a power of 1. It can be written in the form y' + p(x)y = g(x), where y is the function, y' is its derivative, p(x) is a function of x, and g(x) is a function of x.
To show that a function is a solution to a linear DE, you need to substitute the function and its derivatives into the DE and verify that the resulting equation is true. This means that the function satisfies the original DE.
The process for solving a linear DE involves finding the general solution, which is the most general form of the solution that satisfies the DE. This is done by separating variables, integrating, and then solving for the constant of integration. The general solution will also include an arbitrary constant, and this can be used to find a particular solution that satisfies any additional initial conditions.
Yes, a function can be a solution to a linear DE without satisfying the initial conditions. The general solution will include an arbitrary constant, which can be used to find a particular solution that satisfies the initial conditions. However, if the initial conditions are not satisfied, the solution may not be unique.
To verify that a given function is the general solution to a linear DE, you can substitute the function into the DE and ensure that the resulting equation is true. Additionally, you can differentiate the function and verify that it satisfies the original DE. If the function satisfies both of these conditions, it can be considered the general solution to the linear DE.