- #1
dionysian
- 53
- 1
In my DE book Zill 7th ed it says that a solution to the equation
[tex] ay^{''} + by^{'} + cy = 0
[/tex]
can be found by "trying" a solution [tex] y = e^{mx} [/tex]
I then see how you take the first and second derivative of [tex] y = e^{mx} [/tex] and plug it into the equation and get the auxilary equation [tex] am^2 + bm + c = 0 [/tex]. But why in the world do we "try" the solution [tex] y = e^{mx} [/tex] in the first place? I can see that choosing [tex] y = e^{mx} [/tex] conveintly let's us solve for [tex] m [/tex] but how do we know that the solution is of the form [tex] y = e^{mx} [/tex] in the first place.
ps. this is my first post and i don't know if my latex is showing up properly (it doesn't seem to be in the preview of my post). If my latex is not showing up properly can someone please tell me what i am doing wrong. Here is a sample of what i am entering for one of my expressions y = e^{mx}.
[tex] ay^{''} + by^{'} + cy = 0
[/tex]
can be found by "trying" a solution [tex] y = e^{mx} [/tex]
I then see how you take the first and second derivative of [tex] y = e^{mx} [/tex] and plug it into the equation and get the auxilary equation [tex] am^2 + bm + c = 0 [/tex]. But why in the world do we "try" the solution [tex] y = e^{mx} [/tex] in the first place? I can see that choosing [tex] y = e^{mx} [/tex] conveintly let's us solve for [tex] m [/tex] but how do we know that the solution is of the form [tex] y = e^{mx} [/tex] in the first place.
ps. this is my first post and i don't know if my latex is showing up properly (it doesn't seem to be in the preview of my post). If my latex is not showing up properly can someone please tell me what i am doing wrong. Here is a sample of what i am entering for one of my expressions y = e^{mx}.
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