Question

a(x)y"+b(x)y'+c(x)y+d(x)=0
Show that if d(x)=0 then y(x) is a solution. Help!!!

Answer 1

If d(x)=0, the differential equation becomes:

a(x)y''(x)+b(x)y'(x)+c(x)y(x)=0, or in an easier notation:

ay''+by'+cy=0 (just remember that everything is a function of x :)

I think you have to specify what kind of y you mean here...
Or is it y=0 (for all x). In that case, y'=0 and y''=0, for all x, so y=0 is a solution.

Answer 2

Yeah it says y(x)=0 is a solution.

Answer 3

YW!

Answer 4

Then it says...Show that assuming y=(lambda)t we are led to the equation...
a(lambda)^2+b(lambda)+c=0

Answer 5

Are \(\lambda\) and t also functions of x?

Answer 6

It says that \[\lambda\] and \[\lambda2\] are solutions to the equation....and that after we solve it, it then becomes \[y=C1e^(\lambda1)t +C2e^(\lambda2)t\]