Proving Equal Roots in ar^2+br+c=0 with L[e^(rt)]

[asdf1]

If ar^2+br+c=0 has equal roots r1, show that
L[e^(rt)]=a(e^(rt))``+b(e^(rt))`+ ce^(rt)=a{(r-r1)^2}e^(rt)

could someone offer some advice?

 

[HallsofIvy]

If ar^2+br+c=0 has equal roots r1, show that
L[e^(rt)]=a(e^(rt))``+b(e^(rt))`+ ce^(rt)=a{(r-r1)^2}e^(rt)
could someone offer some advice?

The only advice I can give is that you go ahead and do what is shown on the left hand side!
Surely, you know what the first and second derivatives of e^rt are!
And, of course, If r₁ is a double root of ar²+ br+ c= 0, then ar²+ br+ c= a(r- r_{1[/sup])(r-r2).}

 

[asdf1]

ok ~ thanks!