Question

how do you get from y=Ae^-4x + Be^-6x
to differential eqution f''(x) +10f(x) +24y=0

Answer 1

What do you mean by how do you 'get to it?'

Answer 2

here we have y=f(x)
and hence y' = f'(x) and y''=f''(x)

Answer 3

diiferntiating y we have ( y=Ae^-4x + Be^-6x)
y'=-4Ae^-4x -6Be^-6x
again on differentiating further we have
y''=16Ae^-4x +36Be^-6x

now we need to eliminate A and B from y=Ae^-4x + Be^-6x and y'=-4Ae^-4x -6Be^-6x
and put the values of A and B in y''=16Ae^-4x +36Be^-6x

otherwise
y''+24y=16Ae^-4x +36Be^-6x +24Ae^-4x + 24Be^-6x
y'' +24y= 10(4Ae^-4x +6Be^-6x) but y'=-4Ae^-4x -6Be^-6x
hence y''+24y=-10y'
or we have y''+24y+10y'=0
or f''(x)+10f'(x)+24y=0

Answer 4

Or since \[y=Ae^{mx}+Be^{nx}\]
That means when we solve a quadratic and got \[r=m \text{ or } r=n => r-m=0 \text{ or } r-n=0\] \[=> (r-m)(r-n)=0 => r^2+(-n-m)r+mn=0\]
Which means we solve the differential equation:
\[y''+(-n-m)y'+mny=0\]