How do I find the particular solution of y''(x) +5y'(x)-6y=0, x=0, y=4, y'(x) =11?

Answer 1

There is no particular solution per se since the DE = 0. If y'(0) = 11, the general solution is

y(x) = 5e^x - e^-6x

Answer 2

You solve the DE y(x)=c1e^-3x+c2e^-2x so y'(x)=-3c1e^-3x-2c2e^-2x

then sub the initial conditions in y(0)=4=c1+c2 and y'(0)=11=-3c1-2c2 and solve for c1 and c2

Got it?

Jim