I have an equation y''+y=0 with y(0)=1 and y'(0)=0
It has general solution of y=Asin(x)+Bcos(x) ?
with condition gives 1 = 0 + B so B=1
If so y' = Acos(x)-sin(x)
with condition gives 0 = A - 0 so A=0
Therefore the solution is y = cos(x)
Am I going about this the write way?
And if so does anything change to the general solution if I am solving
y''+4y=0 with y(0)=1 and y'(0)=0 ?
If you have , you have what is known as the characteristic equation.
In this case it becomes , which clearly has no real solutions. You get , which gives
So the general solution becomes .
In general, if you get a solution to the characteristic equation, in the form
, then the homogenous solution is