Thread: 2nd order DE

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 ?

Thank you!

Originally Posted by Bushy

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?

Yes!

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 ?

Yes -

Originally Posted by Danny

Yes -

Is this because in general differentiating sin(2x) and cos(2x) twice will give me -4sin(2x) and -4cos(2x) ?

Hi!

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

Thank you, this makes sense.