Differential Equations

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution.

 

1. y''+y=0

 

2. u''+7u=0

 

Solve the given initial value problem.

3. y''+9y=0

 

Would someone mind explaining these problems?

Answer 1

I will answer the first one.  The other two can be solved in a similar manner.

 

y" + y = 0 is a second order linear homogeneous differential equation.  If the auxiliary equation for such an equation has the complex roots a + bi and a - bi, then the general solution of the differential equation is Ae^axcos(bx) + Be^axsin(bx).

 

For the differential equation y" + y = 0, the auxiliary equation is m²+1=0, which  has the roots ±i = 0±i

 

General solution:  y = Ae^0xcosx + Be^0xsinx = Acosx + Bsinx

 

 

Answer 2

You have a homogenous 2nd-order equation.  We use the substitution

 

y = e^rt

 

Then taking the 2nd-derivative of this gives us

 

y'' = r²e^rt

 

 

Substituting into the differential equation, we get

 

r²e^rt + e^rt = 0

 

 

e^rt(r² + 1) = 0

 

 

We get the characteristic equation,

 

r² + 1 = 0

 

The roots are then

 

r = 0 ± i

 

 

y = Ae^t cos(t) + iBe^-t sin(t)

 

 

Answer the next two questions the same way.