Solve initial value problem...

Question

solve initial value problem y''-2y'+2y=0, y(0)=2, y'(0)=0

Answer 1

The characteristic equation of the differential equation

y''+2y+2y=0 (Eq. 1)

is

r²+2r+2=0. (Eq. 2)

The solutions of (Eq. 2( are the conjugate cimplex numbers

r₁=a+bi = -(-2 +√(2²-4(1)(2))/2=1+i

and

r₂=a-bi==1-i

The general solution of (Eq. 1) is therefore

y(x)=(C₁cos(bx)+C₂sin(bx))e^ax=(C₁cos(x)+C₂sin(x))e^x, (Eq. 3)

where C₁ and C₂ are real constants that can be found using the initial conditions.

Since

y'(x)=(C₁cos(x)+C₂sin(x))'e^x+(C₁cos(x)+C₂sin(x))(e^x)'

=(-C₁sin(x)+C₂coss(x))e^x+(C₁cos(x)+C₂sin(x))e^x

=[C₁(cos(x)-sin(x))+C₂(cos(x)+sin(x))]e^x (Eq. 4)

Using the initial conditions together with (Eq. 3() and (Eq. 4) we get the system equations in C₁ and C₂

y(0)=C₁=2 (S1)

y'(0)=C₁+C₂=0 (S2)

Plugging (S1) in (S2), we get C₂ =-2

It follows that the solution to the initial value problem is

y(x)=(2cos(x)-2sin(x))e^x=2)=(cos(x)-sin(x))e^x=