Differential Equation........

Chris's characteristic equation results from assuming a solution of the form y(t) = ke^rt. This results in:

r²y + y = 0, which, to be true for all t, means r² + 1 = 0, so r = i and -i.  This means the general solution can be written as the sum: y = k₁e^it + k₂e^-it.  Turning this into trig functions we get y = Asin(t) + Bcos(t).  (The k's, A and B are arbitrary constants thus far, of course).

Solve ( d^2 y(x))/( dx^2) + y(x) = 0, such that y(0) = 2 and y'(0) = 1:
Assume a solution will be proportional to e^(λ x) for some constant λ.
Substitute y(x) = e^(λ x) into the differential equation:
( d^2 )/( dx^2)(e^(λ x)) + e^(λ x) = 0
Substitute ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x):
λ^2 e^(λ x) + e^(λ x) = 0
Factor out e^(λ x):
(λ^2 + 1) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
λ^2 + 1 = 0
Solve for λ:
λ = i or λ = -i
The roots λ = ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) = c_1 e^(i x) + c_2/e^(i x)
Apply Euler's identity e^(α + i β) = e^α cos(β) + i e^α sin(β):
y(x) = c_1 (cos(x) + i sin(x)) + c_2 (cos(x) - i sin(x))
Regroup terms:
y(x) = (c_1 + c_2) cos(x) + i (c_1 - c_2) sin(x)
Redefine c_1 + c_2 as c_1 and i (c_1 - c_2) as c_2, since these are arbitrary constants:
y(x) = c_1 cos(x) + c_2 sin(x)
Solve for the unknown constants using the initial conditions:
Compute ( dy(x))/( dx):
( dy(x))/( dx) = ( d)/( dx)(c_1 cos(x) + c_2 sin(x))
 = -(c_1 sin(x)) + c_2 cos(x)
Substitute y(0) = 2 into y(x) = c_2 sin(x) + c_1 cos(x):
c_1 = 2
Substitute y'(0) = 1 into ( dy(x))/( dx) = c_2 cos(x) - c_1 sin(x):
c_2 = 1
Solve the system:
c_1 = 2
c_2 = 1
Substitute c_1 = 2 and c_2 = 1 into y(x) = c_2 sin(x) + c_1 cos(x):
Answer: |y(x) = 2 cos(x) + sin(x)

y'' + y = 0, y(0)=2, y'(0)=1

Let the characteristic equation be  :  r^2   + 1    =  0

The complex roots are i and − i

The general solution to the differential equation is then......:

y.(t)  = c₁cos(t)  + c₂sin(t)

And since  y(0)=2, we have

2 = c₁cos(0)  + c₂sin(0) 

2 = c₁

And   y'(t)  = −2sin((t) + c₂cos((t)

And  y'(0)=1   .....so.....

y'(0)  = −2sin(0) + c₂cos(0)  =  1  .→ c₂ = 1

So....the specific solution is :

y.(t)  = 2cos(t)  + sin(t)

Melody......see a mehod of solving these "complex root"  form equations here :

http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

See if this helps :  http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx

All physicists will immediately recognise this differential equation as that describing simple harmonic motion, for which the general solution is y(t) = Asin(wt) + Bcos(wt) where A and B are amplitudes (determined by the initial conditions) and w is frequency:

.

Solve the following differential equation:
y'' + y = 0,

y(0)=2,

y'(0)=1.

\(\begin{array}{|rcll|} \hline y(x)'' + y(x) &=& 0 \\ y(x)'' &=& -y(x) \\ \hline \end{array}\)

We search a function which arises twice derived again, but with a negative algebraic sign.

There occurs very fast sin (x) or cos (x).

Example 1:

\(\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) \\ y'(x) &=& c_1 \cdot \cos(x) \\ y''(x) &=& -c_1 \cdot \sin(x) \\\\ \Rightarrow y''(x) &=& -y(x) \\ \hline \end{array}\)

Example 2:

\(\begin{array}{|rcll|} \hline y(x) &=& c_2 \cdot \cos(x) \\ y'(x) &=& -c_2 \cdot \sin(x) \\ y''(x) &=& -c_2 \cdot \cos(x) \\\\ \Rightarrow y''(x) &=& -y(x) \\ \hline \end{array}\)

The solution exists of a functional family and becomes unequivocal by initial conditions.

\(\begin{array}{|rcll|} \hline y_{family}(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \\ y'(x) &=& c_1 \cdot \cos(x) - c_2 \cdot \sin(x) \\ y''(x) &=& -c_1 \cdot \sin(x) - c_2 \cdot \cos(x) \\ &=& - (c_1 \cdot \sin(x) + c_2 \cdot \cos(x)) \\ &=& -y(x) \ \checkmark \\ \hline \end{array}\)

computation of c₁ and c₂

\(\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \quad & | \quad y(0) = 2 \\ 2 &=& c_1 \cdot \sin(0) + c_2 \cdot \cos(0) \\ 2 &=& 0 + c_2 \cdot 1 \\ c_2 &=& 2 \\\\ y'(x) &=& c_1 \cdot \cos(x) - c_2 \cdot \sin(x) \quad & | \quad y'(0) = 1 \\ 1 &=& c_1 \cdot \cos(0) - c_2 \cdot \sin(0) \\ 1 &=& c_1 \cdot 1 - 0 \\ c_1 &=& 1 \\ \hline \end{array}\)

The function y(x):

\(\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \quad & | \quad c_1 = 1 \quad c_2 = 2 \\\\ y(x) &=& 1 \cdot \sin(x) + 2 \cdot \cos(x) \\ \mathbf{y(x)} & \mathbf{=} & \mathbf{ \sin(x) + 2 \cdot \cos(x) } \\ \hline \end{array}\)

Thanks  Alan and Heureka,

I recognised this as simple harmonic motion and I have some intrinsical understanding of your starting points.

I just have no understanding of Chris's first line that  is  r^2  + 1 =0  I just can't see the relevance but it doesn't matter becasue the line can be skipped.  I will look at the sites Chris has sent me too and also look at what our guest (thanks guest) did them maybe it will twig with me :)

Thanks Alan :)