(D^2-3D+2)y=0,y(0)=1,y(0)=1

Solution for (D^2-3D+2)y=0,y(0)=1,y(0)=1 equation:

(^2-3+2)D=0.D(0)=1.D(0)=1

We move all terms to the left:

(^2-3+2)D-(0.D(0))=0

We add all the numbers together, and all the variables

(^2-3+2)D-0=0

We add all the numbers together, and all the variables

(^2-3+2)D=0

We multiply parentheses

D^2-3D+2D=0

We add all the numbers together, and all the variables

D^2-1D=0

a = 1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*1}=\frac{0}{2}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*1}=\frac{2}{2}
=1
$