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