91Ó°ÊÓ

The given initial conditions are $y=−10\text{and}dy/dx=0$, when.$t=0$

General solution of first firm is$y=A{e}^{i\mathrm{Ó\neg }t}+B{e}^{−i\mathrm{Ó\neg }t}$

Use the initial condition $y=−10$and $dy/dx=0$at time$t=0$. Using the first firm of general solutionand its derivative

$\begin{array}{c}y=A{e}^{i\mathrm{Ó\neg }t}+B{e}^{−i\mathrm{Ó\neg }t}\\ \frac{dy}{d\mathrm{Ó\neg }}=i\mathrm{Ó\neg }A{e}^{i\mathrm{Ó\neg }t}−i\mathrm{Ó\neg }B{e}^{−i\mathrm{Ó\neg }t}\end{array}$

When apply the initial condition

$\begin{array}{c}−10=A+B\\ 0=i\mathrm{Ó\neg }(A−B)\end{array}$

The value of constant and by above condition,

$\begin{array}{c}A=B=−\frac{10}{2}\\ =−5\end{array}$

Then the particular solution

$\begin{array}{c}y=\frac{10}{2}(e^{i\mathrm{Ó\neg }t}+e^{−i\mathrm{Ó\neg }t})\\ =−10\cos \mathrm{Ó\neg }t.\end{array}$

Apply the initial condition on the third solution and find its first derivative.

$\begin{array}{c}y=c\sin (\mathrm{Ó\neg }t+\mathrm{γ})\\ \frac{dy}{dx}=c\mathrm{Ó\neg }\cos (\mathrm{Ó\neg }t+\mathrm{γ})\end{array}$

Therefore, it gets

$\begin{array}{c}−10=c\sin \mathrm{γ}\text{ }→\text{ }c=\frac{−10}{\sin \mathrm{γ}}\\ 0=c\sin \mathrm{γ}\text{ }→\text{ }\mathrm{γ}=90\\ \mathrm{γ}=90,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}c=−10\end{array}$

Therefore, the particular solution is$y=10\sin (\mathrm{Ó\neg }t+90).$