Question

In the problem $\mathbf{7}\mathbf{-}\mathbf{10}\mathbf{,}\mathbf{}\mathit{x}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$is a two-parameter family of solutions of the second-order ${\mathit{x}}^{\mathbf{n}}\mathbf{+}\mathit{x}\mathbf{=}\mathbf{0}$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions:

_{$\mathit{x}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{1}$}

Short answer

The solution of the second- order isx = -cost.

Step-by-step solution

Definition of Second-order arithmetic

Second-order arithmeticis an axiomatization allowing quantification of sets of numbers.

Step 2:

From, $x=c_1\cos t+c_2\sin t$we obtain$x^{\text{'}}=-c_1\sin t+c_2\cos t$

$x\left(\frac{\pi }{2}\right)=0\Rightarrow 0=c_1\cos \frac{\pi }{2}+c_2\sin \frac{\pi }{2}$

${\text{=c}}_1{\text{(0) +c}}_2\text{(1)}$

= c₂

Therefore c₂= 0

$x^{\text{'}}\left(\frac{\pi }{2}\right)=1\Rightarrow 1=-c_1\sin \frac{\pi }{2}+c_2\cos \frac{\pi }{2}$

${\text{= -c}}_1{\text{(1) +c}}_2\text{(0)}$

= -c₁

Therefore c₁ = -1

Therefore, x = -cost is a solution.