Question

In the problem 7-10,$x=c_1\cos t+c_2\sin t$is a two-parameter family of solutions of the second-order$x^n+x=0$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions:

$x\left(\frac{\pi }{2}\right)=0,x\text{'}\left(\frac{\pi }{2}\right)=1$

Short answer

The solution of the second- order is_{$x=-\cos t+8\sin t.$}

Step-by-step solution

Definition of Second-order arithmetic

Second-order arithmetic is an axiomatization allowing quantification of sets of numbers.

Solve for constant

Consider the following second order initial value problem.

$x^n+x=0$

Here, it is given that$x=c_1\cos t+c_2\sin t$ is a solution for the above initial value problem.

The objective is to find the solution for the given DE using the initial conditions$x\left(0\right)=-1,x^{\text{'}}\left(0\right)=8.$

Consider the solution of the second order initial value problem,

$x\left(t\right)=c_1\cos t+c_2\sin t$

To find the constant c₁, Use the initial condition, x(0) = -1

x (0) = -1

$\Rightarrow -1=c_1\cos 0+c_2\sin 0$

$-1=c_1\left(1\right)+c_2\left(0\right)$

c₁ = -1

Solve for the equation

To find the constant c₂, differentiate $x\left(t\right)=c_1\cos t+c_2\sin t$with respect to 't'.

$x^{\text{'}}\left(t\right)=-c_1\sin t+c_2\cos t$

Use the initial condition, x'(0) = 8

$\Rightarrow 8=-c_1\sin 0+c_2\cos 0$

$8=-c_1\left(0\right)+c_2\left(1\right)$

= c₂

Put the values c₁= -1 and c₂=8 in $x\left(t\right)=c_1\cos t+c_2\sin t$, then$x\left(t\right)=-\cos t+8\sin t$

Therefore, the required solution is$x\left(t\right)=-\cos t+8\sin t$