Question reguarding initial value problem ODE

Question

$$\mathrm{x = c_1cos( t) + c_2sin(t)}$$ is a two-parameter family of solutions of the second-order DE $$x'' + x = 0.$$ Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions: $$x(0)=1 $$ $$x'(0)=8$$

I found the derivative

$$x' = - c_1\sin (t) + c_2\cos (t)$$

given $$x(0) = - 1 $$

I solved for $c_1$

$$c_1 = - 1$$ when $x=0$

I then plugged in $c_1$ to $x'(0)$ in order to solve for $c_2$, but $\sin(t)$ at $0$ would be zero.

Plugging in $0$ to $\cos(t)$ would just be $c_2$. The Problem gives

$x'(0)=8$, so if $c_2$ is $c_2(1)$, then how can $c_2$ be equal to $7$?

Answer 1

This type of question generally defines a $2\times2$ linear system of equations. However, in this specific example, it so happens that one of the terms in both equations vanishes when evaluated at $0$, which makes our life easier.

We calculate $$ x(0) = c_1\cos(0)+c_2\sin(0) = c_1. $$ Then, because $c_2\sin(0) =0$, we see immediately that $c_1=1=x(0)$.

Next, just as you calculated, $x'(t) = -c_1\sin(t) + c_2\cos(t)$. Now, in a similar manner to the previous case, $c_1\sin(0) = 0$, so evaluating at $t=0$ gives $$ x'(0) = -c_1\sin(0) + c_2 \cos(0) = c_2.$$

From the condition $x'(0) = 8$, it follows that $c_2 = 8$.

Answer 2

Solution of $$x'' + x = 0$$is $$x(t)=a_1 \cos t+a_2 \sin t$$

Here $x'(t)=-a_1 \sin t+a_2 \cos t$

The given conditions are $x(0)=1 \quad \text{and }\quad x'(0)=8$

$x(0)=1\implies a_1\cos 0 + a_2 \sin 0=1\implies a_1=1$

$x'(0)=8\implies -a_1\sin 0 +a_2 \cos 0 =8\implies a_2 =8$

Hence $x(t)= \cos t+8 \sin t$