Question

In this problem, x = c1 cos t + c2 sin 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(π/4) = 2 2 , x'(π/4) = 0

Answer 1

Differentiate the given solution:

$x=C_1\cos(t)+C_2\sin(t) \implies x'=-C_1\sin(t)+C_2\cos(t)$

Now, given that x (π/4) = √2/2 … (I'm assuming there are symbols missing somewhere) … you have

$\dfrac{\sqrt2}2=C_1\cos\left(\dfrac\pi4\right)+C_2\sin\left(\dfrac\pi4\right)$

$\implies\dfrac1{\sqrt2} = \dfrac{C_1}{\sqrt2}+\dfrac{C_2}{\sqrt2}$

$\implies C_1+C_2=1$

Similarly, given that x' (p/4) = 0, you have

$0=-C_1\sin\left(\dfrac\pi4\right)+C_2\cos\left(\dfrac\pi4\right)$

$\implies 0=-\dfrac{C_1}{\sqrt2}+\dfrac{C_2}{\sqrt2}$

$\implies C_1=C_2$

From this result, it follows that

$C_1+C_2=2C_1=1 \implies C_1=C_2=\dfrac12$

So the particular solution to the DE that satisfies the given conditions is

$\boxed{x=\dfrac12\cos(t)+\dfrac12\sin(t)}$