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vova2212 [387]
2 years ago
13

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 solut

ion of the second-order IVP consisting of this differential equation and the given initial conditions. x(π/4) = 2 2 , x'(π/4) = 0
Mathematics
1 answer:
Norma-Jean [14]2 years ago
7 0

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 <em>x</em> (<em>π</em>/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 <em>x'</em> (<em>p</em>/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)}

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