ODEHard

2nd order ODE — harmonic oscillator

Solve the differential equation y″ + 4y = 0 with initial conditions y(0) = 2, y′(0) = 0.

Given data

y″ + 4y = 0y(0) = 2y'(0) = 0

Steps

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Write the associated characteristic equation.
Write your working — the numerical result must appear in the text (logica)
Solve the characteristic equation: λ² = −4.
Apply y(0) = 2 to find C₁.
Calculate y′(x) and apply y′(0) = 0 to find C₂.

Full worked solution

Write the associated characteristic equation.
$\lambda^2 + 4 = 0$
The characteristic equation is $\lambda^2 + 4 = 0$ . Obtained by substituting $y = e^{\lambda x}$ , which gives $\lambda^2 e^{\lambda x} + 4e^{\lambda x} = 0$ .
Solve the characteristic equation: λ² = −4.
$\lambda = \pm 2i$
$\lambda^2 = -4 \rightarrow \lambda = \pm 2i$ . Complex conjugate roots $\lambda = \pm i\omega$ with $\omega = 2$ give the general solution $y = C_1\cos(2x) + C_2\sin(2x)$ .
Apply y(0) = 2 to find C₁.
$y(0) = C_1 \cdot \cos(0) + C_2 \cdot \sin(0) = C_1 = 2$
$y(0) = C_1\cos(0) + C_2\sin(0) = C_1 = 2$ . Using $\cos(0)=1$ , $\sin(0)=0$ , we find $C_1 = \mathbf{2}$ .
Calculate y′(x) and apply y′(0) = 0 to find C₂.
$y'(0) = -2C_1\sin(0) + 2C_2\cos(0) = 2C_2 = 0$
$y'(x) = -2C_1\sin(2x) + 2C_2\cos(2x)$ . At $x=0$ : $y'(0) = 2C_2 = 0 \rightarrow C_2 = \mathbf{0}$ . The solution is $\mathbf{y(x) = 2\cos(2x)}$ .

Result:y(x) = 2·cos(2x).

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