AHL 5.18—Eulers method for 2nd order DEs

Let $y=\frac{dx}{dt}$, so $\frac{d^{2}x}{d{t}^2}=\frac{dy}{dt}$.

The original DE is $\frac{d^{2}x}{d{t}^2}=-3x-2\frac{dx}{dt}$. Replacing $\frac{d^{2}x}{d{t}^2}$ with $\frac{dy}{dt}$ and $\frac{dx}{dt}$ with $y$ gives $\frac{dy}{dt}=-3x-2y$.

No further simplification is needed since the terms are already in $x$ and $y$.

This matches option A exactly.

Swaps the coefficients of $x$ and $y$, giving $-3y-2x$ instead of $-3x-2y$ — a coefficient-swap error.

Has a sign error on the $y$ term, giving $+2y$ instead of $-2y$.

Also swaps coefficients, writing $-2x-3y$ instead of $-3x-2y$.

Only option A preserves both the correct coefficients and correct signs from the original equation.

$y=\frac{dx}{dt}$, so $y_0$ is the initial value of the derivative, and $x_0$ is the initial displacement.

Given $x(0)=2$, this is $x_0=2$. Given $\frac{dx}{dt}(0)=0$, this is $y_0=0$.

$(x_0,y_0)=(2,0)$.

This matches option A.

Swaps $x_0$ and $y_0$, confusing displacement with velocity.

Incorrectly derives $y_0=4$ from the coefficient in the DE, rather than the given derivative condition.

Ignores the given initial displacement entirely, setting both to zero.

Only option A correctly assigns each initial condition to its corresponding variable.