Second order DEs via Euler's method practice questions.

The second order equation y'' = -y + 2y' is written as a first order system with v = y'. Determine the correct system.

• y' = v, v' = -y + 2v
• y' = v, v' = y - 2v
• y' = -v, v' = -y + 2v
• y' = v, v' = -y - 2v

Answer: y' = v, v' = -y + 2v

Let v = y'. Then y' = v and v' = y'' = −y + 2v, reproducing the original equation.

For y'' = -4y with y(0) = 1, y'(0) = 0, one Euler step of size h = 0.1 is taken (v = y'). Calculate y at t = 0.1.

• 0.96
• 1.00
• 1.04
• 0.60

Answer: 1.00

y₁ = y₀ + h·v₀ = 1 + 0.1×0 = 1.00. The velocity is zero at the start, so y is unchanged after one step.

For y'' = -4y with y(0) = 1, y'(0) = 0 and h = 0.1 (v = y'), determine v at t = 0.1 after one Euler step.

• 0
• -0.4
• 0.4
• -4

Answer: -0.4

v₁ = v₀ + h·y''₀ = 0 + 0.1×(−4×1) = −0.4, since y''₀ = −4y₀ = −4.

A projectile model gives y'' = -9.8 with y(0) = 0, y'(0) = 20. Using Euler with h = 0.5 (v = y'), calculate y at t = 0.5.

• 9.8
• 10.0
• 20.0
• 0.0

Answer: 10.0

y₁ = y₀ + h·v₀ = 0 + 0.5×20 = 10.0 m. (The acceleration only affects v this step, not y.)

For the system y' = v, v' = -y, with y(0) = 2, v(0) = 0 and h = 0.2, determine v at t = 0.2 after one Euler step.

• -0.4
• 0.4
• -2
• 0

Answer: -0.4

v₁ = v₀ + h·(−y₀) = 0 + 0.2×(−2) = −0.4.

In Euler's method for a second order DE, the update formulas are yₙ₊₁ = yₙ + h·vₙ and vₙ₊₁ = vₙ + h·f(tₙ, yₙ, vₙ). Identify what f represents.

• The value of y
• The first derivative y' (the velocity)
• The second derivative y''
• The step size

Answer: The second derivative y''

Since v = y', the rate of change of v is v' = y'', so f gives the second derivative y''.