Euler's method for coupled systems Problem set B One question at a time

The variables \(x\) and \(y\) satisfy \(\dfrac{dx}{dt}=y-2t\) and \(\dfrac{dy}{dt}=x+y-t\), with \(x(0)=2\) and \(y(0)=−1\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(y_n-2t_n\right)\), \(y_{n+1}=y_n+0.1\left(x_n+y_n-t_n\right)\)
• First step: t = 0.1, x = 1.9, y = −0.9
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.9, −0.9), (0.2, 1.79, −0.81), (0.3, 1.669, −0.732), (0.4, 1.5358, −0.6683), (0.5, 1.389, −0.6216)
• x(0.5) ≈ 1.39, y(0.5) ≈ −0.622

The variables \(x\) and \(y\) satisfy \(\dfrac{dx}{dt}=x^2-y\) and \(\dfrac{dy}{dt}=xy-1\), with \(x(0)=1\) and \(y(0)=0.5\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(x_n^2-y_n\right)\), \(y_{n+1}=y_n+0.1\left(x_ny_n-1\right)\)
• First step: t = 0.1, x = 1.05, y = 0.45
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.05, 0.45), (0.2, 1.1153, 0.3972), (0.3, 1.1999, 0.3416), (0.4, 1.3097, 0.2825), (0.5, 1.453, 0.2195)
• x(0.5) ≈ 1.45, y(0.5) ≈ 0.220

The variables \(x\) and \(y\) satisfy \(\dfrac{dx}{dt}=y+e^{-2t}\) and \(\dfrac{dy}{dt}=x\sin t-y\), with \(x(0)=0.5\) and \(y(0)=1\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(y_n+e^{-2t_n}\right)\), \(y_{n+1}=y_n+0.1\left(x_n\sin t_n-y_n\right)\)
• First step: t = 0.1, x = 0.7, y = 0.9
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 0.7, 0.9), (0.2, 0.8719, 0.817), (0.3, 1.0206, 0.7526), (0.4, 1.1507, 0.7075), (0.5, 1.2664, 0.6816)
• x(0.5) ≈ 1.27, y(0.5) ≈ 0.682

A lake has prey fish \(x\) and predator fish \(y\) (in hundreds). At time \(t\) months, \(\dfrac{dx}{dt}=0.9x-0.3xy\) and \(\dfrac{dy}{dt}=0.25xy-0.5y\), with \(x(0)=4\) and \(y(0)=2\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(0.9x_n-0.3x_ny_n\right)\), \(y_{n+1}=y_n+0.1\left(0.25x_ny_n-0.5y_n\right)\)
• First step: t = 0.1, x = 4.12, y = 2.1
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 4.12, 2.1), (0.2, 4.2312, 2.2113), (0.3, 4.3314, 2.3346), (0.4, 4.4178, 2.4707), (0.5, 4.488, 2.6201)
• x(0.5) ≈ 4.49 hundred, y(0.5) ≈ 2.62 hundred

Two rooms are heated. Their temperatures above the outside, \(x\) and \(y\) in \(^\circ\text{C}\), at time \(t\) hours satisfy \(\dfrac{dx}{dt}=3+t-0.5x+0.3y\) and \(\dfrac{dy}{dt}=0.3x-0.4y\), with \(x(0)=4\) and \(y(0)=1\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(3+t_n-0.5x_n+0.3y_n\right)\), \(y_{n+1}=y_n+0.1\left(0.3x_n-0.4y_n\right)\)
• First step: t = 0.1, x = 4.13, y = 1.08
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 4.13, 1.08), (0.2, 4.2659, 1.1607), (0.3, 4.4074, 1.2422), (0.4, 4.5543, 1.3248), (0.5, 4.7063, 1.4084)
• x(0.5) ≈ 4.71 °C, y(0.5) ≈ 1.41 °C

Two greenhouses are heated by a daily cycle. Their temperatures above the outside, \(x\) and \(y\) in \(^\circ\text{C}\), at time \(t\) (in suitable units) satisfy \(\dfrac{dx}{dt}=2\cos t-0.4x+0.1y\) and \(\dfrac{dy}{dt}=0.4x-0.3y\), with \(x(0)=1\) and \(y(0)=0\). Work in radians. Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(2\cos t_n-0.4x_n+0.1y_n\right)\), \(y_{n+1}=y_n+0.1\left(0.4x_n-0.3y_n\right)\)
• First step: t = 0.1, x = 1.16, y = 0.04
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.16, 0.04), (0.2, 1.313, 0.0852), (0.3, 1.4573, 0.1352), (0.4, 1.5915, 0.1894), (0.5, 1.7139, 0.2474)
• x(0.5) ≈ 1.71 °C, y(0.5) ≈ 0.247 °C

The numbers of two kinds of bacteria, \(x\) and \(y\) (in millions), satisfy \(\dfrac{dx}{dt}=x(2-y)\) and \(\dfrac{dy}{dt}=y(x-2)\), with \(x(0)=3\) and \(y(0)=1\), where \(t\) is in hours. Use Euler's method with \(h=0.25\) to estimate \(x\) and \(y\) when \(t=1\).

1. (a) Find the estimates of \(x(1)\) and \(y(1)\).

   • \(t_{n+1}=t_n+0.25\), \(x_{n+1}=x_n+0.25\left(x_n(2-y_n)\right)\), \(y_{n+1}=y_n+0.25\left(y_n(x_n-2)\right)\)
   • From the calculator, (t, x, y) for n = 1 to 4: (0.25, 3.75, 1.25), (0.5, 4.4531, 1.7969), (0.75, 4.6793, 2.8989), (1, 3.6278, 4.8406)
   • x(1) ≈ 3.63, y(1) ≈ 4.84

2. (b) Which of the two is larger at \(t=1\)?

   • 3.63 and 4.84
   • y is larger

The system \(\dfrac{dx}{dt}=ay-x+t\), \(\dfrac{dy}{dt}=bx+y\) has \(x(0)=1\) and \(y(0)=2\), where \(a\) and \(b\) are constants. Euler's method with \(h=0.1\) gives \(x_1=1.3\) and \(y_1=2.5\).

1. (a) Find \(a\) and \(b\).

   • x₁ = 1 + 0.1(2a − 1 + 0) = 1.3, so 2a − 1 = 3 and a = 2
   • y₁ = 2 + 0.1(b + 2) = 2.5, so b + 2 = 5 and b = 3

2. (b) Find \(x_2\).

   • Use t₁ = 0.1, x₁ = 1.3, y₁ = 2.5
   • x₂ = 1.3 + 0.1(2 × 2.5 − 1.3 + 0.1) = 1.68