Euler's method for coupled systems Problem set A Worked solutions

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

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(x_n-y_n+t_n\right)\), \(y_{n+1}=y_n+0.1\left(2x_n-t_n\right)\)
• First step: t = 0.1, x = 0.9, y = 2.2
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 0.9, 2.2), (0.2, 0.78, 2.37), (0.3, 0.641, 2.506), (0.4, 0.4845, 2.6042), (0.5, 0.3125, 2.6611)
• x(0.5) ≈ 0.313, y(0.5) ≈ 2.66

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

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(x_ny_n\right)\), \(y_{n+1}=y_n+0.1\left(x_n-y_n^2\right)\)
• First step: t = 0.1, x = 1.1, y = 1
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.1, 1), (0.2, 1.21, 1.01), (0.3, 1.3322, 1.029), (0.4, 1.4693, 1.0563), (0.5, 1.6245, 1.0917)
• x(0.5) ≈ 1.62, y(0.5) ≈ 1.09

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

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(y_n\cos t_n\right)\), \(y_{n+1}=y_n+0.1\left(-x_n\sin t_n\right)\)
• First step: t = 0.1, x = 2.1, y = 1
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 2.1, 1), (0.2, 2.1995, 0.979), (0.3, 2.2955, 0.9353), (0.4, 2.3848, 0.8675), (0.5, 2.4647, 0.7746)
• x(0.5) ≈ 2.46, y(0.5) ≈ 0.775

Two species of plankton compete in a pond. Their numbers \(x\) and \(y\) (in thousands) at time \(t\) days satisfy \(\dfrac{dx}{dt}=0.8x-0.1x^2-0.3xy\) and \(\dfrac{dy}{dt}=0.6y-0.1y^2-0.1xy\), with \(x(0)=2\) and \(y(0)=3\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(0.8x_n-0.1x_n^2-0.3x_ny_n\right)\), \(y_{n+1}=y_n+0.1\left(0.6y_n-0.1y_n^2-0.1x_ny_n\right)\)
• First step: t = 0.1, x = 1.94, y = 3.03
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.94, 3.03), (0.2, 1.8812, 3.0612), (0.3, 1.8236, 3.0936), (0.4, 1.767, 3.1271), (0.5, 1.7113, 3.1617)
• x(0.5) ≈ 1.71 thousand, y(0.5) ≈ 3.16 thousand

Two tanks hold salt solution. The mass of salt \(x\) kg in tank 1 and \(y\) kg in tank 2 after \(t\) minutes satisfy \(\dfrac{dx}{dt}=2+0.5t-0.4x+0.1y\) and \(\dfrac{dy}{dt}=0.4x-0.6y\), with \(x(0)=1\) and \(y(0)=2\). Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(2+0.5t_n-0.4x_n+0.1y_n\right)\), \(y_{n+1}=y_n+0.1\left(0.4x_n-0.6y_n\right)\)
• First step: t = 0.1, x = 1.18, y = 1.92
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 1.18, 1.92), (0.2, 1.357, 1.852), (0.3, 1.5312, 1.7952), (0.4, 1.7029, 1.7487), (0.5, 1.8723, 1.7119)
• x(0.5) ≈ 1.87 kg, y(0.5) ≈ 1.71 kg

Two lakes receive pollution. The amounts \(x\) and \(y\) (in tonnes) in the lakes at time \(t\) months satisfy \(\dfrac{dx}{dt}=3\sin t-0.5x+0.3y\) and \(\dfrac{dy}{dt}=0.5x-0.4y\), with \(x(0)=0\) and \(y(0)=1\). Work in radians. Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) when \(t=0.5\).

• \(t_{n+1}=t_n+0.1\), \(x_{n+1}=x_n+0.1\left(3\sin t_n-0.5x_n+0.3y_n\right)\), \(y_{n+1}=y_n+0.1\left(0.5x_n-0.4y_n\right)\)
• First step: t = 0.1, x = 0.03, y = 0.96
• From the calculator, (t, x, y) for n = 1 to 5: (0.1, 0.03, 0.96), (0.2, 0.0873, 0.9231), (0.3, 0.1702, 0.8905), (0.4, 0.277, 0.8634), (0.5, 0.4059, 0.8427)
• x(0.5) ≈ 0.406 tonnes, y(0.5) ≈ 0.843 tonnes

The populations of two species of beetle, \(x\) and \(y\) (in thousands), satisfy \(\dfrac{dx}{dt}=x(1-y)\) and \(\dfrac{dy}{dt}=y(x-1)\), with \(x(0)=1.5\) and \(y(0)=0.5\), where \(t\) is in years. 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 \(y(1)\).

   • \(t_{n+1}=t_n+0.25\), \(x_{n+1}=x_n+0.25\left(x_n(1-y_n)\right)\), \(y_{n+1}=y_n+0.25\left(y_n(x_n-1)\right)\)
   • From the calculator, (t, x, y) for n = 1 to 4: (0.25, 1.6875, 0.5625), (0.5, 1.8721, 0.6592), (0.75, 2.0316, 0.8029), (1, 2.1317, 1.01)
   • x(1) ≈ 2.13, y(1) ≈ 1.01

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

   • 2.13 and 1.01
   • x is larger

The system \(\dfrac{dx}{dt}=ax+y+t\), \(\dfrac{dy}{dt}=x-by\) has \(x(0)=2\) and \(y(0)=1\), where \(a\) and \(b\) are constants. Euler's method with \(h=0.1\) gives \(x_1=2.35\) and \(y_1=1.12\).

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

   • x₁ = 2 + 0.1(2a + 1 + 0) = 2.35, so 2a + 1 = 3.5 and a = 1.25
   • y₁ = 1 + 0.1(2 − b) = 1.12, so 2 − b = 1.2 and b = 0.8

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

   • Use t₁ = 0.1, x₁ = 2.35, y₁ = 1.12
   • x₂ = 2.35 + 0.1(1.25 × 2.35 + 1.12 + 0.1) = 2.76575 ≈ 2.77