Numerical Analysis/ODE Exercises

Solve the following differential equations using the given method

Euler's Method

1) Solve differential equation y' = 3t²y in [0,1] using Euler's method with n=10,y(0)=1.

Solution:

We need to find the solution ODE y' = 3t²y using Euler's method.

y_n+1=y_n+hf(t_n,y_n)

We divide time span with number points to find the step size h.

h=tmax-tmin/n=1-0/10=0.1

y₁=y₀+hf(t₀,y₀)

=1.000000+.1x3x0²x1

=1.000000

y₂=y₁+hf(t₁,y₁)

=1.000000+.1x3x0.1²x1.000000

=1.003000

y₃=y₂+hf(t₂,y₂)

=1.003000+.1x3x0.2²x1.003000

=1.0150360

y₄=y₃+hf(t₃,y₃)

=1.01503600+.1x3x0.3²x1.0150360

=1.0424420

y₅=y₄+hf(t₄,y₄)

=1.0424420+.1x3x0.4²x1.0424420

=1.0924792

y₆=y₅+hf(t₅,y₅)

=1.0924792+.1x3x0.5²x1.0924792

=1.1744151

y₇=y₆+hf(t₆,y₆)

=1.1744151+.1x3x0.6²x1.1744151

=1.3012520

y₈=y₇+hf(t₇,y₇)

=1.3012520+.1x3x0.7²x1.3012520

=1.4925360

y₉=y₈+hf(t₈,y₈)

=1.4925360+.1x3x0.8²x1.4925360

=1.7791029

y₁₀=y₉+hf(t₉,y₉)

=1.77910290+.1x3x0.9²x1.7791029

=2.2114249

 

Runge-Kutta Second Order Method (RK2)

2) Solve differential equation y' = 2ty in [0,1] using RK2 method y_n+1=y_n+h/4(k₁+3K₂) where k₁=f(t_n,y_n),k₂=f(t_n+2/3h,y_n+2/3hk₁)with n=5,y(0)=1.

Solution:

We need to find the solution ODE y' = 2ty using RK second order method.

y_n+1=y_n+h/4(k₁+3K₂)

k₁=f(t_n,y_n)

k₂=f(t_n+2/3h,y_n+2/3hk₁)

y₁=y₀+h/4(k₁+3K₂)

k₁=f(t₀,y₀)

=2x0.0x1

=0

k₂=f(t₀+2/3h,y₀+2/3hk₁)

=2x2/3x0.2x1

=0.26667

y₁=y₀+h/4(k₁+3K₂)

=1+0.2/4(0+3x0.26667)

=1.04

y₂=y₁+h/4(k₁+3K₂)

k₁=f(t₁,y₁)

=2x0.2x1.04

=0.416

k₂=f(t₁+2/3h,y₁+2/3hk₁)

=f(0.2+0.13333,1.04+0.0554667)

=0.730304

y₂=y₁+h/4(k₁+3K₂)

=1.04+.2/4(0.416+3x0.730304)

=1.17035

y₃=y₂+h/4(k₁+3K₂)

k₁=f(t₂,y₂)

=2x0.4x1.17035

=0.93648

k₂=f(t₂+2/3h,y₂+2/3hk₁)

=f(0.4+2/3(0.2),1.17035+2/3(0.2)(0.93648))

=1.34316

y₃=y₂+h/4(k₁+3K₂)

=1.17035+0.2/4(0.93648+3(1.34318))

=1.41865

y₄=y₃+h/4(k₁+3K₂)

k₁=f(t₃,y₃)

=f(0.6,1.41865)

= 1.70238

k₂=f(t₃+2/3h,y₃+2/3xhxk₁)

=f(.6+.133333,1.41865+2/3(0.2)(1.70238))

=1.64563

y₄=y₃+h/4(k₁+3K₂)

=1.41865+.2/4(1.72038+3(1.64563))

=1.75061

y₅=y₄+h/4(k₁+3K)

k₁=f(t₄,y₄)

=f(0.8,1.75061)

= 2.800098

k₂=f(t₄+2/3h,y₄+2/3hk₁)

=f(0.8+.133333,1.75061+2/3(0.2)(2.800098))

=3.96472

y₅=y₄+h/4(k₁+3K₂)

=1.75061+.2/4(2.800098+3(3.96472))

=2.48532

y₆=y₅+h/4(k₁+3K₂)

k₁=f(t₅,y₅)

=f(1,2.48532)

= 4.97064

k₂=f(t₅+2/3h,y₅+2/3hk₁)

=f(1+.133333,2.48532+2/3(0.2)(4.97064))

=7.1356

y₆=y₅+h/4(k₁+3K₂)

=2.48532+.2/4(4.97064+3(7.1356))

=3.80420

 

Adams-Bashforth Two Step

3) Solve differential equation y' = 2ty in [0,1] using Two step Adams Bashforth method with n=5,y(0)=1.

Solution:

We need to find the solution ODE y' = 2ty using Two step Adams Bashforth method.

y_n+2 = y_n+1 + 3/2 x hf(t_n+1,y_n+1)-1/2 x hf(t_n,y_n)

Calculating h=1-0/5=0.2

We need two approxmations for calculating y₂

Calculating y₁ using RK second order method.

y₁=1.04

y₂ = y₁ + 3/2 x hf(t₁,y₁)-1/2 x hf(t₀,y₀)

=1.04+3/2(.2)f(0.2,1.04)-1/2(0.2)f(0.0,1)

=1.1648

y₃ = y₂ + 3/2 x hf(t₂,y₂)-1/2 x hf(t₁,y₁)

=1.1648+3/2(.2)f(0.4,1.1648)-1/2(0.2)f(0.2,1.04)

=1.40275

y₄ = y₃ + 3/2 x hf(t₂,y₂)-1/2 x hf(t₂,y₂)

=1.40275+3/2(.2)f(0.6,1.40275)-1/2(0.2)f(0.4,1.1648)

=1.81456

y₅ = y₄ + 3/2 x hf(t₄,y₄)-1/2 x hf(t₃,y₃)

=1.81456+3/2(.2)f(0.8,1.81456)-1/2(0.2)f(0.6,1.40275)

=2.21333

y₆ = y₅ + 3/2 x hf(t₅,y₅)-1/2 x hf(t₄,y₄)

=2.21333+3/2(.2)f(1,2.21333)-1/2(0.2)f(0.8,1.81456)

=3.25098

Runge-Kutta Fourth Order (RK4)

4) Solve the ODE $y'=2ty$ on the interval $[0,1]$, using three steps of the Runge-Kutta 4 method with $n=2$ and $y(0)=1$.

Solution:

The RK4 method is given by

$y_{n+1}=y_{n}+1/6(k_{1}+2k_{2}+2k_{3}+k_{4}),\,$

(1)

where

$k_{1}=hf(t_{,}y_{n}),$

$k_{2}=hf(t_{n}+h/2,y_{n}+k_{1}/2),$

$k_{3}=hf(t_{n}+h/2,y_{n}+k_{2}/2),$ and

$k_{4}=hf(t_{n}+h/2,y_{n}+k_{3}).$

The step size $h$, is found using the interval $[0,1]$ and $n=2,$ and is given by

$h={\frac {(1-0)}{n}}={\frac {(1-0)}{2}}=0.5.$

And we know that $y_{0}=y(0)=1,$ so now we can now use the RK4 formula from Eqn.(1) to find each following iteration.

$y_{1}:$

We must first solve for k₁, k₂, k₃, and k₄.

k₁ = (.5)(2)(0)(1) = 0;

k₂ = (.5)(2)(.25)(1) = .25;

k₃ = (.5)(.25)(1.125) = .140625;

k₄ = (.5)(.5)(1.140625) = .28515625.

Applying these k, and y₀ to Eqn.(1), we find

y₁ = 1 + 1/6(0 + (2)(.25) + (2)(.140625) + .28515625) = 1.177734375.

$y_{2}:$

Similarly, we find k₁, k₂, k₃, and k₄.

k₁ = (.5)(.5)(1.177734375) = .294433593;

k₂ = (.5)(.75)(1.324951172) = .496856689

k₃ = (.5)(.75)(1.42616272) = .534811019;

k₄ = (.5)(1)(1.712545394) = .856272697.

Applying these k, and y₁ to Eqn.(1), we find

y₂ = 1.177734375 + 1/6(.294433593 + (2)(.496856689) + (2)(.534811019) + .856272697) = 1.713407993

$y_{3}:$

Repeating in a similar fashion, we find y₃.

k₁ = (.5)(1)(1.713407993) = .856703996;

k₂ = (.5)(1.25)(2.141759991) = 1.338599995;

k₃ = (.5)(1.25)(2.38270799) = 1.489192494;

k₄ = (.5)(1.5)(3.202600487) = 2.401950365.

Applying these k, and y₂ to Eqn.(1), we find

y₃ = 1.713407993 + 1/6(.856703996 + (2)(1.338599995) = (2)(1.489192494) + 2.401950365) = 3.19911455,

with $y_{3}=3.19911455$ as our final answer.