# Checking the Solutions for First Order Differential Equations

Hi all,

I just want to make sure, whether the solution of x(t) and y(t) are correct in t term.

If the solutions are incorrect, it is SymPy fault.

Because if I put t>0 I get x(t) < 0 and I think it should not be possible. x(t) is the horizontal coordinate while y(t) is the vertical coordinate of a projectile motion.

the code:

using SymPy, Plots
@syms t, g, r, A, u, h
@syms v()
@syms w()
@syms x()
@syms y()

# Computing IVP with SymPy
diffeqv = Eq(v(t).diff(t), -r*v(t)); string(diffeqv)
diffeqw = Eq(w(t).diff(t), -g -r*w(t)); string(diffeqw)

# To solve the ODE, pass it and the function to solve with dsolve.
vt = dsolve(diffeqv, v(t))
wt = dsolve(diffeqw, w(t))

sol_vt = rhs(vt)
sol_wt = rhs(wt)

# To solve the Initial Value Problem, initial condition v(0) = u cos (A)
ivp_vt = dsolve(diffeqv, v(t), ics = Dict(v(0) => u*cos(A) ))
ivp_wt = dsolve(diffeqw, w(t), ics = Dict(w(0) => u*sin(A) ))

# Manual way
# Find the constant Câ
c1 = free_symbols(vt)[1]
# To solve the Initial Value Problem, initial condition v(0) = u cos (A)
C1 = solve(sol_vt(t=>0) ~ u*cos(A), c1)
#ivp_vt = sol_vt(c1=>C1[1])

# Part (b)
diffeqx = Eq(x(t).diff(t), rhs(ivp_vt));
diffeqy = Eq(y(t).diff(t), rhs(ivp_wt));

# To solve the ODE, pass it and the function to solve with dsolve.
xt = dsolve(diffeqx, x(t))
yt = dsolve(diffeqy, y(t))

sol_xt = rhs(xt)
sol_yt = rhs(yt)
ivp_xt = dsolve(diffeqx, x(t), ics = Dict(x(0) => 0 ))
ivp_yt = dsolve(diffeqy, y(t), ics = Dict(y(0) => h ))

println("Modeling with First Order Equations with Julia")
println("A baseball is thrown with the initial speed of u and initial angle of elevation is A")
println("v(t) is the horizontal component of the velocity of the thrown baseball")
println("w(t) is the vertical component of the velocity of the thrown baseball")
println("x(t) is the horizontal coordinate of the thrown baseball at time t")
println("y(t) is the vertical coordinate of the thrown baseball at time t")

sympy.pretty_print(ivp_vt)
sympy.pretty_print(ivp_wt)
println("")
println("The horizontal and vertical coordinates : ")
sympy.pretty_print(ivp_xt)
sympy.pretty_print(ivp_yt)


To check via PrettyTables:

using SymPy, Plots, LaTeXStrings, Plots.PlotMeasures, PrettyTables

gr()
u = 125
r = 1/5
h = 3
A = 15
g = 32

#int = range(0, stop=10, length=300)
int = -3Ï:1:0.2Ï
x(t) = (u/r)*cos(A) - (u/r)*exp(-r*t)*cos(A)
# 625*cos(A)-625*exp(-r*t)*cos(A)
y(t) = -(g*t/r) + (g + h*r^2 + r*u*sin(A))/(r^2) - (u*sin(A)/r + (g/r^2))*exp(-r*t)

println(pretty_table(String,
vcat([[string(t) x(t) y(t)] for t in -4:0.5:4]...);


The solutions of x(t) and y(t) from the bookâ solution manual:

I get the desired plot by choosing A=20,45,70

using SymPy, Plots, LaTeXStrings, Plots.PlotMeasures

gr()
u = 125
r = 1/5
h = 3
A1 = 20
A2 = 45
A3 = 70
g = 32

#int = range(0, stop=10, length=300)
int = 0:0.1:2Ï
x1(t) = (u/r)*cos(A1) - (u/r)*exp(-r*t)*cos(A1)
x2(t) = (u/r)*cos(A2) - (u/r)*exp(-r*t)*cos(A2)
x3(t) = (u/r)*cos(A3) - (u/r)*exp(-r*t)*cos(A3)

y1(t) = -(g*t/r) + (g + h*r^2 + r*u*sin(A1))/(r^2) - (u*sin(A1)/r + (g/r^2))*exp(-r*t)
y2(t) = -(g*t/r) + (g + h*r^2 + r*u*sin(A2))/(r^2) - (u*sin(A2)/r + (g/r^2))*exp(-r*t)
y3(t) = -(g*t/r) + (g + h*r^2 + r*u*sin(A3))/(r^2) - (u*sin(A3)/r + (g/r^2))*exp(-r*t)

# note the dot, which means compute x for all values in the int range
plot(x1.(int),y1.(int), ylims=(0,200),
label=L"A=20",
legend=:topright)
plot!(x2.(int),y2.(int), ylims=(0,200),
label=L"A=45",
legend=:topright)
plot!(x3.(int),y3.(int), ylims=(0,200),
label=L"A=70", left_margin=10mm,
legend=:topright)
xlabel!(L"x(t) = \frac{u}{r} \cos \ A (1 - e^{-rt}) ")
ylabel!(L"y(t) = - \frac{gt}{r} + \frac{g + ur \ \sin \ A + hr^{2}}{r^{2}} - (\frac{u}{r}")
#=
@syms t
solve(diff(y(t),t)) # to find extreme point

=#



I think you should realize that cos() etc are working using radians, not degrees.

1 Like

Yes, this reminds me how I should use pi instead of anglesâŠ Thanks.