Code for solving first-order ode with symbolic initial condition

I am very new to MATLAB, and writing MATLAB code for solving the problem #20 of Section 1.1 in Kreyszig (2011).

Problem #20 is as follows:

Exponentialdecay. Subsonicflight. The efficiency of the engines of subsonic airplanes depends on air pressure and is usually maximum near 35,000 ft. Find the air pressure at this height. Physical information. The rate of change is proportional to the pressure. At 18,000 ft it is half its value at sea level.

I wrote the following code for solving the problem, but get a wrong answer.

clear
syms y(t) k
ode = diff(y) == k*y

ode(t) = 

cond = y(18000) == y(0)/2

cond = 

ySol(t) = dsolve(ode, cond)

ySol(t) = 

0

I got a trivial solution.

How to get a non-trivial solution of this problem?

If I solve the problem with pencil and eraser, the non-trivial solution is .

Answer 1

[Torsten]

syms y(h) k
ode = diff(y) == k*y;
sol_ode_general = dsolve(ode)

sol_ode_general = 

vars = symvar(sol_ode_general)

vars = 

eqn1 = subs(sol_ode_general,vars(2),0) == 1.013e5;                %pressure in Pa, distance in m
eqn2 = subs(sol_ode_general,vars(2),18000*0.3048) == 1.013e5/2;   %pressure in Pa, distance in m
[sol.C_1 sol.k] = solve([eqn1,eqn2],[vars(1),vars(3)]);
sol_ode_specific = subs(sol_ode_general,[vars(1) vars(3)],[sol.C_1,sol.k])

sol_ode_specific = 

Pressure_at_36000_feet = subs(sol_ode_specific,vars(2),36000*0.3048)     % pressure in Pa, distance in m

ans = 

25325

[DH]

Many thanks.

One more question:

Can you tell me why the following code also yields the [ empty sym ] ?

My guess is that the dsolve() function uses only one initial condition for a first-degree ODE.

To me, cond = [y(0) == y0, y(18000) == y0/2] seems to be a single initial condition since this is the same as cond = y(18000) == y(0)/2 . :)

syms y(t) k y0
ode = diff(y) == k*y

ode(t) = 

cond = [y(0) == y0, y(18000) == y0/2]

cond = 

sol = dsolve(ode, cond)

Warning: Unable to find symbolic solution.

 
sol =
 
[ empty sym ]
 

[John D'Errico]

@동현

You cannot use dsolve to solve a problem with one degree of freedom, but then specifying two conditions. This seems to be true, even though one of the conditions would effectively resolve the first.

syms y(t) k y0
ode = diff(y) == k*y

ode(t) = 

ysol = dsolve(ode,y(0) == y0)

ysol = 

So that solves the ODE, as a function of the one initial condition.

Now we can introduce the second condition, to then resolve k.

ksol = solve(subs(ysol,t,18000) == y0/2,k)

ksol = 

[Walter Roberson]

dsolve() is only designed to solve for one initial condition per level of derivative. It constructs the general formula solution(t) == f(t) + C and then takes the time of the first initial condition, such as the 0 of y(0), subsitutes that in to get solution(0) == f(0) + C, substitutes in the known value of the initial conditon, getting y0 = f(0) + C and solves for C, and back-substitutes that to get solution(t) = f(t) + specific_constant . Then when the second initial condition y(18000) is hit, there is no remaining constant to solve for in solution(t) = f(t) + specific_constant -- at least not a constant at the same level, and if y0 happened to involve multiple unknown variables, the second initial condition would not know which one it is intended to solve for.

dsolve() is not designed to specifically look for periodic solutions. If you have unknowns in the equations whose value is to be determined by the multiple initial conditions, then solve for one initial condition using dsolve() and after that step through one initial condition at a time solve() for each appropriate variable.