[?] Solve a system of DEs with initial conditions?

Questions
,
1

So I need to solve a system of 3 DEs, I have 3 unknown functions and 4 initial conditions which is enough to find particular solutions.

I have tried doing it myself and I came up with a solution:

SqfY5rNdUXQ.jpg
SqfY5rNdUXQ.jpg720×960 156 KB

Then I tried using the DSolve[] function but it keeps returning an error.

ScreenShot2018-06-05at10.33.49PM.png
ScreenShot2018-06-05at10.33.49PM.png3100×1580 112 KB

What’s the problem? How do I do it?

2

In[1]:= Needs[“VariationalMethods`”]
L = z’^2 - 2*(xCos[x] + Sin[x])z’[x] - y[x]^2 +
l[x](z[x] - y’[x] - xSin);
eq1 = EulerEquations[L, y, x];
eq2 = EulerEquations[L, z, x];
eq3 = {-x*Sin + z - y’ == 0, y[0] == 1, y[Pi/2] == 0,
z[0] == 0, z[Pi/2] == -1};

In[6]:= s = DSolve[{eq1, eq2, eq3}, {l, y, z}, x]

Out[6]= {{l -> 
   Function[{x}, -(1/8) E^(-(\[Pi]/2) - 
      x) (-2 E^(\[Pi]/2) \[Pi] + 2 E^\[Pi] \[Pi] + 2 E^(2 x) \[Pi] + 
       2 E^(\[Pi]/2 + 2 x) \[Pi] - 4 E^(\[Pi]/2 + x) Cos[x] + 
       2 E^x \[Pi] Cos[x] + 2 E^(\[Pi] + x) \[Pi] Cos[x] + 
       4 E^(\[Pi]/2 + x) Cos[x] Cos[2 x] - 
       16 E^(\[Pi]/2 + x) Sin[x] - 2 E^x \[Pi] Sin[x] - 
       4 E^(\[Pi]/2 + x) \[Pi] Sin[x] + 
       2 E^(\[Pi] + x) \[Pi] Sin[x] - 2 E^(\[Pi]/2) Cos[x] Sin[x] + 
       2 E^(\[Pi]/2 + 2 x) Cos[x] Sin[x] + 
       4 E^(\[Pi]/2 + x) Cos[x] Sin[x]^2 + E^(\[Pi]/2) Sin[2 x] - 
       E^(\[Pi]/2 + 2 x) Sin[2 x] + 
       2 E^(\[Pi]/2 + x) Sin[x] Sin[2 x])], 
  y -> Function[{x}, -(1/16) E^(-(\[Pi]/2) - 
      x) (2 E^(\[Pi]/2) \[Pi] - 2 E^\[Pi] \[Pi] + 2 E^(2 x) \[Pi] + 
       2 E^(\[Pi]/2 + 2 x) \[Pi] - 16 E^(\[Pi]/2 + x) Cos[x] - 
       2 E^x \[Pi] Cos[x] - 4 E^(\[Pi]/2 + x) \[Pi] Cos[x] + 
       2 E^(\[Pi] + x) \[Pi] Cos[x] - 4 E^(\[Pi]/2 + x) Sin[x] - 
       2 E^x \[Pi] Sin[x] - 2 E^(\[Pi] + x) \[Pi] Sin[x] + 
       2 E^(\[Pi]/2) Cos[x] Sin[x] + 
       2 E^(\[Pi]/2 + 2 x) Cos[x] Sin[x] + 
       4 E^(\[Pi]/2 + x) Cos[x]^2 Sin[x] - 
       4 E^(\[Pi]/2 + x) Cos[2 x] Sin[x] - E^(\[Pi]/2) Sin[2 x] - 
       E^(\[Pi]/2 + 2 x) Sin[2 x] + 
       2 E^(\[Pi]/2 + x) Cos[x] Sin[2 x])], 
  z -> Function[{x}, -(1/16) E^(-(\[Pi]/2) - 
      x) (-2 E^(\[Pi]/2) \[Pi] + 2 E^\[Pi] \[Pi] + 2 E^(2 x) \[Pi] + 
       2 E^(\[Pi]/2 + 2 x) \[Pi] + 4 E^(\[Pi]/2 + x) Cos[x] - 
       2 E^x \[Pi] Cos[x] - 2 E^(\[Pi] + x) \[Pi] Cos[x] - 
       4 E^(\[Pi]/2 + x) Cos[x] Cos[2 x] + 
       16 E^(\[Pi]/2 + x) Sin[x] + 2 E^x \[Pi] Sin[x] + 
       4 E^(\[Pi]/2 + x) \[Pi] Sin[x] - 
       2 E^(\[Pi] + x) \[Pi] Sin[x] - 16 E^(\[Pi]/2 + x) x Sin[x] - 
       2 E^(\[Pi]/2) Cos[x] Sin[x] + 
       2 E^(\[Pi]/2 + 2 x) Cos[x] Sin[x] - 
       4 E^(\[Pi]/2 + x) Cos[x] Sin[x]^2 + E^(\[Pi]/2) Sin[2 x] - 
       E^(\[Pi]/2 + 2 x) Sin[2 x] - 
       2 E^(\[Pi]/2 + x) Sin[x] Sin[2 x])]}}
3

Thanks a lot, Alexander!