solve a partial differential equation

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Yuchi Kang
Yuchi Kang 2018-10-20
编辑： Stephan 2018-10-21
Hi！ I want to solve a second partial differential equation.
The general solution is
Which command should I use to get the general solution? I think dsolve doesn't work well. Thanks for helping me.

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Stephan
Stephan 2018-10-21
Hi,
the following code solves the equation with the initial conditions:
y(0)=0
dy/dx(0)=0
Then it rewrites the result in terms of sin ans cos:
syms y(x) x E I N P c l Dyt(x)
eqn = E*I*diff(y,x,2)+N*y == -P*c/l*x;
[eqns, vars] = reduceDifferentialOrder(eqn,y);
conds = [y(0)==0,Dyt(0)==0];
sol = dsolve(eqns,conds);
sol_sin = simplify(rewrite(sol.y,'sin'))
I think that was the question.
Best regards
Stephan
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Stephan
Stephan 2018-10-21
编辑：Stephan 2018-10-21
yes - this is what subs is for - i would suggest to this before reducing the differential order:
syms y(x) x E I N P c l Dyt(x) k
eqn = diff(y,x,2)+N/E*I*y == -P*c/(l*E*I)*x;
eqn = subs(eqn,(N/E*I),k^2)
[eqns, vars] = reduceDifferentialOrder(eqn,y);
conds = [y(0)==0,Dyt(0)==0];
sol = dsolve(eqns,conds)
sol_sin = rewrite(sol.y,'sin')
which gives:
sol_sin =
- ((P*c*1i)/(2*E*I*k^3*l) + (P*c*(- 1 + k*x*1i)*(sin(k*x)*1i - 2*sin((k*x)/2)^2 + 1)*1i)/(2*E*I*k^3*l))*(sin(k*x)*1i + 2*sin((k*x)/2)^2 - 1) - ((P*c*1i)/(2*E*I*k^3*l) + (P*c*(1 + k*x*1i)*(sin(k*x)*1i + 2*sin((k*x)/2)^2 - 1)*1i)/(2*E*I*k^3*l))*(sin(k*x)*1i - 2*sin((k*x)/2)^2 + 1)
You can also get rid of the imaginary parts of the results by making some assumptions. For example i suspect E to be youngs modul, which should be a real number and positive always:
syms y(x) x E I N P c l Dyt(x) k
eqn = diff(y,x,2)+N/E*I*y == -P*c/(l*E*I)*x;
eqn = subs(eqn,(N/E*I),k^2)
[eqns, vars] = reduceDifferentialOrder(eqn,y);
conds = [y(0)==0,Dyt(0)==0];
assume([x P c l k],'real');
assumeAlso(c>=0 & l>=0 & k>=0);
assumptions
sol = dsolve(eqns,conds)
sol_sin = rewrite(sol.y,'sin')
results in:
sol_sin =
(P*c*sin(k*x))/(E*I*k^3*l) - (P*c*x)/(E*I*k^2*l)
You should check if the assumptions i made are correct in your case and modify them if needed.
Stephan
Stephan 2018-10-21
If this answer was helpful to you, please accept it.

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