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Question: solve integro-differential equation

Question:solve integro-differential equation

mehdi jafari 769 Maple
ode dsolve numeric differential-equations integro-differential-equation
November 19 2020
1

how this integro-differential equation can be solved?
any assumption or suggestion is appreciated. tnx in advance

EDITED: parameters are added.


 

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restart
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w[c]:=0.01:delta:=5:Omega:=5:w[0]:=1:gamma_0:=1:k[b]:=1:T:=100:nu[n]:=2*Pi*n*k[b]*T;

200*Pi*n

 (1)
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F:=2*gamma_0*w[c]^2*exp(-w[c]*tau)*sin(delta/Omega*(sin(Omega*t)-sin(Omega*(t-tau)))+w[0]*tau):
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k1:=2*k[b]*T*w[c]^2*sum((w[c]*exp(-w[c]*tau)-abs(nu[n])*exp(-abs(nu[n])*tau))/(w[c]^2-nu[n]^2),n=-infinity..infinity):
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G:=cos(delta/Omega*(sin(Omega*t)-sin(Omega*(t-tau)))+w[0]*tau)*k1:
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eq:=diff(rho(t),t)+Int(G*rho(tau),tau=0..t)=1/2*Int(G-F,tau=0..t);

diff(rho(t), t)+Int(0.200e-1*cos(sin(5*t)-sin(5*t-5*tau)+tau)*(sum((0.1000000000e-1*exp(-0.1000000000e-1*tau)-628.3185308*abs(n)*exp(-628.3185308*abs(n)*tau))/(-394784.1762*n^2+0.1000000000e-3), n = -infinity .. infinity))*rho(tau), tau = 0 .. t) = (1/2)*(Int(0.200e-1*cos(sin(5*t)-sin(5*t-5*tau)+tau)*(sum((0.1000000000e-1*exp(-0.1000000000e-1*tau)-628.3185308*abs(n)*exp(-628.3185308*abs(n)*tau))/(-394784.1762*n^2+0.1000000000e-3), n = -infinity .. infinity))-0.2e-3*exp(-0.1e-1*tau)*sin(sin(5*t)-sin(5*t-5*tau)+tau), tau = 0 .. t))

 (2)
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