# SECTION 3.5# Example 1 de := L*diff(I(t),t) + R*I(t) = E(t);

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# The above result is due to the fact that I is a reserved symbol# in Maple, as can be seen by: ?II^2;

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# As explained in the above help, this can be changed by, for example,interface(imaginaryunit = j);

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j^2;

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I^2;

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de := L*diff(I(t),t) + R*I(t) = E(t);

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# Solution by using an integration factor (first put into standard form):de1 := de/L;

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mu := exp(int(R/L,t));

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de2 := mu*de1;

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# Check that LHS = derivitive of mu*I(t);diff(mu*I(t),t) - lhs(de2);

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simplify(%);

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# Solve by integrating both sides:mu*I = Int(rhs(de2),t);

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sol := I = rhs(%)/mu;

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subs({R=1,L=.01,E(t)=sin(100*t)},sol);

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sol := I = value(rhs(%))+K;

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subs({t=0,I=0},%);

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solve(%,K);

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subs(K=%,sol);

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# Solution using dsolve:subs({R=1,L=.01,E(t)=sin(100*t)},de);

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de3 := %;

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dsolve(de3);

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dsolve({de3,I(0)=0});

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# Example 2de := R*diff(q(t),t) + q(t)/C = V;

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# Solution by integrating factor (first put into standard form):de1 := de/R;

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mu := exp(int(1/(R*C),t));

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de2 := mu*de1;

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# Check:diff(mu*q(t),t) - lhs(de2);

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simplify(%);

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# Solve by integrating both sides:mu*q = Int(rhs(de2),t);

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sol := q = (value(rhs(%)) + K)/mu;

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subs({q=Q,t=0},sol);

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solve(%,K);

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subs(K=%,sol);

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# Solution using dsolve:dsolve(de);

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dsolve({de,q(0)=Q});

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