# SECTION 4.5# Example 1 de := diff(y(t),t,t) + 3*diff(y(t),t) + 2*y(t) = f;

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y1 := 3*t/2 - 9/4;

NiM+JSN5MUcsJiooIiIkIiIiIiIjISIiJSJ0R0YoRigjIiIqIiIlRio=

subs(y(t) = y1,de);

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expand(%);

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y2 := exp(3*t)/2;

NiM+JSN5MkcsJComIyIiIiIiI0YoLSUkZXhwRzYjLCQqJiIiJEYoJSJ0R0YoRihGKEYo

subs(y(t) = y2,de);

NiMvLCgtJSVkaWZmRzYkLCQqJiMiIiIiIiNGKy0lJGV4cEc2IywkKiYiIiRGKyUidEdGK0YrRitGKy0lIiRHNiRGM0YsRisqJkYyRistRiY2JEYoRjNGK0YrRi1GKyUiZkc=

expand(%);

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yp := A*y1 + B*y2;

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subs(y(t) = yp, f = 3*t + 10*exp(3*t),de);

NiMvLCotJSVkaWZmRzYkLCYqJiUiQUciIiIsJiooIiIkRisiIiMhIiIlInRHRitGKyMiIioiIiVGMEYrRisqJiNGK0YvRisqJiUiQkdGKy0lJGV4cEc2IywkKiZGLkYrRjFGK0YrRitGK0YrLSUiJEc2JEYxRi9GKyomRi5GKy1GJjYkRihGMUYrRisqKEYvRitGKkYrRixGK0YrRjdGKywmKiZGLkYrRjFGK0YrKiYiIzVGK0Y5RitGKw==

expand(%);

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eq1 := 3*A = 3; eq2 := 10*B = 10;

NiM+JSRlcTFHLywkKiYiIiQiIiIlIkFHRilGKUYoNiM+JSRlcTJHLywkKiYiIzUiIiIlIkJHRilGKUYo

solve({eq1,eq2},{A,B});

NiM8JC8lIkFHIiIiLyUiQkdGJg==

subs(%,yp);

NiMsKCooIiIkIiIiIiIjISIiJSJ0R0YmRiYjIiIqIiIlRigqJiNGJkYnRiYtJSRleHBHNiMsJComRiVGJkYpRiZGJkYmRiY=

subs(y(t) = yp, f = -9*t + 20*exp(3*t),de);

NiMvLCotJSVkaWZmRzYkLCYqJiUiQUciIiIsJiooIiIkRisiIiMhIiIlInRHRitGKyMiIioiIiVGMEYrRisqJiNGK0YvRisqJiUiQkdGKy0lJGV4cEc2IywkKiZGLkYrRjFGK0YrRitGK0YrLSUiJEc2JEYxRi9GKyomRi5GKy1GJjYkRihGMUYrRisqKEYvRitGKkYrRixGK0YrRjdGKywmKiZGM0YrRjFGK0YwKiYiIz9GK0Y5RitGKw==

expand(%);

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eq1 := 3*A = -9; eq2 := 10*B = 20;

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solve({eq1,eq2},{A,B});

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subs(%,yp);

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# Example 2

ysol := t^2 + c1*exp(t) + c2*exp(-t);

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subs(t=0,ysol);

NiMsJiomJSNjMUciIiItJSRleHBHNiMiIiFGJkYmKiYlI2MyR0YmRidGJkYm

eq1 := % = 1;

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diff(ysol,t);

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subs(t=0,%);

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eq2 := simplify(%) = 0;

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solve({eq1,eq2},{c1,c2});

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subs(%,ysol);

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# Example 3de := diff(y(t),t,t) + 2*diff(y(t),t) + 2*y(t) = 5*sin(t) + 5*cos(t);

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yp := A*sin(t) + B*cos(t);

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subs(y(t) = yp,de);

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expand(%);

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eq1 := A - 2*B = 5;

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eq2 := B + 2*A = 5;

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solve({eq1,eq2},{A,B});

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subs(%,yp);

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yp := %;

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ysol := exp(-t)*(c1*cos(t)+c2*sin(t)) + yp;

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subs(t=0,ysol);

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eq1 := simplify(%) = 1;

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diff(ysol,t);

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subs(t=0,%);

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eq2 := simplify(%) = 2;

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solve({eq1,eq2},{c1,c2});

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subs(%,ysol);

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# Solution using dsolve:dsolve(de);

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dsolve({de,y(0)=1,D(y)(0)=2});

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# Example 4de := diff(y(t),t,t) - y(t) = 8*t*exp(t) + 2*exp(t);

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yp := t*(A1*t + A0)*exp(t);

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subs(y(t) = yp,de);

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expand(%);

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%/exp(t);

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eq := simplify(%);

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subs(t=0,%);

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eq1 := %;

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diff(eq,t);

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eq2 := %;

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solve({eq1,eq2},{A0,A1});

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subs(%,yp);

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# Example 5de := diff(y(t),t,t) + 2*diff(y(t),t) + 2*y(t) = 5*exp(-t)*sin(t) + 5*t^3*exp(-t)*cos(t);

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P := r^2 + 2*r + 2 = 0;

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solve(P,r);

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dsolve(de);

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# Is the above solution consistent with the form of yp given in the text? # Can some terms in the above solution be combined with terms of the homgeneous solution, ie, terms # with _C1 and _C2? Is the form of yp unique? # Example 6de := diff(y(t),t,t,t) + 2*diff(y(t),t,t) + diff(y(t),t) = 5*exp(-t)*sin(t) + 3 + 7*t*exp(-t);

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dsolve(de);

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