pag. 401, #25. Initial value problem by Laplace transform.DO #26. We use an alias to make the Maple output shorter.alias(Y(s)=laplace(y(t),t,s)):
with(inttrans):We enter the differential equation and initial values. (Note the use of Maple's composition operator @ to enter the initial value of the second derivative.)diffeq:=diff(y(t),t$3)-diff(y(t),t$2)+diff(y(t),t)-y(t)=0;
inits:=y(0)=1,D(y)(0)=1,(D@@2)(y)(0)=3;laplace(diffeq,t,s);
subs(inits,%);
expand(%);
algeqn:=collect(%,Y(s));Note that the differential equation has been transformed into an algebraic equation. We can then use algebra or Maple's solve command to solve for Y(s).solve(algeqn,Y(s));Finally, we take the inverse Laplace transform of Y(s).sol:=y(t)=invlaplace(%,s,t);Checking:subs(sol,diffeq):
expand(%);subs(t=0,rhs(sol)):
simplify(%);diff(rhs(sol),t):
subs(t=0,%):
simplify(%);diff(rhs(sol),t$2):
subs(t=0,%):
simplify(%);pag. 401, #37. Initial value problem by Laplace transform, non-constant coefficients.DO #36restart:
alias(Y(s)=laplace(y(t),t,s)):
with(inttrans):
diffeq:=t*diff(y(t),t$2)-2*diff(y(t),t)+t*y(t)=0;
inits:=y(0)=1,D(y)(0)=0;laplace(diff(y(t),t$2),t,s);
expand(%);
l1:=-diff(%,s);l2:=laplace(diff(y(t),t),t,s);
laplace(y(t),t,s);l3:=-diff(%,s);
subs(inits,l1-2*l2+l3)=0;
expand(%);newdiffeq:=collect(%,diff(Y(s),s));Note that the differential equation has been transformed into a new differential equation. We can then use Maple's dsolve command to solve for Y(s). (Since Y(s) is an alias, rather than a real function, Maple will not work correctly if we try to write a differential equation viewing Y(s) as a function of s, so we rewrite the equation in terms of Z(s).)(-s^2-1)*diff(Z(s),s)-4*s*Z(s)+3=0;
dsolve((-s^2-1)*diff(Z(s),s)-4*s*Z(s)+3=0,Z(s));Finally, we take the inverse Laplace transform of Y(s).convert(rhs(%),parfrac,s);
Z(s):=expand(%);Since the first term is the derivative of 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 , which is the Laplace transform of 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 , the inverse Laplace transform of the first term is LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Jy1GLDYlUSJ0RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVExJkludmlzaWJsZVRpbWVzO0YnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZRLUYjNiYtRiw2JVEkc2luRicvRjVGQkY+LUY7Ni1RMCZBcHBseUZ1bmN0aW9uO0YnRj5GQEZDRkVGR0ZJRktGTUZPRlItSShtZmVuY2VkR0YkNiQtRiM2JEYxRj5GPkY+RitGPkYrRj4= . By the hint in the book, the inverse Laplace transform of the second term is 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 , and the inverse Laplace transform of 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 is LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSRjb3NGJy8lJ2l0YWxpY0dRJmZhbHNlRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVEwJkFwcGx5RnVuY3Rpb247RidGNy8lJmZlbmNlR0Y2LyUqc2VwYXJhdG9yR0Y2LyUpc3RyZXRjaHlHRjYvJSpzeW1tZXRyaWNHRjYvJShsYXJnZW9wR0Y2LyUubW92YWJsZWxpbWl0c0dGNi8lJ2FjY2VudEdGNi8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRk4tSShtZmVuY2VkR0YkNiQtRiM2JC1GLDYlUSJ0RicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0Y3RjdGK0Y3 . Hence the solution issol:=y(t)=t*sin(t)+_C1/2*(sin(t)-t*cos(t))+cos(t);Checking:subs(sol,diffeq):
expand(%);subs(t=0,rhs(sol)):
simplify(%);diff(rhs(sol),t):
subs(t=0,%):
simplify(%);pag. 423, #13. Laplace transform of a convolution.DO #14.restart:
f:=Int((t-v)*exp(3*v),v=0..t);This integral is the convolution of t and exp(3*t). Hence the Laplace transform is the product of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbW5HRiQ2JFEiMUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJ0Y0LUYjNiYtSSNtaUdGJDYjUSFGJy1GIzYkLUklbXN1cEdGJDYlLUY6NiVRInNGJy8lJ2l0YWxpY0dRJXRydWVGJy9GNVEnaXRhbGljRictRjE2JFEiMkYnRjQvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnRjRGOUY0LyUubGluZXRoaWNrbmVzc0dRIjFGJy8lK2Rlbm9tYWxpZ25HUSdjZW50ZXJGJy8lKW51bWFsaWduR0ZVLyUpYmV2ZWxsZWRHUSZmYWxzZUYnRjQ= and 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 .1/s^2*1/(s-3);We can compare this result to the output of Maple's laplace command.with(inttrans):
laplace(f,t,s);pag. 423, #23. Transfer function, impulse response.DO #24.restart:
with(inttrans):
diffeq:=diff(y(t),t$2)+9*y(t)=g(t);
inits1:=y(0)=2,D(y)(0)=-3;
inits2:=y(0)=0,D(y)(0)=0;alias(Y(s)=laplace(y(t),t,s),G(s)=laplace(g(t),t,s)):The transfer function is calculated as the ratio of Y(s)/G(s) under the assumption that the initial conditions are all 0.laplace(diffeq,t,s);
subs(inits2,%);
expand(%);
collect(%,Y(s));Hence, the transfer function H(s) is given bysolve(%,Y(s))/G(s):
H(s):=expand(%);
The impulse response function h(t) is the inverse Laplace transform of the transfer function.h(t):=invlaplace(H(s),s,t);We add the convolution of g(t) and the impulse response function to a particular solution of the homogeneous initial value problem having non-zero initial values to get the solution to the nonhomogeneous problem with the same initial values but with right hand side g(t).(The point of this maneuver is to separate the effects of the non-zero initial values from those of the non-zero input function g.)yp:=dsolve({lhs(diffeq)=0,inits1},y(t));nivsol:=Int(subs(t=t-v,h(t))*subs(t=v,g(t)),v=0..t)+rhs(yp);We can compare our result to that of dsolve with the laplace option.dsolve({diffeq,inits1},y(t),method=laplace);pag. 431, #21. Initial value problem with impulse, plot.DO #22.restart: with(inttrans):
diffeq:=diff(y(t),t$2)+y(t)=Dirac(t-2*Pi);
inits:=y(0)=0,D(y)(0)=1;alias(Y(s)=laplace(y(t),t,s)):
laplace(diffeq,t,s);
subs(inits,%);
collect(%,Y(s));
solve(%,Y(s));
expand(%);
Y(s)=simplify(%,exp);Multiplication by 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 translates f(t)*Heaviside(t) right by LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictSSNtb0dGJDYtUTEmSW52aXNpYmxlVGltZXM7RidGNS8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPi8lKXN0cmV0Y2h5R0Y+LyUqc3ltbWV0cmljR0Y+LyUobGFyZ2VvcEdGPi8lLm1vdmFibGVsaW1pdHNHRj4vJSdhY2NlbnRHRj4vJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZNLUYsNiVRJSZwaTtGJy8lJ2l0YWxpY0dGPkY1RjVGK0Y1 , i.e., we get f(t-2*Pi)*Heaviside(t-2*Pi). Hence, the inverse Laplace transform of the second term is sin(t-2*Pi)*Heaviside(t-2*Pi). We can compare this to the output of Maple's dsolve with the laplace option.sol:=dsolve({diffeq,inits},y(t),method=laplace);We plot the solution.plot(rhs(sol),t=-5..15);