How can I solve this ODE $ frac {d^2x}{dt^2} + 8frac {dx}{dt} + 25x = 10u(t)$?












1












$begingroup$


The problem its that I don't know how to treat the $ 10u(t) $ to obtain the particular solution since I don't know what the $ u(t) $ function represents.
I've already have the complementary solution of the homogeneous equation associated which is: $ X_c(t)=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t} $ .



Here again, I leave the equation to solve.
$$ frac {d^2x}{dt^2} + 8frac {dx}{dt} + 25x = 10u(t)$$



Thank you.



EDIT:



Thanks SplitInfinity, Yes, the $ u(t) $ was the Step Function and something I forgot to mention that the problem says: consider initial conditions in $ 0 $. So the step function states:
$$ u(t)=
begin{cases}
0 & t<0 \
1 & t ge 0
end{cases} $$
in this case $ t=1 $, therefore the equation ends up like this
$$ begin{align}
x''+8x''+25x&=10(1) \
x''+8x''+25x&=10
end{align}$$



because the non-homogenous part its a linear polinomynal the $X_p$ (proposal solution) need to be in this way:
$$ X_p=A \
X'_p=0 \
X''_p=0 $$
So the equation ends up like:
$$ begin{align}
x_p''+8x'+25x_p & =10 \
0+8(0)+25(A) & = 10 \
A & = frac {10}{25} \
A & = frac 25
end{align}$$



Now the General Slution is given by $ X=X_c+X_p $ wich is:
$$ X=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t}+frac 25 $$
To find out the $C_1 text{ and } C_2$ values we need to apply the initial values, wich are:
$$ x(0)=0 \
x'(0)=0 $$
in order to do so, we derivate $X$
$$ X'=C_1e^{-4t}(-3sin{3t}-4cos{3t})+C_2e^{-4t}(3cos{3t}-4sin{3t}) $$
Now, applying initial values to X:
$$ require{cancel} begin{align}
X(0) & = C_1cancelto{1}{e^{-4(0)}}cancelto{1}{cos{3(0)}}+cancelto{0}{C_2e^{-4(0)}sin{3(0)}}+frac25 \
0 &= C_1 + frac 25 \
C_1 &= -frac25
end{align} $$



Now to $ X' $:
$$ begin{align}
X'(0) &= C_1cancelto{1}{e^{-4(0)}}(cancelto{0}{-3sin{3(0)}})+C_2cancelto{1}{e^{-4(0)}}(3cancelto{1}{cos{3(0)}}-cancelto{0}{4sin{3(0)}}) \
0 &= C_1(-4)+C_2(3) \
0 &= -4C_1 + 3C_2 \
3_C2 &= 4C_1 \
text{Replacing the $C_1$ value} \
C_2 &= frac{4(-frac25)}{3} \
C_2 &= -frac8{15}
end{align} $$



Replacing the constants $ C_1 text{ and } C_2 $ in $X$ we found the solution:
$$ begin{align}
X &= -frac25 e^{-4t}cos{3t}-frac8{15}e^{-4t}sin{3t}+frac25 \
X &= frac1{15} left[ e^{-4}(-6cos{3t}-8sin{3t})+6 right]
end{align} $$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $u(t)$ represents the unit step function.
    $endgroup$
    – SplitInfinity
    Mar 5 '16 at 20:26












  • $begingroup$
    Thank you @SplitInfinity that help a lot in the understanding the solution.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:42
















1












$begingroup$


The problem its that I don't know how to treat the $ 10u(t) $ to obtain the particular solution since I don't know what the $ u(t) $ function represents.
I've already have the complementary solution of the homogeneous equation associated which is: $ X_c(t)=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t} $ .



Here again, I leave the equation to solve.
$$ frac {d^2x}{dt^2} + 8frac {dx}{dt} + 25x = 10u(t)$$



Thank you.



EDIT:



Thanks SplitInfinity, Yes, the $ u(t) $ was the Step Function and something I forgot to mention that the problem says: consider initial conditions in $ 0 $. So the step function states:
$$ u(t)=
begin{cases}
0 & t<0 \
1 & t ge 0
end{cases} $$
in this case $ t=1 $, therefore the equation ends up like this
$$ begin{align}
x''+8x''+25x&=10(1) \
x''+8x''+25x&=10
end{align}$$



because the non-homogenous part its a linear polinomynal the $X_p$ (proposal solution) need to be in this way:
$$ X_p=A \
X'_p=0 \
X''_p=0 $$
So the equation ends up like:
$$ begin{align}
x_p''+8x'+25x_p & =10 \
0+8(0)+25(A) & = 10 \
A & = frac {10}{25} \
A & = frac 25
end{align}$$



Now the General Slution is given by $ X=X_c+X_p $ wich is:
$$ X=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t}+frac 25 $$
To find out the $C_1 text{ and } C_2$ values we need to apply the initial values, wich are:
$$ x(0)=0 \
x'(0)=0 $$
in order to do so, we derivate $X$
$$ X'=C_1e^{-4t}(-3sin{3t}-4cos{3t})+C_2e^{-4t}(3cos{3t}-4sin{3t}) $$
Now, applying initial values to X:
$$ require{cancel} begin{align}
X(0) & = C_1cancelto{1}{e^{-4(0)}}cancelto{1}{cos{3(0)}}+cancelto{0}{C_2e^{-4(0)}sin{3(0)}}+frac25 \
0 &= C_1 + frac 25 \
C_1 &= -frac25
end{align} $$



Now to $ X' $:
$$ begin{align}
X'(0) &= C_1cancelto{1}{e^{-4(0)}}(cancelto{0}{-3sin{3(0)}})+C_2cancelto{1}{e^{-4(0)}}(3cancelto{1}{cos{3(0)}}-cancelto{0}{4sin{3(0)}}) \
0 &= C_1(-4)+C_2(3) \
0 &= -4C_1 + 3C_2 \
3_C2 &= 4C_1 \
text{Replacing the $C_1$ value} \
C_2 &= frac{4(-frac25)}{3} \
C_2 &= -frac8{15}
end{align} $$



Replacing the constants $ C_1 text{ and } C_2 $ in $X$ we found the solution:
$$ begin{align}
X &= -frac25 e^{-4t}cos{3t}-frac8{15}e^{-4t}sin{3t}+frac25 \
X &= frac1{15} left[ e^{-4}(-6cos{3t}-8sin{3t})+6 right]
end{align} $$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $u(t)$ represents the unit step function.
    $endgroup$
    – SplitInfinity
    Mar 5 '16 at 20:26












  • $begingroup$
    Thank you @SplitInfinity that help a lot in the understanding the solution.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:42














1












1








1


1



$begingroup$


The problem its that I don't know how to treat the $ 10u(t) $ to obtain the particular solution since I don't know what the $ u(t) $ function represents.
I've already have the complementary solution of the homogeneous equation associated which is: $ X_c(t)=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t} $ .



Here again, I leave the equation to solve.
$$ frac {d^2x}{dt^2} + 8frac {dx}{dt} + 25x = 10u(t)$$



Thank you.



EDIT:



Thanks SplitInfinity, Yes, the $ u(t) $ was the Step Function and something I forgot to mention that the problem says: consider initial conditions in $ 0 $. So the step function states:
$$ u(t)=
begin{cases}
0 & t<0 \
1 & t ge 0
end{cases} $$
in this case $ t=1 $, therefore the equation ends up like this
$$ begin{align}
x''+8x''+25x&=10(1) \
x''+8x''+25x&=10
end{align}$$



because the non-homogenous part its a linear polinomynal the $X_p$ (proposal solution) need to be in this way:
$$ X_p=A \
X'_p=0 \
X''_p=0 $$
So the equation ends up like:
$$ begin{align}
x_p''+8x'+25x_p & =10 \
0+8(0)+25(A) & = 10 \
A & = frac {10}{25} \
A & = frac 25
end{align}$$



Now the General Slution is given by $ X=X_c+X_p $ wich is:
$$ X=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t}+frac 25 $$
To find out the $C_1 text{ and } C_2$ values we need to apply the initial values, wich are:
$$ x(0)=0 \
x'(0)=0 $$
in order to do so, we derivate $X$
$$ X'=C_1e^{-4t}(-3sin{3t}-4cos{3t})+C_2e^{-4t}(3cos{3t}-4sin{3t}) $$
Now, applying initial values to X:
$$ require{cancel} begin{align}
X(0) & = C_1cancelto{1}{e^{-4(0)}}cancelto{1}{cos{3(0)}}+cancelto{0}{C_2e^{-4(0)}sin{3(0)}}+frac25 \
0 &= C_1 + frac 25 \
C_1 &= -frac25
end{align} $$



Now to $ X' $:
$$ begin{align}
X'(0) &= C_1cancelto{1}{e^{-4(0)}}(cancelto{0}{-3sin{3(0)}})+C_2cancelto{1}{e^{-4(0)}}(3cancelto{1}{cos{3(0)}}-cancelto{0}{4sin{3(0)}}) \
0 &= C_1(-4)+C_2(3) \
0 &= -4C_1 + 3C_2 \
3_C2 &= 4C_1 \
text{Replacing the $C_1$ value} \
C_2 &= frac{4(-frac25)}{3} \
C_2 &= -frac8{15}
end{align} $$



Replacing the constants $ C_1 text{ and } C_2 $ in $X$ we found the solution:
$$ begin{align}
X &= -frac25 e^{-4t}cos{3t}-frac8{15}e^{-4t}sin{3t}+frac25 \
X &= frac1{15} left[ e^{-4}(-6cos{3t}-8sin{3t})+6 right]
end{align} $$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




The problem its that I don't know how to treat the $ 10u(t) $ to obtain the particular solution since I don't know what the $ u(t) $ function represents.
I've already have the complementary solution of the homogeneous equation associated which is: $ X_c(t)=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t} $ .



Here again, I leave the equation to solve.
$$ frac {d^2x}{dt^2} + 8frac {dx}{dt} + 25x = 10u(t)$$



Thank you.



EDIT:



Thanks SplitInfinity, Yes, the $ u(t) $ was the Step Function and something I forgot to mention that the problem says: consider initial conditions in $ 0 $. So the step function states:
$$ u(t)=
begin{cases}
0 & t<0 \
1 & t ge 0
end{cases} $$
in this case $ t=1 $, therefore the equation ends up like this
$$ begin{align}
x''+8x''+25x&=10(1) \
x''+8x''+25x&=10
end{align}$$



because the non-homogenous part its a linear polinomynal the $X_p$ (proposal solution) need to be in this way:
$$ X_p=A \
X'_p=0 \
X''_p=0 $$
So the equation ends up like:
$$ begin{align}
x_p''+8x'+25x_p & =10 \
0+8(0)+25(A) & = 10 \
A & = frac {10}{25} \
A & = frac 25
end{align}$$



Now the General Slution is given by $ X=X_c+X_p $ wich is:
$$ X=C_1e^{-4t}cos {3t} + C_2e^{-4t}sin {3t}+frac 25 $$
To find out the $C_1 text{ and } C_2$ values we need to apply the initial values, wich are:
$$ x(0)=0 \
x'(0)=0 $$
in order to do so, we derivate $X$
$$ X'=C_1e^{-4t}(-3sin{3t}-4cos{3t})+C_2e^{-4t}(3cos{3t}-4sin{3t}) $$
Now, applying initial values to X:
$$ require{cancel} begin{align}
X(0) & = C_1cancelto{1}{e^{-4(0)}}cancelto{1}{cos{3(0)}}+cancelto{0}{C_2e^{-4(0)}sin{3(0)}}+frac25 \
0 &= C_1 + frac 25 \
C_1 &= -frac25
end{align} $$



Now to $ X' $:
$$ begin{align}
X'(0) &= C_1cancelto{1}{e^{-4(0)}}(cancelto{0}{-3sin{3(0)}})+C_2cancelto{1}{e^{-4(0)}}(3cancelto{1}{cos{3(0)}}-cancelto{0}{4sin{3(0)}}) \
0 &= C_1(-4)+C_2(3) \
0 &= -4C_1 + 3C_2 \
3_C2 &= 4C_1 \
text{Replacing the $C_1$ value} \
C_2 &= frac{4(-frac25)}{3} \
C_2 &= -frac8{15}
end{align} $$



Replacing the constants $ C_1 text{ and } C_2 $ in $X$ we found the solution:
$$ begin{align}
X &= -frac25 e^{-4t}cos{3t}-frac8{15}e^{-4t}sin{3t}+frac25 \
X &= frac1{15} left[ e^{-4}(-6cos{3t}-8sin{3t})+6 right]
end{align} $$






ordinary-differential-equations




ordinary-differential-equations






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edited Mar 13 '16 at 1:51







Daniel Colon

















asked Mar 5 '16 at 20:04









Daniel ColonDaniel Colon

613




613












  • $begingroup$
    $u(t)$ represents the unit step function.
    $endgroup$
    – SplitInfinity
    Mar 5 '16 at 20:26












  • $begingroup$
    Thank you @SplitInfinity that help a lot in the understanding the solution.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:42


















  • $begingroup$
    $u(t)$ represents the unit step function.
    $endgroup$
    – SplitInfinity
    Mar 5 '16 at 20:26












  • $begingroup$
    Thank you @SplitInfinity that help a lot in the understanding the solution.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:42
















$begingroup$
$u(t)$ represents the unit step function.
$endgroup$
– SplitInfinity
Mar 5 '16 at 20:26






$begingroup$
$u(t)$ represents the unit step function.
$endgroup$
– SplitInfinity
Mar 5 '16 at 20:26














$begingroup$
Thank you @SplitInfinity that help a lot in the understanding the solution.
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:42




$begingroup$
Thank you @SplitInfinity that help a lot in the understanding the solution.
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:42










2 Answers
2






active

oldest

votes


















1












$begingroup$

Solving this with Laplace Transform, with $t>0$:



$$x''(t)+8x'(t)+25x=10u(t)Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)+8x'(t)+25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+mathcal{L}_{t}left[8x'(t)right]_{(s)}+mathcal{L}_{t}left[25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+8mathcal{L}_{t}left[x'(t)right]_{(s)}+25mathcal{L}_{t}left[x(t)right]_{(s)}=10mathcal{L}_{t}left[u(t)right]_{(s)}Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8left(sx(s)-x(0)right)+25x(s)=10u(s)Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]-sx(0)-x'(0)-8x(0)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]=10u(s)+sx(0)+x'(0)+8x(0)Longleftrightarrow$$
$$x(s)=frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}Longleftrightarrow$$
$$mathcal{L}_{s}^{-1}left[x(s)right]_{(t)}=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}$$



Notice, when $u(t)=theta(t)$ with $theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:



$$u(s)=frac{1}{s}$$



So:



$$x(t)=mathcal{L}_{s}^{-1}left[frac{frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=frac{12+2e^{-4t}left(3left(5x(0)-2right)cos(3t)+left(20x(0)+5x'(0)-8right)sin(3t)right)}{30}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:48










  • $begingroup$
    @DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
    $endgroup$
    – Jan
    Mar 13 '16 at 10:55










  • $begingroup$
    Why isn't this marked as the answer?
    $endgroup$
    – DavidG
    Dec 26 '18 at 4:39



















0












$begingroup$

The homogeneous solution is
$$y_-(t)=e^{-4t}(A_-cos(3t)+B_-sin(3t).$$



A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=frac25$, so that the full solution family is
$$
y_+(t)=frac25+e^{-4t}(A_+cos(3t)+B_+sin(3t).
$$



If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-;=;y_+(0)=frac25+A_+\
y_-'(0)=-4A_-+3B_-;=;y_+'(0)=-4A_++3B_+\
implies B_-=frac8{15}+B_+
$$



which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-cos(3t)+B_-sin(3t)+frac2{15}u(t)left(3-e^{-4t}(3cos(3t)+4sin(3t))right)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:45











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Solving this with Laplace Transform, with $t>0$:



$$x''(t)+8x'(t)+25x=10u(t)Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)+8x'(t)+25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+mathcal{L}_{t}left[8x'(t)right]_{(s)}+mathcal{L}_{t}left[25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+8mathcal{L}_{t}left[x'(t)right]_{(s)}+25mathcal{L}_{t}left[x(t)right]_{(s)}=10mathcal{L}_{t}left[u(t)right]_{(s)}Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8left(sx(s)-x(0)right)+25x(s)=10u(s)Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]-sx(0)-x'(0)-8x(0)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]=10u(s)+sx(0)+x'(0)+8x(0)Longleftrightarrow$$
$$x(s)=frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}Longleftrightarrow$$
$$mathcal{L}_{s}^{-1}left[x(s)right]_{(t)}=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}$$



Notice, when $u(t)=theta(t)$ with $theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:



$$u(s)=frac{1}{s}$$



So:



$$x(t)=mathcal{L}_{s}^{-1}left[frac{frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=frac{12+2e^{-4t}left(3left(5x(0)-2right)cos(3t)+left(20x(0)+5x'(0)-8right)sin(3t)right)}{30}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:48










  • $begingroup$
    @DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
    $endgroup$
    – Jan
    Mar 13 '16 at 10:55










  • $begingroup$
    Why isn't this marked as the answer?
    $endgroup$
    – DavidG
    Dec 26 '18 at 4:39
















1












$begingroup$

Solving this with Laplace Transform, with $t>0$:



$$x''(t)+8x'(t)+25x=10u(t)Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)+8x'(t)+25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+mathcal{L}_{t}left[8x'(t)right]_{(s)}+mathcal{L}_{t}left[25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+8mathcal{L}_{t}left[x'(t)right]_{(s)}+25mathcal{L}_{t}left[x(t)right]_{(s)}=10mathcal{L}_{t}left[u(t)right]_{(s)}Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8left(sx(s)-x(0)right)+25x(s)=10u(s)Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]-sx(0)-x'(0)-8x(0)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]=10u(s)+sx(0)+x'(0)+8x(0)Longleftrightarrow$$
$$x(s)=frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}Longleftrightarrow$$
$$mathcal{L}_{s}^{-1}left[x(s)right]_{(t)}=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}$$



Notice, when $u(t)=theta(t)$ with $theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:



$$u(s)=frac{1}{s}$$



So:



$$x(t)=mathcal{L}_{s}^{-1}left[frac{frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=frac{12+2e^{-4t}left(3left(5x(0)-2right)cos(3t)+left(20x(0)+5x'(0)-8right)sin(3t)right)}{30}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:48










  • $begingroup$
    @DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
    $endgroup$
    – Jan
    Mar 13 '16 at 10:55










  • $begingroup$
    Why isn't this marked as the answer?
    $endgroup$
    – DavidG
    Dec 26 '18 at 4:39














1












1








1





$begingroup$

Solving this with Laplace Transform, with $t>0$:



$$x''(t)+8x'(t)+25x=10u(t)Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)+8x'(t)+25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+mathcal{L}_{t}left[8x'(t)right]_{(s)}+mathcal{L}_{t}left[25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+8mathcal{L}_{t}left[x'(t)right]_{(s)}+25mathcal{L}_{t}left[x(t)right]_{(s)}=10mathcal{L}_{t}left[u(t)right]_{(s)}Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8left(sx(s)-x(0)right)+25x(s)=10u(s)Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]-sx(0)-x'(0)-8x(0)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]=10u(s)+sx(0)+x'(0)+8x(0)Longleftrightarrow$$
$$x(s)=frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}Longleftrightarrow$$
$$mathcal{L}_{s}^{-1}left[x(s)right]_{(t)}=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}$$



Notice, when $u(t)=theta(t)$ with $theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:



$$u(s)=frac{1}{s}$$



So:



$$x(t)=mathcal{L}_{s}^{-1}left[frac{frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=frac{12+2e^{-4t}left(3left(5x(0)-2right)cos(3t)+left(20x(0)+5x'(0)-8right)sin(3t)right)}{30}$$






share|cite|improve this answer









$endgroup$



Solving this with Laplace Transform, with $t>0$:



$$x''(t)+8x'(t)+25x=10u(t)Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)+8x'(t)+25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+mathcal{L}_{t}left[8x'(t)right]_{(s)}+mathcal{L}_{t}left[25x(t)right]_{(s)}=mathcal{L}_{t}left[10u(t)right]_{(s)}Longleftrightarrow$$
$$mathcal{L}_{t}left[x''(t)right]_{(s)}+8mathcal{L}_{t}left[x'(t)right]_{(s)}+25mathcal{L}_{t}left[x(t)right]_{(s)}=10mathcal{L}_{t}left[u(t)right]_{(s)}Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8left(sx(s)-x(0)right)+25x(s)=10u(s)Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]-sx(0)-x'(0)-8x(0)=10u(s)Longleftrightarrow$$
$$x(s)left[s^2+8s+25right]=10u(s)+sx(0)+x'(0)+8x(0)Longleftrightarrow$$
$$x(s)=frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}Longleftrightarrow$$
$$mathcal{L}_{s}^{-1}left[x(s)right]_{(t)}=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=mathcal{L}_{s}^{-1}left[frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}$$



Notice, when $u(t)=theta(t)$ with $theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:



$$u(s)=frac{1}{s}$$



So:



$$x(t)=mathcal{L}_{s}^{-1}left[frac{frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}right]_{(t)}Longleftrightarrow$$
$$x(t)=frac{12+2e^{-4t}left(3left(5x(0)-2right)cos(3t)+left(20x(0)+5x'(0)-8right)sin(3t)right)}{30}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 5 '16 at 20:38









JanJan

21.8k31240




21.8k31240








  • 1




    $begingroup$
    Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:48










  • $begingroup$
    @DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
    $endgroup$
    – Jan
    Mar 13 '16 at 10:55










  • $begingroup$
    Why isn't this marked as the answer?
    $endgroup$
    – DavidG
    Dec 26 '18 at 4:39














  • 1




    $begingroup$
    Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:48










  • $begingroup$
    @DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
    $endgroup$
    – Jan
    Mar 13 '16 at 10:55










  • $begingroup$
    Why isn't this marked as the answer?
    $endgroup$
    – DavidG
    Dec 26 '18 at 4:39








1




1




$begingroup$
Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:48




$begingroup$
Thank you @Jan the Laplace solution is going to be helpful in the next series of problems that I have to resolve. Would you recommend me a book that explain in a very clear way the Laplace topic? Thank you again.
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:48












$begingroup$
@DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
$endgroup$
– Jan
Mar 13 '16 at 10:55




$begingroup$
@DanielColon You're welcome! I'm an engineer so I use books that are explaining engineering so look at the internet your will find some good books
$endgroup$
– Jan
Mar 13 '16 at 10:55












$begingroup$
Why isn't this marked as the answer?
$endgroup$
– DavidG
Dec 26 '18 at 4:39




$begingroup$
Why isn't this marked as the answer?
$endgroup$
– DavidG
Dec 26 '18 at 4:39











0












$begingroup$

The homogeneous solution is
$$y_-(t)=e^{-4t}(A_-cos(3t)+B_-sin(3t).$$



A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=frac25$, so that the full solution family is
$$
y_+(t)=frac25+e^{-4t}(A_+cos(3t)+B_+sin(3t).
$$



If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-;=;y_+(0)=frac25+A_+\
y_-'(0)=-4A_-+3B_-;=;y_+'(0)=-4A_++3B_+\
implies B_-=frac8{15}+B_+
$$



which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-cos(3t)+B_-sin(3t)+frac2{15}u(t)left(3-e^{-4t}(3cos(3t)+4sin(3t))right)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:45
















0












$begingroup$

The homogeneous solution is
$$y_-(t)=e^{-4t}(A_-cos(3t)+B_-sin(3t).$$



A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=frac25$, so that the full solution family is
$$
y_+(t)=frac25+e^{-4t}(A_+cos(3t)+B_+sin(3t).
$$



If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-;=;y_+(0)=frac25+A_+\
y_-'(0)=-4A_-+3B_-;=;y_+'(0)=-4A_++3B_+\
implies B_-=frac8{15}+B_+
$$



which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-cos(3t)+B_-sin(3t)+frac2{15}u(t)left(3-e^{-4t}(3cos(3t)+4sin(3t))right)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:45














0












0








0





$begingroup$

The homogeneous solution is
$$y_-(t)=e^{-4t}(A_-cos(3t)+B_-sin(3t).$$



A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=frac25$, so that the full solution family is
$$
y_+(t)=frac25+e^{-4t}(A_+cos(3t)+B_+sin(3t).
$$



If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-;=;y_+(0)=frac25+A_+\
y_-'(0)=-4A_-+3B_-;=;y_+'(0)=-4A_++3B_+\
implies B_-=frac8{15}+B_+
$$



which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-cos(3t)+B_-sin(3t)+frac2{15}u(t)left(3-e^{-4t}(3cos(3t)+4sin(3t))right)
$$






share|cite|improve this answer









$endgroup$



The homogeneous solution is
$$y_-(t)=e^{-4t}(A_-cos(3t)+B_-sin(3t).$$



A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=frac25$, so that the full solution family is
$$
y_+(t)=frac25+e^{-4t}(A_+cos(3t)+B_+sin(3t).
$$



If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-;=;y_+(0)=frac25+A_+\
y_-'(0)=-4A_-+3B_-;=;y_+'(0)=-4A_++3B_+\
implies B_-=frac8{15}+B_+
$$



which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-cos(3t)+B_-sin(3t)+frac2{15}u(t)left(3-e^{-4t}(3cos(3t)+4sin(3t))right)
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 7 '16 at 17:48









LutzLLutzL

58.5k42054




58.5k42054












  • $begingroup$
    Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:45


















  • $begingroup$
    Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
    $endgroup$
    – Daniel Colon
    Mar 13 '16 at 1:45
















$begingroup$
Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:45




$begingroup$
Thank you @LutzL it was very helpful seeing that I was in the right way, only get me a little confuse the + and _ signs in the typesetting
$endgroup$
– Daniel Colon
Mar 13 '16 at 1:45


















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