Mixing
Solve the ODE $ddot{y}+2lambdadot{y} + n^2y=fcos(wt)$ for given boundary conditions.
$begingroup$
I'm having trouble understanding the given solution to this problem.
Solve the ODE $$ddot{y}+2lambdadot{y} + n^2y=fcos(wt)$$ with $y(0)=dot{y}(0)=0$, where $f$, $n$, $w$ and $lambda$ are positive constants with $lambda <<n$.
Solution
The 3rd line onward is what I am having difficulty understanding. My attempt is below.
Attempt
$y_h=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))$ where $y_h$ is the homogeneous solution.
For the particular solution guess $y_p=B_1cos(wt)+B_2sin(wt)$, so $dot{y}_p=-B_1 wsin(wt)+B_2 wcos(wt)$, and $ddot{y}_p=-B_1 w^2cos(wt)-B_2 w^2sin(wt)$.
Substitute these into the ODE to get
$$ cos(wt)[B_1 (n^2-w^2)+2lambda w B_2]+sin(wt)[B_2(n^2-w^2)-2lambda w B_1]=fcos(wt).$$
Then $$B_1 (n^2- w^2)+2lambda w B_2 =f quad&quad B_2(n^2-w^2)-2lambda w B_1=0$$
The 2nd equation gives $B_1=B_2dfrac{(n^2-w^2)}{2lambda w}$. Substituting into the 1st equation gives $B_2=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}$, which implies $B_1=dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}$.
Substituting these back into the original guess we get
$$y_p=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
So the general solution is
$$y=y_h+y_p=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))+dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
...and now I just feel lost (partly due to being rusty and the solution being written in a foreign style).
ordinary-differential-equations
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the given solution to this problem.
Solve the ODE $$ddot{y}+2lambdadot{y} + n^2y=fcos(wt)$$ with $y(0)=dot{y}(0)=0$, where $f$, $n$, $w$ and $lambda$ are positive constants with $lambda <<n$.
Solution
The 3rd line onward is what I am having difficulty understanding. My attempt is below.
Attempt
$y_h=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))$ where $y_h$ is the homogeneous solution.
For the particular solution guess $y_p=B_1cos(wt)+B_2sin(wt)$, so $dot{y}_p=-B_1 wsin(wt)+B_2 wcos(wt)$, and $ddot{y}_p=-B_1 w^2cos(wt)-B_2 w^2sin(wt)$.
Substitute these into the ODE to get
$$ cos(wt)[B_1 (n^2-w^2)+2lambda w B_2]+sin(wt)[B_2(n^2-w^2)-2lambda w B_1]=fcos(wt).$$
Then $$B_1 (n^2- w^2)+2lambda w B_2 =f quad&quad B_2(n^2-w^2)-2lambda w B_1=0$$
The 2nd equation gives $B_1=B_2dfrac{(n^2-w^2)}{2lambda w}$. Substituting into the 1st equation gives $B_2=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}$, which implies $B_1=dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}$.
Substituting these back into the original guess we get
$$y_p=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
So the general solution is
$$y=y_h+y_p=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))+dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
...and now I just feel lost (partly due to being rusty and the solution being written in a foreign style).
ordinary-differential-equations
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the given solution to this problem.
Solve the ODE $$ddot{y}+2lambdadot{y} + n^2y=fcos(wt)$$ with $y(0)=dot{y}(0)=0$, where $f$, $n$, $w$ and $lambda$ are positive constants with $lambda <<n$.
Solution
The 3rd line onward is what I am having difficulty understanding. My attempt is below.
Attempt
$y_h=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))$ where $y_h$ is the homogeneous solution.
For the particular solution guess $y_p=B_1cos(wt)+B_2sin(wt)$, so $dot{y}_p=-B_1 wsin(wt)+B_2 wcos(wt)$, and $ddot{y}_p=-B_1 w^2cos(wt)-B_2 w^2sin(wt)$.
Substitute these into the ODE to get
$$ cos(wt)[B_1 (n^2-w^2)+2lambda w B_2]+sin(wt)[B_2(n^2-w^2)-2lambda w B_1]=fcos(wt).$$
Then $$B_1 (n^2- w^2)+2lambda w B_2 =f quad&quad B_2(n^2-w^2)-2lambda w B_1=0$$
The 2nd equation gives $B_1=B_2dfrac{(n^2-w^2)}{2lambda w}$. Substituting into the 1st equation gives $B_2=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}$, which implies $B_1=dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}$.
Substituting these back into the original guess we get
$$y_p=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
So the general solution is
$$y=y_h+y_p=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))+dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
...and now I just feel lost (partly due to being rusty and the solution being written in a foreign style).
ordinary-differential-equations
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
$endgroup$
I'm having trouble understanding the given solution to this problem.
Solve the ODE $$ddot{y}+2lambdadot{y} + n^2y=fcos(wt)$$ with $y(0)=dot{y}(0)=0$, where $f$, $n$, $w$ and $lambda$ are positive constants with $lambda <<n$.
Solution
The 3rd line onward is what I am having difficulty understanding. My attempt is below.
Attempt
$y_h=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))$ where $y_h$ is the homogeneous solution.
For the particular solution guess $y_p=B_1cos(wt)+B_2sin(wt)$, so $dot{y}_p=-B_1 wsin(wt)+B_2 wcos(wt)$, and $ddot{y}_p=-B_1 w^2cos(wt)-B_2 w^2sin(wt)$.
Substitute these into the ODE to get
$$ cos(wt)[B_1 (n^2-w^2)+2lambda w B_2]+sin(wt)[B_2(n^2-w^2)-2lambda w B_1]=fcos(wt).$$
Then $$B_1 (n^2- w^2)+2lambda w B_2 =f quad&quad B_2(n^2-w^2)-2lambda w B_1=0$$
The 2nd equation gives $B_1=B_2dfrac{(n^2-w^2)}{2lambda w}$. Substituting into the 1st equation gives $B_2=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}$, which implies $B_1=dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}$.
Substituting these back into the original guess we get
$$y_p=dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
So the general solution is
$$y=y_h+y_p=e^{-lambda t}(A_1cos(sigma t)+A_2sin(sigma t))+dfrac{2lambda w f}{(n^2-w^2)^2+4lambda^2w^2}cos(wt)+dfrac{f(n^2-w^2)}{(n^2-w^2)^2+4lambda^2w^2}sin(wt).$$
...and now I just feel lost (partly due to being rusty and the solution being written in a foreign style).
ordinary-differential-equations
ordinary-differential-equations
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
asked Jan 5 at 2:57
Ryan TandyRyan Tandy
16410
16410
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The solutions are both correct, if you were to insert the constants $B_1,B_2$ correctly back into $y_p$. The difference of the forms is "hidden" in the not largely discussed nature of $alpha$.
The cited solution considers the original equation as the real part of the complex equation
$$
ddot z+2λdot z+n^2z=f,e^{iomega t}.
$$
This equation now only has a single exponential as inhomogeneity, so that the standard procedure of the method of undetermined coefficients easily applies, trying $z_p(t)=C,e^{iomega t}$ one finds $$C=frac{f}{-ω^2+2iλω+n^2}.$$
At this point, instead of using the polar decomposition of $C=Rcdot e^{iα}$, you could as well just compute the real part of the solution found,
$$
y_p(t)=Re(z_p(t))=Releft(frac{f((n^2-ω^2)-2iλω)(cos ωt+isin ωt)}{(n^2-ω^2)^2+(2λω)^2}right)
\
=frac{f((n^2-ω^2)cos ωt+2λωsin ωt)}{(n^2-ω^2)^2+(2λω)^2}
$$
which is your solution with the constants inserted correctly.
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
$endgroup$
add a comment |
$begingroup$
This kind of DE (linear with constant coefficients) is well suited to be solved with the called operational methods like the Laplace transform method. After applying the transform we have:
$$
mathcal{L}left(ddot{y}+2lambdadot{y} + n^2yright)=mathcal{L}left(fcos(wt)right)
$$
or
$$
s^2Y(s)-sy_0-dot y_0+2lambda(s Y(s)-y_0)+n^2Y(s) = frac{f s}{s^2+w^2}
$$
or
$$
Y(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}+frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
here we have
$$
Y_h(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}\
Y_p(s) = frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
$Y_h(s)$ is the homogeneous solution which in this case is null due to the initial conditions $dot y_0=y_0= 0$ so here
$$
y(t) = mathcal{L}^{-1}left(frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}right)
$$
that can be easily determined using the anti-transform tables.
NOTE
$$
frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)} = frac{a_1 s+ b_1}{s^2+w^2}+frac{a_2s+b_2}{s^2+2lambda s+n^2}
$$
and
$$
mathcal{L}^{-1}left(frac{a_1 s+ b_1}{s^2+w^2}right) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)\
mathcal{L}^{-1}left(frac{a_2s+b_2}{s^2+2lambda s+n^2}right) = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
Here $phi(t)$ is the Heavside step function. We can observe also that
$$
y_p(t) = y_{ss}(t)+y_{tr}(t)
$$
with
$$
y_{ss}(t) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)
$$
the steady state response to the forcing input $(fcos(wt))$
and
$$
y_{tr} = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
the transitory response to the forcing input
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062363%2fsolve-the-ode-ddoty2-lambda-doty-n2y-f-coswt-for-given-boundary-con%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The solutions are both correct, if you were to insert the constants $B_1,B_2$ correctly back into $y_p$. The difference of the forms is "hidden" in the not largely discussed nature of $alpha$.
The cited solution considers the original equation as the real part of the complex equation
$$
ddot z+2λdot z+n^2z=f,e^{iomega t}.
$$
This equation now only has a single exponential as inhomogeneity, so that the standard procedure of the method of undetermined coefficients easily applies, trying $z_p(t)=C,e^{iomega t}$ one finds $$C=frac{f}{-ω^2+2iλω+n^2}.$$
At this point, instead of using the polar decomposition of $C=Rcdot e^{iα}$, you could as well just compute the real part of the solution found,
$$
y_p(t)=Re(z_p(t))=Releft(frac{f((n^2-ω^2)-2iλω)(cos ωt+isin ωt)}{(n^2-ω^2)^2+(2λω)^2}right)
\
=frac{f((n^2-ω^2)cos ωt+2λωsin ωt)}{(n^2-ω^2)^2+(2λω)^2}
$$
which is your solution with the constants inserted correctly.
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
$endgroup$
add a comment |
$begingroup$
The solutions are both correct, if you were to insert the constants $B_1,B_2$ correctly back into $y_p$. The difference of the forms is "hidden" in the not largely discussed nature of $alpha$.
The cited solution considers the original equation as the real part of the complex equation
$$
ddot z+2λdot z+n^2z=f,e^{iomega t}.
$$
This equation now only has a single exponential as inhomogeneity, so that the standard procedure of the method of undetermined coefficients easily applies, trying $z_p(t)=C,e^{iomega t}$ one finds $$C=frac{f}{-ω^2+2iλω+n^2}.$$
At this point, instead of using the polar decomposition of $C=Rcdot e^{iα}$, you could as well just compute the real part of the solution found,
$$
y_p(t)=Re(z_p(t))=Releft(frac{f((n^2-ω^2)-2iλω)(cos ωt+isin ωt)}{(n^2-ω^2)^2+(2λω)^2}right)
\
=frac{f((n^2-ω^2)cos ωt+2λωsin ωt)}{(n^2-ω^2)^2+(2λω)^2}
$$
which is your solution with the constants inserted correctly.
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
$endgroup$
add a comment |
$begingroup$
The solutions are both correct, if you were to insert the constants $B_1,B_2$ correctly back into $y_p$. The difference of the forms is "hidden" in the not largely discussed nature of $alpha$.
The cited solution considers the original equation as the real part of the complex equation
$$
ddot z+2λdot z+n^2z=f,e^{iomega t}.
$$
This equation now only has a single exponential as inhomogeneity, so that the standard procedure of the method of undetermined coefficients easily applies, trying $z_p(t)=C,e^{iomega t}$ one finds $$C=frac{f}{-ω^2+2iλω+n^2}.$$
At this point, instead of using the polar decomposition of $C=Rcdot e^{iα}$, you could as well just compute the real part of the solution found,
$$
y_p(t)=Re(z_p(t))=Releft(frac{f((n^2-ω^2)-2iλω)(cos ωt+isin ωt)}{(n^2-ω^2)^2+(2λω)^2}right)
\
=frac{f((n^2-ω^2)cos ωt+2λωsin ωt)}{(n^2-ω^2)^2+(2λω)^2}
$$
which is your solution with the constants inserted correctly.
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
$endgroup$
The solutions are both correct, if you were to insert the constants $B_1,B_2$ correctly back into $y_p$. The difference of the forms is "hidden" in the not largely discussed nature of $alpha$.
The cited solution considers the original equation as the real part of the complex equation
$$
ddot z+2λdot z+n^2z=f,e^{iomega t}.
$$
This equation now only has a single exponential as inhomogeneity, so that the standard procedure of the method of undetermined coefficients easily applies, trying $z_p(t)=C,e^{iomega t}$ one finds $$C=frac{f}{-ω^2+2iλω+n^2}.$$
At this point, instead of using the polar decomposition of $C=Rcdot e^{iα}$, you could as well just compute the real part of the solution found,
$$
y_p(t)=Re(z_p(t))=Releft(frac{f((n^2-ω^2)-2iλω)(cos ωt+isin ωt)}{(n^2-ω^2)^2+(2λω)^2}right)
\
=frac{f((n^2-ω^2)cos ωt+2λωsin ωt)}{(n^2-ω^2)^2+(2λω)^2}
$$
which is your solution with the constants inserted correctly.
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
edited Jan 5 at 11:12
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
answered Jan 5 at 10:50
LutzLLutzL
57.3k42054
57.3k42054
add a comment |
add a comment |
$begingroup$
This kind of DE (linear with constant coefficients) is well suited to be solved with the called operational methods like the Laplace transform method. After applying the transform we have:
$$
mathcal{L}left(ddot{y}+2lambdadot{y} + n^2yright)=mathcal{L}left(fcos(wt)right)
$$
or
$$
s^2Y(s)-sy_0-dot y_0+2lambda(s Y(s)-y_0)+n^2Y(s) = frac{f s}{s^2+w^2}
$$
or
$$
Y(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}+frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
here we have
$$
Y_h(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}\
Y_p(s) = frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
$Y_h(s)$ is the homogeneous solution which in this case is null due to the initial conditions $dot y_0=y_0= 0$ so here
$$
y(t) = mathcal{L}^{-1}left(frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}right)
$$
that can be easily determined using the anti-transform tables.
NOTE
$$
frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)} = frac{a_1 s+ b_1}{s^2+w^2}+frac{a_2s+b_2}{s^2+2lambda s+n^2}
$$
and
$$
mathcal{L}^{-1}left(frac{a_1 s+ b_1}{s^2+w^2}right) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)\
mathcal{L}^{-1}left(frac{a_2s+b_2}{s^2+2lambda s+n^2}right) = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
Here $phi(t)$ is the Heavside step function. We can observe also that
$$
y_p(t) = y_{ss}(t)+y_{tr}(t)
$$
with
$$
y_{ss}(t) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)
$$
the steady state response to the forcing input $(fcos(wt))$
and
$$
y_{tr} = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
the transitory response to the forcing input
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
$endgroup$
add a comment |
$begingroup$
This kind of DE (linear with constant coefficients) is well suited to be solved with the called operational methods like the Laplace transform method. After applying the transform we have:
$$
mathcal{L}left(ddot{y}+2lambdadot{y} + n^2yright)=mathcal{L}left(fcos(wt)right)
$$
or
$$
s^2Y(s)-sy_0-dot y_0+2lambda(s Y(s)-y_0)+n^2Y(s) = frac{f s}{s^2+w^2}
$$
or
$$
Y(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}+frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
here we have
$$
Y_h(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}\
Y_p(s) = frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
$Y_h(s)$ is the homogeneous solution which in this case is null due to the initial conditions $dot y_0=y_0= 0$ so here
$$
y(t) = mathcal{L}^{-1}left(frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}right)
$$
that can be easily determined using the anti-transform tables.
NOTE
$$
frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)} = frac{a_1 s+ b_1}{s^2+w^2}+frac{a_2s+b_2}{s^2+2lambda s+n^2}
$$
and
$$
mathcal{L}^{-1}left(frac{a_1 s+ b_1}{s^2+w^2}right) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)\
mathcal{L}^{-1}left(frac{a_2s+b_2}{s^2+2lambda s+n^2}right) = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
Here $phi(t)$ is the Heavside step function. We can observe also that
$$
y_p(t) = y_{ss}(t)+y_{tr}(t)
$$
with
$$
y_{ss}(t) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)
$$
the steady state response to the forcing input $(fcos(wt))$
and
$$
y_{tr} = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
the transitory response to the forcing input
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
$endgroup$
add a comment |
$begingroup$
This kind of DE (linear with constant coefficients) is well suited to be solved with the called operational methods like the Laplace transform method. After applying the transform we have:
$$
mathcal{L}left(ddot{y}+2lambdadot{y} + n^2yright)=mathcal{L}left(fcos(wt)right)
$$
or
$$
s^2Y(s)-sy_0-dot y_0+2lambda(s Y(s)-y_0)+n^2Y(s) = frac{f s}{s^2+w^2}
$$
or
$$
Y(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}+frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
here we have
$$
Y_h(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}\
Y_p(s) = frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
$Y_h(s)$ is the homogeneous solution which in this case is null due to the initial conditions $dot y_0=y_0= 0$ so here
$$
y(t) = mathcal{L}^{-1}left(frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}right)
$$
that can be easily determined using the anti-transform tables.
NOTE
$$
frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)} = frac{a_1 s+ b_1}{s^2+w^2}+frac{a_2s+b_2}{s^2+2lambda s+n^2}
$$
and
$$
mathcal{L}^{-1}left(frac{a_1 s+ b_1}{s^2+w^2}right) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)\
mathcal{L}^{-1}left(frac{a_2s+b_2}{s^2+2lambda s+n^2}right) = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
Here $phi(t)$ is the Heavside step function. We can observe also that
$$
y_p(t) = y_{ss}(t)+y_{tr}(t)
$$
with
$$
y_{ss}(t) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)
$$
the steady state response to the forcing input $(fcos(wt))$
and
$$
y_{tr} = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
the transitory response to the forcing input
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
$endgroup$
This kind of DE (linear with constant coefficients) is well suited to be solved with the called operational methods like the Laplace transform method. After applying the transform we have:
$$
mathcal{L}left(ddot{y}+2lambdadot{y} + n^2yright)=mathcal{L}left(fcos(wt)right)
$$
or
$$
s^2Y(s)-sy_0-dot y_0+2lambda(s Y(s)-y_0)+n^2Y(s) = frac{f s}{s^2+w^2}
$$
or
$$
Y(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}+frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
here we have
$$
Y_h(s) = frac{(s+2lambda)y_0+dot y_0}{s^2+2lambda s+n^2}\
Y_p(s) = frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}
$$
$Y_h(s)$ is the homogeneous solution which in this case is null due to the initial conditions $dot y_0=y_0= 0$ so here
$$
y(t) = mathcal{L}^{-1}left(frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)}right)
$$
that can be easily determined using the anti-transform tables.
NOTE
$$
frac{f s}{(s^2+w^2)(s^2+2lambda s+n^2)} = frac{a_1 s+ b_1}{s^2+w^2}+frac{a_2s+b_2}{s^2+2lambda s+n^2}
$$
and
$$
mathcal{L}^{-1}left(frac{a_1 s+ b_1}{s^2+w^2}right) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)\
mathcal{L}^{-1}left(frac{a_2s+b_2}{s^2+2lambda s+n^2}right) = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
Here $phi(t)$ is the Heavside step function. We can observe also that
$$
y_p(t) = y_{ss}(t)+y_{tr}(t)
$$
with
$$
y_{ss}(t) = frac 1 w(a_1 w cos (w t)+b_1 sin (w t))phi(t)
$$
the steady state response to the forcing input $(fcos(wt))$
and
$$
y_{tr} = frac{e^{-lambda t} left((b_2-a_2 lambda ) sinh left(t sqrt{lambda ^2-n^2}right)+a_2 sqrt{(lambda -n) (lambda +n)} cosh left(t sqrt{lambda
^2-n^2}right)right)}{sqrt{(lambda -n) (lambda +n)}}phi(t)
$$
the transitory response to the forcing input
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
edited Jan 5 at 11:02
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
answered Jan 5 at 10:23
CesareoCesareo
8,6343516
8,6343516
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062363%2fsolve-the-ode-ddoty2-lambda-doty-n2y-f-coswt-for-given-boundary-con%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
