Showing that a second order differential equation has unique bounded solution
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I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:
$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$
The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!
differential-equations upper-lower-bounds
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I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:
$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$
The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!
differential-equations upper-lower-bounds
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:
$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$
The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!
differential-equations upper-lower-bounds
I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:
$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$
The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!
differential-equations upper-lower-bounds
differential-equations upper-lower-bounds
edited Nov 17 at 3:42
Tianlalu
2,8811935
2,8811935
asked Nov 17 at 3:35
Leighmm
32
32
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1 Answer
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You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.
add a comment |
up vote
1
down vote
accepted
You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.
You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.
answered Nov 17 at 3:44
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
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