Mixing
Values of Constants for an Exponentially Decaying General Solution
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I have a doubt about the following question:
For the differential equation with constant coefficients
$$y'' + by' + cy = 0 $$
the general solution $y(x)$ approaches zero as $x$ approaches $infty$ if
(a) $b gt 0, c gt 0 $
(b) $b gt 0, c lt 0 $
(c) $b lt 0, c gt 0 $
(d) $b lt 0, c lt 0 $
Doubt:
On solving the differential equation by forming the auxiliary equation, I get the general solution (assuming $sqrtb^2-4c gt 0$) as: $$y=c_1e^xfrac-b + sqrtb^2 - 4c2 + c_2e^xfrac-b - sqrtb^2 - 4c2$$
I tried doing $limlimits_x to infty y$, which equated to zero (I might be stupid by doing so). But I have no idea on answering the actual question. I would appreciate if someone could help me and let me know if I am missing anything out.
Edit 1: The solution according to the answer key is (a).
differential-equations
asked Jul 26 at 18:14
Pranav Dinesh
75
add a comment |Â
up vote
0
down vote
favorite
I have a doubt about the following question:
For the differential equation with constant coefficients
$$y'' + by' + cy = 0 $$
the general solution $y(x)$ approaches zero as $x$ approaches $infty$ if
(a) $b gt 0, c gt 0 $
(b) $b gt 0, c lt 0 $
(c) $b lt 0, c gt 0 $
(d) $b lt 0, c lt 0 $
Doubt:
On solving the differential equation by forming the auxiliary equation, I get the general solution (assuming $sqrtb^2-4c gt 0$) as: $$y=c_1e^xfrac-b + sqrtb^2 - 4c2 + c_2e^xfrac-b - sqrtb^2 - 4c2$$
I tried doing $limlimits_x to infty y$, which equated to zero (I might be stupid by doing so). But I have no idea on answering the actual question. I would appreciate if someone could help me and let me know if I am missing anything out.
Edit 1: The solution according to the answer key is (a).
differential-equations
asked Jul 26 at 18:14
Pranav Dinesh
75
1
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a doubt about the following question:
For the differential equation with constant coefficients
$$y'' + by' + cy = 0 $$
the general solution $y(x)$ approaches zero as $x$ approaches $infty$ if
(a) $b gt 0, c gt 0 $
(b) $b gt 0, c lt 0 $
(c) $b lt 0, c gt 0 $
(d) $b lt 0, c lt 0 $
Doubt:
On solving the differential equation by forming the auxiliary equation, I get the general solution (assuming $sqrtb^2-4c gt 0$) as: $$y=c_1e^xfrac-b + sqrtb^2 - 4c2 + c_2e^xfrac-b - sqrtb^2 - 4c2$$
I tried doing $limlimits_x to infty y$, which equated to zero (I might be stupid by doing so). But I have no idea on answering the actual question. I would appreciate if someone could help me and let me know if I am missing anything out.
Edit 1: The solution according to the answer key is (a).
differential-equations
asked Jul 26 at 18:14
Pranav Dinesh
75
I have a doubt about the following question:
For the differential equation with constant coefficients
$$y'' + by' + cy = 0 $$
the general solution $y(x)$ approaches zero as $x$ approaches $infty$ if
(a) $b gt 0, c gt 0 $
(b) $b gt 0, c lt 0 $
(c) $b lt 0, c gt 0 $
(d) $b lt 0, c lt 0 $
Doubt:
On solving the differential equation by forming the auxiliary equation, I get the general solution (assuming $sqrtb^2-4c gt 0$) as: $$y=c_1e^xfrac-b + sqrtb^2 - 4c2 + c_2e^xfrac-b - sqrtb^2 - 4c2$$
I tried doing $limlimits_x to infty y$, which equated to zero (I might be stupid by doing so). But I have no idea on answering the actual question. I would appreciate if someone could help me and let me know if I am missing anything out.
Edit 1: The solution according to the answer key is (a).
differential-equations
asked Jul 26 at 18:14
Pranav Dinesh
75
edited Jul 26 at 18:49
asked Jul 26 at 18:14
Pranav Dinesh
75
asked Jul 26 at 18:14
Pranav Dinesh
75
asked Jul 26 at 18:14
Pranav Dinesh
75
75
1
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37
add a comment |Â
1
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37
1
1
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
down vote
According to your solution,
$$
C_1e^frac x2(-b+sqrtb^2-4c)+C_2e^frac x2(-b-sqrtb^2-4c)
$$
If
a) we have $-b < 0;$ and $sqrtb^2-4c < b;$ with $y$ exponentially decaying. For $c le fracb^24;$ and if $c > fracb^24;$ then we will have
$$
y = e^-fracbx2left(C_1cosomega t + C_2sinomega tright)
$$
hence $b > 0, c > 0;Rightarrow lim_xtoinfty = 0$
b) If $b > 0, c < 0Rightarrow sqrtb^2-4c > b$ and then the exponential $e^(-fracb2+fracsqrtb^2-4c2)x$ grows positively hence $b > 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
c)
If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0;Rightarrow lim_xtoinfty = pminfty$
d) This item is treated in c) hence $b < 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
answered Jul 26 at 18:51
Cesareo
5,5912412
add a comment |Â
up vote
1
down vote
The general solution to your differential equation is $$y= C_1e^lambda_1x+ C_2e^lambda_2x$$ where $lambda _1$ and $lambda_2$ are roots of $$lambda ^2 +blambda +c =0$$
Note that the sum of your eigenvalues is $lambda _1+ lambda _2=-b$ and the product is $lambda _1lambda _2=c$
For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
According to your solution,
$$
C_1e^frac x2(-b+sqrtb^2-4c)+C_2e^frac x2(-b-sqrtb^2-4c)
$$
If
a) we have $-b < 0;$ and $sqrtb^2-4c < b;$ with $y$ exponentially decaying. For $c le fracb^24;$ and if $c > fracb^24;$ then we will have
$$
y = e^-fracbx2left(C_1cosomega t + C_2sinomega tright)
$$
hence $b > 0, c > 0;Rightarrow lim_xtoinfty = 0$
b) If $b > 0, c < 0Rightarrow sqrtb^2-4c > b$ and then the exponential $e^(-fracb2+fracsqrtb^2-4c2)x$ grows positively hence $b > 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
c)
If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0;Rightarrow lim_xtoinfty = pminfty$
d) This item is treated in c) hence $b < 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
answered Jul 26 at 18:51
Cesareo
5,5912412
add a comment |Â
up vote
1
down vote
According to your solution,
$$
C_1e^frac x2(-b+sqrtb^2-4c)+C_2e^frac x2(-b-sqrtb^2-4c)
$$
If
a) we have $-b < 0;$ and $sqrtb^2-4c < b;$ with $y$ exponentially decaying. For $c le fracb^24;$ and if $c > fracb^24;$ then we will have
$$
y = e^-fracbx2left(C_1cosomega t + C_2sinomega tright)
$$
hence $b > 0, c > 0;Rightarrow lim_xtoinfty = 0$
b) If $b > 0, c < 0Rightarrow sqrtb^2-4c > b$ and then the exponential $e^(-fracb2+fracsqrtb^2-4c2)x$ grows positively hence $b > 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
c)
If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0;Rightarrow lim_xtoinfty = pminfty$
d) This item is treated in c) hence $b < 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
answered Jul 26 at 18:51
Cesareo
5,5912412
add a comment |Â
up vote
1
down vote
up vote
1
down vote
According to your solution,
$$
C_1e^frac x2(-b+sqrtb^2-4c)+C_2e^frac x2(-b-sqrtb^2-4c)
$$
If
a) we have $-b < 0;$ and $sqrtb^2-4c < b;$ with $y$ exponentially decaying. For $c le fracb^24;$ and if $c > fracb^24;$ then we will have
$$
y = e^-fracbx2left(C_1cosomega t + C_2sinomega tright)
$$
hence $b > 0, c > 0;Rightarrow lim_xtoinfty = 0$
b) If $b > 0, c < 0Rightarrow sqrtb^2-4c > b$ and then the exponential $e^(-fracb2+fracsqrtb^2-4c2)x$ grows positively hence $b > 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
c)
If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0;Rightarrow lim_xtoinfty = pminfty$
d) This item is treated in c) hence $b < 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
answered Jul 26 at 18:51
Cesareo
5,5912412
According to your solution,
$$
C_1e^frac x2(-b+sqrtb^2-4c)+C_2e^frac x2(-b-sqrtb^2-4c)
$$
If
a) we have $-b < 0;$ and $sqrtb^2-4c < b;$ with $y$ exponentially decaying. For $c le fracb^24;$ and if $c > fracb^24;$ then we will have
$$
y = e^-fracbx2left(C_1cosomega t + C_2sinomega tright)
$$
hence $b > 0, c > 0;Rightarrow lim_xtoinfty = 0$
b) If $b > 0, c < 0Rightarrow sqrtb^2-4c > b$ and then the exponential $e^(-fracb2+fracsqrtb^2-4c2)x$ grows positively hence $b > 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
c)
If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0;Rightarrow lim_xtoinfty = pminfty$
d) This item is treated in c) hence $b < 0, c < 0;Rightarrow lim_xtoinfty = pminfty$
answered Jul 26 at 18:51
Cesareo
5,5912412
answered Jul 26 at 18:51
Cesareo
5,5912412
answered Jul 26 at 18:51
Cesareo
5,5912412
answered Jul 26 at 18:51
Cesareo
5,5912412
5,5912412
add a comment |Â
add a comment |Â
up vote
1
down vote
The general solution to your differential equation is $$y= C_1e^lambda_1x+ C_2e^lambda_2x$$ where $lambda _1$ and $lambda_2$ are roots of $$lambda ^2 +blambda +c =0$$
Note that the sum of your eigenvalues is $lambda _1+ lambda _2=-b$ and the product is $lambda _1lambda _2=c$
For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
add a comment |Â
up vote
1
down vote
The general solution to your differential equation is $$y= C_1e^lambda_1x+ C_2e^lambda_2x$$ where $lambda _1$ and $lambda_2$ are roots of $$lambda ^2 +blambda +c =0$$
Note that the sum of your eigenvalues is $lambda _1+ lambda _2=-b$ and the product is $lambda _1lambda _2=c$
For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The general solution to your differential equation is $$y= C_1e^lambda_1x+ C_2e^lambda_2x$$ where $lambda _1$ and $lambda_2$ are roots of $$lambda ^2 +blambda +c =0$$
Note that the sum of your eigenvalues is $lambda _1+ lambda _2=-b$ and the product is $lambda _1lambda _2=c$
For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
The general solution to your differential equation is $$y= C_1e^lambda_1x+ C_2e^lambda_2x$$ where $lambda _1$ and $lambda_2$ are roots of $$lambda ^2 +blambda +c =0$$
Note that the sum of your eigenvalues is $lambda _1+ lambda _2=-b$ and the product is $lambda _1lambda _2=c$
For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
edited Jul 26 at 20:00
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:26
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
add a comment |Â
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
The answer is correct as per the key. But we haven't covered the method of using eigenvalues in differential equations, although I did learn it in my linear algebra class. So, I would appreciate if you could explain it briefly.
– Pranav Dinesh
Jul 26 at 18:52
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
Mohammed +1 for your nice answer ....in your comment you forgot the variable $$y=C_1 e^lambda _1x + C_2 e^lambda _2x$$
– Isham
Jul 26 at 19:22
1
1
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@Isham Thanks for the comment. You are correct, we need the independent variable there. Thanks again.
– Mohammad Riazi-Kermani
Jul 26 at 19:46
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
@PranavDinesh The roots of the polynomial $λ^2+bλ+c=0$ are called eigenvalues. The general solution to your differential equation is the $y=C_1e^λ_1x+C_2e^λ_2 x$You want both eigenvalues to be negative so the solution to approaches zero.
– Mohammad Riazi-Kermani
Jul 26 at 19:50
add a comment |Â
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1
If the roots of the characteristic equation are real: are they both positive, both negative of one of each? What happens if the roots are complex? What happens in each of those cases?
– Doug M
Jul 26 at 18:21
Well, to be honest, I have no idea on what happens in those cases. I considered all the cases: distinct-real (discriminant $D gt 0 $), repeated ( $D = 0$) and complex ($D lt 0$) and the limit condition (i.e. limx→∞y = 0) applies in all the cases.
– Pranav Dinesh
Jul 26 at 18:37