Mixing
Linear system of differential equations given a solution
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I am asked to Give an example of a linear system for which $(e^{−t} , alpha)$ is a solution for every constant $alpha$ and to give the general solution for this system. I came up with a linear system based on my understanding,but it seems wrong to me, and I would like some guidance on how to approach this type of problem
From what I understand we can broadly describe a linear system of differential equations as
$$x_1' = f(t,x_1,x_2)\ x_2' = f(t,x_1, x_2)$$
and so since I am given the solution $(e^{-t}, alpha)=(x_1(t), x_2(t))$, A linear system for which this is a solution can be given by
$$x_1' = -e^{-t}\ x_2' = 0$$
but to me this appears too simple and incorrect. If so how do I go about approaching such a problem?
ordinary-differential-equations systems-of-equations
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
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add a comment |
$begingroup$
I am asked to Give an example of a linear system for which $(e^{−t} , alpha)$ is a solution for every constant $alpha$ and to give the general solution for this system. I came up with a linear system based on my understanding,but it seems wrong to me, and I would like some guidance on how to approach this type of problem
From what I understand we can broadly describe a linear system of differential equations as
$$x_1' = f(t,x_1,x_2)\ x_2' = f(t,x_1, x_2)$$
and so since I am given the solution $(e^{-t}, alpha)=(x_1(t), x_2(t))$, A linear system for which this is a solution can be given by
$$x_1' = -e^{-t}\ x_2' = 0$$
but to me this appears too simple and incorrect. If so how do I go about approaching such a problem?
ordinary-differential-equations systems-of-equations
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
$endgroup$
$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
1
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49
add a comment |
$begingroup$
I am asked to Give an example of a linear system for which $(e^{−t} , alpha)$ is a solution for every constant $alpha$ and to give the general solution for this system. I came up with a linear system based on my understanding,but it seems wrong to me, and I would like some guidance on how to approach this type of problem
From what I understand we can broadly describe a linear system of differential equations as
$$x_1' = f(t,x_1,x_2)\ x_2' = f(t,x_1, x_2)$$
and so since I am given the solution $(e^{-t}, alpha)=(x_1(t), x_2(t))$, A linear system for which this is a solution can be given by
$$x_1' = -e^{-t}\ x_2' = 0$$
but to me this appears too simple and incorrect. If so how do I go about approaching such a problem?
ordinary-differential-equations systems-of-equations
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
$endgroup$
I am asked to Give an example of a linear system for which $(e^{−t} , alpha)$ is a solution for every constant $alpha$ and to give the general solution for this system. I came up with a linear system based on my understanding,but it seems wrong to me, and I would like some guidance on how to approach this type of problem
From what I understand we can broadly describe a linear system of differential equations as
$$x_1' = f(t,x_1,x_2)\ x_2' = f(t,x_1, x_2)$$
and so since I am given the solution $(e^{-t}, alpha)=(x_1(t), x_2(t))$, A linear system for which this is a solution can be given by
$$x_1' = -e^{-t}\ x_2' = 0$$
but to me this appears too simple and incorrect. If so how do I go about approaching such a problem?
ordinary-differential-equations systems-of-equations
ordinary-differential-equations systems-of-equations
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
edited Jan 24 at 23:13
Richard Villalobos
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
asked Jan 24 at 22:41
Richard VillalobosRichard Villalobos
1807
1807
$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
1
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49
add a comment |
$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
1
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49
$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
1
1
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49
add a comment |
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$begingroup$
It seems that you think that $(e^{-t})'=-te^{-t}$.
$endgroup$
– John B
Jan 24 at 22:56
$begingroup$
apoligies, I will correct my error
$endgroup$
– Richard Villalobos
Jan 24 at 23:13
1
$begingroup$
Good, so you can consider the equation $x_1'=-x_1$.
$endgroup$
– John B
Jan 24 at 23:46
$begingroup$
Ah thank you, I understand now! I was complicating things for myself by assuming that the specific solution I was given could describe the entire system instead of being more general. Thank you!
$endgroup$
– Richard Villalobos
Jan 24 at 23:49