Mixing
Understanding a non-autonomous ODE
Clash Royale CLAN TAG#URR8PPP
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Consider $x'(t) = a(t)x$
a) Find a formula involving integrals for the solution of this system.
b) Prove that your formula gives the general solution of this system.
I am new to ODE's and we have gone over integrating the system...
$x'/x = a(t)$ in order to attempt to isolate x in terms of $t$. I thus get $log(x) = 0.5at^2+C ; text(a constant)$, which yields $x(t) = e^(.5at^2+C)$, not sure if I'm right here. I'm also not sure why it says "involving integrals," making it sound like the solution needs integral signs in it?
For part b, I have no idea how to even start it, not sure what I need to show.
Any help appreciated!
differential-equations
edited 2 hours ago
Robert Lewis
39.8k22459
asked 3 hours ago
MathGuyForLife
414
add a comment |Â
up vote
2
down vote
favorite
Consider $x'(t) = a(t)x$
a) Find a formula involving integrals for the solution of this system.
b) Prove that your formula gives the general solution of this system.
I am new to ODE's and we have gone over integrating the system...
$x'/x = a(t)$ in order to attempt to isolate x in terms of $t$. I thus get $log(x) = 0.5at^2+C ; text(a constant)$, which yields $x(t) = e^(.5at^2+C)$, not sure if I'm right here. I'm also not sure why it says "involving integrals," making it sound like the solution needs integral signs in it?
For part b, I have no idea how to even start it, not sure what I need to show.
Any help appreciated!
differential-equations
edited 2 hours ago
Robert Lewis
39.8k22459
asked 3 hours ago
MathGuyForLife
414
1
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider $x'(t) = a(t)x$
a) Find a formula involving integrals for the solution of this system.
b) Prove that your formula gives the general solution of this system.
I am new to ODE's and we have gone over integrating the system...
$x'/x = a(t)$ in order to attempt to isolate x in terms of $t$. I thus get $log(x) = 0.5at^2+C ; text(a constant)$, which yields $x(t) = e^(.5at^2+C)$, not sure if I'm right here. I'm also not sure why it says "involving integrals," making it sound like the solution needs integral signs in it?
For part b, I have no idea how to even start it, not sure what I need to show.
Any help appreciated!
differential-equations
edited 2 hours ago
Robert Lewis
39.8k22459
asked 3 hours ago
MathGuyForLife
414
Consider $x'(t) = a(t)x$
a) Find a formula involving integrals for the solution of this system.
b) Prove that your formula gives the general solution of this system.
I am new to ODE's and we have gone over integrating the system...
$x'/x = a(t)$ in order to attempt to isolate x in terms of $t$. I thus get $log(x) = 0.5at^2+C ; text(a constant)$, which yields $x(t) = e^(.5at^2+C)$, not sure if I'm right here. I'm also not sure why it says "involving integrals," making it sound like the solution needs integral signs in it?
For part b, I have no idea how to even start it, not sure what I need to show.
Any help appreciated!
differential-equations
differential-equations
edited 2 hours ago
Robert Lewis
39.8k22459
asked 3 hours ago
MathGuyForLife
414
edited 2 hours ago
Robert Lewis
39.8k22459
asked 3 hours ago
MathGuyForLife
414
edited 2 hours ago
Robert Lewis
39.8k22459
edited 2 hours ago
Robert Lewis
39.8k22459
edited 2 hours ago
Robert Lewis
39.8k22459
39.8k22459
asked 3 hours ago
MathGuyForLife
414
asked 3 hours ago
MathGuyForLife
414
asked 3 hours ago
MathGuyForLife
414
414
1
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago
add a comment |Â
1
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago
1
1
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago
add a comment |Â
2 Answers
2
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oldest
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up vote
2
down vote
From
$x'(t) = a(t)x(t) tag 1$
we write
$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$
whence
$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$
or
$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$
or
$x(t) = x(t_0) exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) ne 0$. If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$. Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply. Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.
answered 1 hour ago
Robert Lewis
39.8k22459
add a comment |Â
up vote
1
down vote
If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
From
$x'(t) = a(t)x(t) tag 1$
we write
$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$
whence
$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$
or
$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$
or
$x(t) = x(t_0) exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) ne 0$. If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$. Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply. Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.
answered 1 hour ago
Robert Lewis
39.8k22459
add a comment |Â
up vote
2
down vote
From
$x'(t) = a(t)x(t) tag 1$
we write
$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$
whence
$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$
or
$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$
or
$x(t) = x(t_0) exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) ne 0$. If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$. Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply. Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.
answered 1 hour ago
Robert Lewis
39.8k22459
add a comment |Â
up vote
2
down vote
up vote
2
down vote
From
$x'(t) = a(t)x(t) tag 1$
we write
$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$
whence
$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$
or
$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$
or
$x(t) = x(t_0) exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) ne 0$. If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$. Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply. Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.
answered 1 hour ago
Robert Lewis
39.8k22459
From
$x'(t) = a(t)x(t) tag 1$
we write
$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$
whence
$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$
or
$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$
or
$x(t) = x(t_0) exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) ne 0$. If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$. Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply. Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.
answered 1 hour ago
Robert Lewis
39.8k22459
answered 1 hour ago
Robert Lewis
39.8k22459
answered 1 hour ago
Robert Lewis
39.8k22459
answered 1 hour ago
Robert Lewis
39.8k22459
39.8k22459
add a comment |Â
add a comment |Â
up vote
1
down vote
If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
up vote
1
down vote
If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$
answered 1 hour ago
Claude Leibovici
114k1155129
If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$
answered 1 hour ago
Claude Leibovici
114k1155129
answered 1 hour ago
Claude Leibovici
114k1155129
answered 1 hour ago
Claude Leibovici
114k1155129
answered 1 hour ago
Claude Leibovici
114k1155129
114k1155129
add a comment |Â
add a comment |Â
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1
The right hand side is wrong. $a(t)$ is a function of $t$, so its integral is not $0.5 at^2$ (that is the integral of $a$ times $t$).
– rpa
2 hours ago
I edited your question to get the $LaTeX$ working properly. Remember to enclose your math with "$" signs; thus "$ int_a^b f(t) ;dt$" produces $int_a^b f(t) ; dt$. Cheers!
– Robert Lewis
2 hours ago
Also, rpa is right!
– Robert Lewis
2 hours ago