Show that all solutions of $x'(t) =small begin{cases}-x(t)log(x(t)^2)& x(t)neq 0 \0 &...
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$$x'(t) = begin{cases}-x(t)log(x(t)^2) ,& x(t)neq 0 \hspace{2,6cm} 0 ,&x(t)=0\end{cases}$$
Show that all solutions of $x'(t)$ can be written as
$\1.; x(t)=0 quad forall t in mathbb{R} \2.;x(t)=exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R}\3.;x(t)=-exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R} $
Case 1 is clear and case 2 is only solving the ODE. I´m not sure with case 3. Will I get this case because i have to regard $|x| $ in $log(x^2)$ for solving the ODE?
differential-equations
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
asked Nov 21 at 18:10
mathbob
598
add a comment |
up vote
1
down vote
favorite
$$x'(t) = begin{cases}-x(t)log(x(t)^2) ,& x(t)neq 0 \hspace{2,6cm} 0 ,&x(t)=0\end{cases}$$
Show that all solutions of $x'(t)$ can be written as
$\1.; x(t)=0 quad forall t in mathbb{R} \2.;x(t)=exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R}\3.;x(t)=-exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R} $
Case 1 is clear and case 2 is only solving the ODE. I´m not sure with case 3. Will I get this case because i have to regard $|x| $ in $log(x^2)$ for solving the ODE?
differential-equations
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
asked Nov 21 at 18:10
mathbob
598
1
Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
1
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$x'(t) = begin{cases}-x(t)log(x(t)^2) ,& x(t)neq 0 \hspace{2,6cm} 0 ,&x(t)=0\end{cases}$$
Show that all solutions of $x'(t)$ can be written as
$\1.; x(t)=0 quad forall t in mathbb{R} \2.;x(t)=exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R}\3.;x(t)=-exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R} $
Case 1 is clear and case 2 is only solving the ODE. I´m not sure with case 3. Will I get this case because i have to regard $|x| $ in $log(x^2)$ for solving the ODE?
differential-equations
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
asked Nov 21 at 18:10
mathbob
598
$$x'(t) = begin{cases}-x(t)log(x(t)^2) ,& x(t)neq 0 \hspace{2,6cm} 0 ,&x(t)=0\end{cases}$$
Show that all solutions of $x'(t)$ can be written as
$\1.; x(t)=0 quad forall t in mathbb{R} \2.;x(t)=exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R}\3.;x(t)=-exp(c*exp(-2t))quad forall t in mathbb{R} ; text{ and } ; c in mathbb{R} $
Case 1 is clear and case 2 is only solving the ODE. I´m not sure with case 3. Will I get this case because i have to regard $|x| $ in $log(x^2)$ for solving the ODE?
differential-equations
differential-equations
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
asked Nov 21 at 18:10
mathbob
598
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
asked Nov 21 at 18:10
mathbob
598
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
edited Nov 23 at 18:05
Lorenzo B.
1,6622519
1,6622519
asked Nov 21 at 18:10
mathbob
598
asked Nov 21 at 18:10
mathbob
598
asked Nov 21 at 18:10
mathbob
598
598
1
Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
1
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02
add a comment |
1
Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
1
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02
1
1
Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
1
1
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02
add a comment |
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Yes. At some point in solving the ODE you have to choose whether to take $x$ to be positive or negative. 2 and 3 are just these two solutions.
– whpowell96
Nov 21 at 22:35
Yeah thx, i did it this way.
– mathbob
Nov 21 at 22:44
1
Well it boils down to the fact that $x$ can only be positive, negative, or $0$ and will never change sign. The 0 case is obvious. if $x>0$, then the solution approaches monotonically $x=1$. To see this, look at the sign of the derivative in this range. The same thing happens with $x<0$ at $x=-1$.
– whpowell96
Nov 21 at 23:02