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How to rewrite differential equation $x' = x-y-xsqrt{x^2+y^2}, y' = x+y-ysqrt{x^2+y^2}$
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I have to rewrite the differential equation:
$$x' = x-y-xsqrt{x^2+y^2}$$
$$ y' = x+y-ysqrt{x^2+y^2}$$
in the form $u' = f(u)$ with $f : mathbb{R}^2 to mathbb{R}^2$
How can I do this? Can somebody explain me how I can rewrite this. Because I don't know how to start in combination with that square root
ordinary-differential-equations
asked Jan 26 at 13:40
PaulPaul
12
$endgroup$
add a comment |
$begingroup$
I have to rewrite the differential equation:
$$x' = x-y-xsqrt{x^2+y^2}$$
$$ y' = x+y-ysqrt{x^2+y^2}$$
in the form $u' = f(u)$ with $f : mathbb{R}^2 to mathbb{R}^2$
How can I do this? Can somebody explain me how I can rewrite this. Because I don't know how to start in combination with that square root
ordinary-differential-equations
asked Jan 26 at 13:40
PaulPaul
12
$endgroup$
$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
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Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
1
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
1
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05
add a comment |
$begingroup$
I have to rewrite the differential equation:
$$x' = x-y-xsqrt{x^2+y^2}$$
$$ y' = x+y-ysqrt{x^2+y^2}$$
in the form $u' = f(u)$ with $f : mathbb{R}^2 to mathbb{R}^2$
How can I do this? Can somebody explain me how I can rewrite this. Because I don't know how to start in combination with that square root
ordinary-differential-equations
asked Jan 26 at 13:40
PaulPaul
12
$endgroup$
I have to rewrite the differential equation:
$$x' = x-y-xsqrt{x^2+y^2}$$
$$ y' = x+y-ysqrt{x^2+y^2}$$
in the form $u' = f(u)$ with $f : mathbb{R}^2 to mathbb{R}^2$
How can I do this? Can somebody explain me how I can rewrite this. Because I don't know how to start in combination with that square root
ordinary-differential-equations
ordinary-differential-equations
asked Jan 26 at 13:40
PaulPaul
12
asked Jan 26 at 13:40
PaulPaul
12
asked Jan 26 at 13:40
PaulPaul
12
asked Jan 26 at 13:40
PaulPaul
12
asked Jan 26 at 13:40
PaulPaul
12
12
$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
$begingroup$
Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
1
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
1
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05
add a comment |
$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
$begingroup$
Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
1
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
1
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05
$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
$begingroup$
Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
$begingroup$
Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
1
1
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
1
1
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05
add a comment |
1 Answer
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$begingroup$
$$frac{x'}{x}-frac{y'}{y}=-frac{y}{x}-frac{x}{y}.$$
Let $frac{y}{x}=u.$
Now, we obtain:
$$left(ln|u|right)'=u+frac{1}{u}.$$
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$$frac{x'}{x}-frac{y'}{y}=-frac{y}{x}-frac{x}{y}.$$
Let $frac{y}{x}=u.$
Now, we obtain:
$$left(ln|u|right)'=u+frac{1}{u}.$$
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$frac{x'}{x}-frac{y'}{y}=-frac{y}{x}-frac{x}{y}.$$
Let $frac{y}{x}=u.$
Now, we obtain:
$$left(ln|u|right)'=u+frac{1}{u}.$$
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$frac{x'}{x}-frac{y'}{y}=-frac{y}{x}-frac{x}{y}.$$
Let $frac{y}{x}=u.$
Now, we obtain:
$$left(ln|u|right)'=u+frac{1}{u}.$$
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$$frac{x'}{x}-frac{y'}{y}=-frac{y}{x}-frac{x}{y}.$$
Let $frac{y}{x}=u.$
Now, we obtain:
$$left(ln|u|right)'=u+frac{1}{u}.$$
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 14:14
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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$begingroup$
Try taking $u = (u_1,u_2)$ with $u_1 = x$ and $u_2 = y$. Now $f_1$ will be the right hand side of the first equation and so on. As for solving it: try to rewrite it in terms of polar coordinates $x(t) = r(t)sin(theta(t))$ and $y(t) = r(t)sin(theta(t))$. The resulting equations for $r$ and $theta$ are much easier to solve.
$endgroup$
– Winther
Jan 26 at 13:47
$begingroup$
Ah yes, polar coordinates is a great way to rewrite this system of differential equations. Thank you @Winther
$endgroup$
– Paul
Jan 26 at 13:52
1
$begingroup$
@Moo Yes. Thanks!
$endgroup$
– Winther
Jan 26 at 13:56
1
$begingroup$
Or with $z=x+iy$ you can write $z'=(1+i-|z|)z$.
$endgroup$
– LutzL
Jan 26 at 14:05