Mixing
Solve differential equation $xyy'=x^4+y^4$
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
322110
$endgroup$
add a comment |
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
322110
$endgroup$
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
322110
$endgroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
322110
asked Jan 12 at 17:01
ThomThom
322110
asked Jan 12 at 17:01
ThomThom
322110
asked Jan 12 at 17:01
ThomThom
322110
asked Jan 12 at 17:01
ThomThom
322110
322110
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
1 Answer
1
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$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
58116
$endgroup$
add a comment |
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$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
58116
$endgroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
58116
answered Jan 12 at 19:15
ChristophChristoph
58116
answered Jan 12 at 19:15
ChristophChristoph
58116
answered Jan 12 at 19:15
ChristophChristoph
58116
58116
add a comment |
add a comment |
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$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49