Mixing
How to find integrating factor for this differential equation
$begingroup$
Solve this differential equation $dx/dt=x^2t^3+xt$
This is not exact.I don't find any way to solve this by integrating factor
ordinary-differential-equations
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
$endgroup$
add a comment |
$begingroup$
Solve this differential equation $dx/dt=x^2t^3+xt$
This is not exact.I don't find any way to solve this by integrating factor
ordinary-differential-equations
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
$endgroup$
1
$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47
add a comment |
$begingroup$
Solve this differential equation $dx/dt=x^2t^3+xt$
This is not exact.I don't find any way to solve this by integrating factor
ordinary-differential-equations
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
$endgroup$
Solve this differential equation $dx/dt=x^2t^3+xt$
This is not exact.I don't find any way to solve this by integrating factor
ordinary-differential-equations
ordinary-differential-equations
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
asked Jan 25 at 14:07
Epsilon DeltaEpsilon Delta
876
876
1
$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47
add a comment |
1
$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47
1
1
$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47
$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
$endgroup$
add a comment |
$begingroup$
$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
$endgroup$
add a comment |
$begingroup$
$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
$endgroup$
$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
answered Mar 11 at 17:10
JJacquelinJJacquelin
45k21855
45k21855
add a comment |
add a comment |
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$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47