A simple looking DE
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While working on a physics problem, I have encountered the following DE.
$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.
I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.
ordinary-differential-equations
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
$endgroup$
add a comment |
$begingroup$
While working on a physics problem, I have encountered the following DE.
$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.
I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.
ordinary-differential-equations
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
$endgroup$
$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46
add a comment |
$begingroup$
While working on a physics problem, I have encountered the following DE.
$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.
I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.
ordinary-differential-equations
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
$endgroup$
While working on a physics problem, I have encountered the following DE.
$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.
I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
asked Jan 15 at 17:37
Esha ManideepEsha Manideep
34
34
$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46
add a comment |
$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46
$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46
$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46
add a comment |
1 Answer
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$begingroup$
Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
add a comment |
$begingroup$
Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
add a comment |
$begingroup$
Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
edited Jan 16 at 8:08
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 15 at 17:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
add a comment |
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
1
1
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19
add a comment |
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$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46