Second order non-linear differential equation involving log

[5hassay]

EDIT: my problem is solved, thank you to those who helped

Homework Statement

Solve:

x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})

Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order non-linear equations by doing tricks such as substituting u = y^{\prime}, factoring, and then dealing with the two cases that result.

Homework Equations

None.

The Attempt at a Solution

I tried u := y^{\prime}, so u^{\prime} = y^{\prime \prime}, and so substitution gives

x u^{\prime} = u \log(\frac{u}{x})

but I can't seem to be able to do anything with that.

I also I noticed that

y^{\prime \prime} = y^{\prime} x^{-1} \log(y^{\prime} x^{-1})

which I thought was interesting. But I couldn't really do anything with it.

Thank you.

 

[SteamKing]

How about log (y'/x) = log (y') - log (x)?

 

[Dick]

Homework Statement

Solve:

x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})

Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order non-linear equations by doing tricks such as substituting u = y^{\prime}, factoring, and then dealing with the two cases that result.

Homework Equations

None.

The Attempt at a Solution

I tried u := y^{\prime}, so u^{\prime} = y^{\prime \prime}, and so substitution gives

x u^{\prime} = u \log(\frac{u}{x})

but I can't seem to be able to do anything with that.

I also I noticed that

y^{\prime \prime} = y^{\prime} x^{-1} \log(y^{\prime} x^{-1})

which I thought was interesting. But I couldn't really do anything with it.

Thank you.

You want to try and find a variable change that will make it separable. Since you have u/x in the log, substituting xv=u might be a good thing to try.

 

[Chestermiller]

Homework Statement

Solve:

x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})

Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order non-linear equations by doing tricks such as substituting u = y^{\prime}, factoring, and then dealing with the two cases that result.

Homework Equations

None.

The Attempt at a Solution

I tried u := y^{\prime}, so u^{\prime} = y^{\prime \prime}, and so substitution gives

x u^{\prime} = u \log(\frac{u}{x})

but I can't seem to be able to do anything with that.

I also I noticed that

y^{\prime \prime} = y^{\prime} x^{-1} \log(y^{\prime} x^{-1})

which I thought was interesting. But I couldn't really do anything with it.

Thank you.

You have x u^{\prime} = u \log(\frac{u}{x}). This can be rewritten as
\frac{d \ln u}{d\ln x}=\ln(u)-\ln(x)
Let ln(u) = w and ln(x) = z

 

[5hassay]

You want to try and find a variable change that will make it separable. Since you have u/x in the log, substituting xv=u might be a good thing to try.

thanks for the help.

OK, that was productive. Here's what I did:

v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime}

Substituting,

x^2 v^{\prime} + y^{\prime} = y^{\prime} \log(v) \Longrightarrow x v^{\prime} + v = v \log(v)
\Longrightarrow \frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow (v - v \log(v)) dx + x dv = 0

M := v - v \log(v)
N := x

M_v = 1 - \log(v) - 1 = -\log(v) \neq 1 = N_x

so the equation is not exact. But

\displaystyle{\frac{N_x - M_v}{M} = \frac{1 + log(v)}{v(1 - \log(v))} \neq f(v)}
\displaystyle{\frac{M_v - N_x}{N} = -\frac{\log(v) + 1}{x} \neq f(x)}
\displaystyle{\frac{N_x - M_v}{x M - v N} = -\frac{1 + \log(v)}{x v \log(v)} \neq f(xv)}

so I am stuck (one of those works when we are given a separable equation in my course).

 

[Dick]

thanks for the help.

OK, that was productive. Here's what I did:

v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime}

Substituting,

x^2 v^{\prime} + y^{\prime} = y^{\prime} \log(v) \Longrightarrow x v^{\prime} + v = v \log(v)
\Longrightarrow \frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow (v - v \log(v)) dx + x dv = 0

M := v - v \log(v)
N := x

M_v = 1 - \log(v) - 1 = -\log(v) \neq 1 = N_x

so the equation is not exact. But

\displaystyle{\frac{N_x - M_v}{M} = \frac{1 + log(v)}{v(1 - \log(v))} \neq f(v)}
\displaystyle{\frac{M_v - N_x}{N} = -\frac{\log(v) + 1}{x} \neq f(x)}
\displaystyle{\frac{N_x - M_v}{x M - v N} = -\frac{1 + \log(v)}{x v \log(v)} \neq f(xv)}

so I am stuck (one of those works when we are given a separable equation in my course).

xv'+v=vlog(v) is already separable. Just separate it and go from there.

 

[5hassay]

You have x u^{\prime} = u \log(\frac{u}{x}). This can be rewritten as
\frac{d \ln u}{d\ln x}=\ln(u)-\ln(x)
Let ln(u) = w and ln(x) = z

thanks for the help.

I don't know how to evaluate

\frac{d \log(u)}{d \log(x)}

I understand that I would be differentiating \log(u) as a function of \log(x), but I don't see how that is a function of \log(x).

 

[5hassay]

xv'+v=vlog(v) is already separable. Just separate it and go from there.

true. OK, then

\displaystyle{\frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow \int{\frac{1}{v} \cdot \frac{1}{\log(v) - 1}} dv = \int{\frac{1}{x}}} dx
\displaystyle{\Longrightarrow \log(\log(v) - 1) = \log(x) + C \Longrightarrow \log(v) - 1 = C x \Longrightarrow v = \frac{y^{\prime}}{x} = e^{C x + 1}}
\displaystyle{\Longrightarrow y = \int{x e^{C x + 1}} dx = x \frac{1}{C} e^{C x + 1} - \frac{1}{C} \int{e^{C x + 1}} dx = \frac{1}{C^2} e^{C x + 1} (C x - 1) + D}

where I just re-used C for e^C.

I haven't verified it yet by substituting

 

[Dick]

true. OK, then

\displaystyle{\frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow \int{\frac{1}{v} \cdot \frac{1}{\log(v) - 1}} dv = \int{\frac{1}{x}}} dx
\displaystyle{\Longrightarrow \log(\log(v) - 1) = \log(x) + C \Longrightarrow \log(v) - 1 = C x \Longrightarrow v = \frac{y^{\prime}}{x} = e^{C x + 1}}
\displaystyle{\Longrightarrow y = \int{x e^{C x + 1}} dx = x \frac{1}{C} e^{C x + 1} - \int{e^{C x + 1}} dx = \frac{1}{C} e^{C x + 1} (x - 1) + D}

where I just re-used C for e^C.

I haven't verified it yet by substituting

Looks fine for v, but I think you dropped a C when you were integrating by parts.

 

[Chestermiller]

thanks for the help.

I don't know how to evaluate

\frac{d \log(u)}{d \log(x)}

I understand that I would be differentiating \log(u) as a function of \log(x), but I don't see how that is a function of \log(x).

If u is a function of x, then it is a function of ln(x).

 

[5hassay]

Looks fine for v, but I think you dropped a C when you were integrating by parts.

awesome. Also, I also think you are correct about the missing C, thank you--I shall update that post. Thank you for your help!

 

[Dick]

awesome. Also, I also think you are correct about the missing C, thank you--I shall update that post. Thank you for your help!

Very welcome! If you want some extra practice you should try doing it Chestermiller's way as well, it's very nice if you don't mind keeping track of a few extra substitution variables.