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Solve complicated second order differential equation


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Hello everyone,

 

I have been having a hard time with this equation...

 image.png.3e981ecea4568784dbb34c51bc6ba435.png

I would really like to know if anybody has an idea on how to solve it...

 

All the suggestions are very much appreciated :)

 

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It would be a lot easier to understand and be sure of your notation if you provided a symbols key.

is FcH+c meant to be  (F) (cH+)c  ?

Have you plotted a direction or characteristic field for the equation?

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The following 6 symbols below are constants:

image.png.25dc4f062819621693324188a05a0cde.png  ... image.png.a36a2c60f2e946f8846ae3b0d25331d5.png ... image.png.7747609ae106d22aa6bf0ab5fc06582f.png ... F... R... T...

And what I need is to solve the equation for image.png.c7c97beea08762a7ae066a9364e9089c.png, which varies with r.

 

Regarding your questions, I haven't plotted the characteristic field as I never heard about it...

Only image.png.c7c97beea08762a7ae066a9364e9089c.png is a function of r...

 

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4 hours ago, Miguel_s said:

The following 6 symbols below are constants:

image.png.25dc4f062819621693324188a05a0cde.png  ... image.png.a36a2c60f2e946f8846ae3b0d25331d5.png ... image.png.7747609ae106d22aa6bf0ab5fc06582f.png ... F... R... T...

And what I need is to solve the equation for image.png.c7c97beea08762a7ae066a9364e9089c.png, which varies with r.

 

Regarding your questions, I haven't plotted the characteristic field as I never heard about it...

Only image.png.c7c97beea08762a7ae066a9364e9089c.png is a function of r... 

 

Since you only have two variables I agree with mathematic that this is an ODE.

I would proceed as follows:

Replacing all your constants and collecting them together, you only need 3 constants.

I have started with four, A, B, C and D and finally added E = B/A


[math] - \frac{d}{{dr}}\left( {\varepsilon r\frac{{d\varphi }}{{dr}}} \right) = B{e^{ - \left( {C\varphi  + D} \right)}}[/math]


[math] - \varepsilon \frac{d}{{dr}}\left( {r\frac{{d\varphi }}{{dr}}} \right) = B{e^{ - \left( {C\varphi  + D} \right)}}[/math]


[math]A\frac{d}{{dr}}\left( {r\frac{{d\varphi }}{{dr}}} \right) = B{e^{ - \left( {C\varphi  + D} \right)}}[/math]


[math]A\left( {\frac{{d\varphi }}{{dr}} + r\frac{{{d^2}\varphi }}{{d{r^2}}}} \right) = B{e^{ - \left( {C\varphi  + D} \right)}}[/math]


[math]\left( {\frac{{d\varphi }}{{dr}} + r\frac{{{d^2}\varphi }}{{d{r^2}}}} \right) = E{e^{ - \left( {C\varphi  + D} \right)}}[/math]

 

 

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17 minutes ago, studiot said:

Since you only have two variables I agree with mathematic that this is an ODE.

I would proceed as follows:

Replacing all your constants and collecting them together, you only need 3 constants.

I have started with four, A, B, C and D and finally added E = B/A


ddr(εrdφdr)=Be(Cφ+D)


εddr(rdφdr)=Be(Cφ+D)


Addr(rdφdr)=Be(Cφ+D)


A(dφdr+rd2φdr2)=Be(Cφ+D)


(dφdr+rd2φdr2)=Ee(Cφ+D)

 

 

It definitely looks better. From that stage on is where I can't just go through... I have tried variable substitution before and nothing...

 

If you have any idea I would really appreciate  

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