Solution for a second order differential equation

[anooja559]

Hi,
Could you please help me to solve a second-order differential equation given below
∂M/r∂r+∂²M/∂r² = A

[Moderator's note: Moved from a technical forum and thus no template.]

 

[WWGD]

Hi,
Could you please help me to solve a second-order differential equation given below
∂M/r∂r+∂²M/∂r² = A

Hi Anooja, what have you done so far? Where are you stuck?

 

[anooja559]

Hi Anooja, what have you done so far? Where are you stuck?

Hi WWGD,
Thank you so much.
So I integrated the equation
∫(∂M/r∂r+∂²M/∂r² )= ∫A ----(1)
ie ∫∂M/r∂r+∫∂²M/∂r² = ∫A
1/r*M-∫-1/r²*M + ∂M/∂r = A*r+C1 (integral by parts)
2*M/r+∂M/∂r = A*r+C1 ----(2)
Integrating eq(2) again

∫2*M/r+∂M/∂r = ∫A*r+C1

Which gives,
2*M*ln(r)+M = A*r²+C1*r+C2

However, the reference shows the solution as M = Ar²+C1*ln(r)+C2

 

[willem2]

Integrating eq(2) again

∫2*M/r+∂M/∂r = ∫A*r+C1

Which gives,
2*M*ln(r)+M = A*r2+C1*r+C2

∫2*M/r isn't equal to 2M∫ (1/r).
I can't really see how to do it, except by just guessing $M = a r^b$

The reference solution isn't completely correct, it should be $$M = \frac{ Ar^2}{4} +C_1 ln(r) +C_2$$

 

[anooja559]

∫2*M/r isn't equal to 2M∫ (1/r).
I can't really see how to do it, except by just guessing $M = a r^b$

The reference solution isn't completely correct, it should be $$M = \frac{ Ar^2}{4} +C_1 ln(r) +C_2$$

Dear Willem,
That is great, Thanks a lot, Could you please elaborate on how to get this

 

[vela]

If you let $y = \frac{dM}{dr}$, you have the first-order differential equation
$$ y' + \frac 1r y = A.$$ Can you solve that one?