# Solution for a second order differential equation

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Hi,
∂M/r∂r+∂2M/∂r2 = A

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WWGD
Gold Member
Hi,
∂M/r∂r+∂2M/∂r2 = A
Hi Anooja, what have you done so far? Where are you stuck?

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

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 $$M = \frac{ Ar^2}{4} +C_1 ln(r) +C_2$$

∫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
Staff Emeritus
$$y' + \frac 1r y = A.$$ Can you solve that one?