I'm not going to provide a full solution here, only point out an important facet of any solution you may contemplate.
Consider the following composite wall:
It has $3$ conduction zones, as well as $1$ inside convection zone and $1$ outside convection zone.
We assume steady state, so all temperatures are invariant to time. The red curve is the temperature profile throughout the wall.
It can then be shown that the heat flux $\dot{Q}$ flowing through the wall is given by:
$$\dot{Q}=UA\left(T_e-T_o\right)$$
($^{\dagger}$ see note)
where $U$ the overall heat transfer coefficient is given by:
$$\frac1U=\frac{1}{h_e}+\sum_i \frac{t_i}{\lambda_i}+\frac{1}{h_o}$$
Here, the $h$ are convection coefficients (see Newton's law of cooling), $t_i$ the thicknesses of the wall components and their respective heat conductivities $\lambda_i$.
Note that this doesn't take into account any thermal inertias but those don't feature in a steady state regime.
Without knowing $U$ no solution, transient or steady state, can be developed.
($^{\dagger}$ and not $T_{10}$ as written in the figure)
What differential equation does the system satisfy?
One possible model is as follows.
Assume an object with internal heat generation $P$, inside uniform temperature $T_e$, mass $M$ and heat capacity $M$, then for an infinitesimal time interval the heat balance is:
$$Mc\mathrm{d}T_e(t)=P\mathrm{d}t-UA\left(T_e(t)-T_o\right)\mathrm{d}t$$
$$Mc\frac{\mathrm{d}T_e(t)}{\mathrm{d}t}=P-UA\left(T_e(t)-T_o\right)$$
Substitute: $\left(T_e(t)-T_o\right)=\Theta$, then:
$$\mathrm{d}T_e(t)=\mathrm{d}\Theta$$
So:
$$Mc\frac{\mathrm{d}\Theta}{\mathrm{d}t}+UA\Theta=P$$
$$\Theta'+\alpha \Theta=\frac{P}{Mc}$$
where $\alpha=\frac{UA}{Mc}$
The ODE solves to:
$$\Theta=c_1\exp(-\alpha t)+\frac{P}{UA}$$
Initial condition:
$$\Theta(0)=T(0)-T_o$$
$$T(0)-T_o=c_1+\frac{P}{UA}$$
$$c_1=T(0)-T_o-\frac{P}{UA}$$
$$\boxed{T(t)-T_o=\left[T(0)-T_o-\frac{P}{UA}\right]\exp(-\alpha t)+\frac{P}{UA}}$$
Note that for $t \to +\infty$, $T(\infty)-T_o=\frac{P}{UA}$, the steady state temperature.
Another approach is the Fourier heat equation with load:
$$\frac{\partial T}{\partial t}=\alpha \nabla^2 +\dot{Q}$$
which for the steady state reduces to:
$$\alpha \nabla^2 +\dot{Q}=0$$
Assuming a square/rectangular geometry:
$$\alpha \left(T_{xx}+T_{yy}+T_{zz} \right)+\dot{Q}=0$$
which requires three boundary conditions.
Simplifying further, by assuming the object is a perfect cube, then:
$$T_{xx}=T_{yy}=T_{zz}$$
So:
$$3\alpha T_{xx}+\dot{Q}=0$$
No partials are needed here, so:
$$3\alpha \frac{\mathrm{d}^2T}{\mathrm{d}x^2}+\dot{Q}=0$$
Set a corner of the cube at the origin of the Cartesian coordinate system and call the side of the cube $L$.
Assume the heat loss from the cube's surfaces is by convection only, then we can write:
$$k\left(\frac{\mathrm{d}T}{\mathrm{d}x}\right)_{x=0}=h(T(0)-T_o)$$
and:
$$k\left(\frac{\mathrm{d}T}{\mathrm{d}x}\right)_{x=L}=h(T(L)-T_o)$$
Make a substitution:
$$u=T-T_o$$
Then our ODE and BCs become:
$$3\alpha u''+\dot{Q}=0$$
$$u'(0)=\frac{h}{k}u(0)$$
$$u'(L)=\frac{h}{k}u(L)$$
The ODE solves easily:
$$u'=-\frac{\dot{Q}}{3\alpha}x+c_1$$
$$u=-\frac{\dot{Q}}{6\alpha}x^2+c_1 x+c_2$$
Now use the BCs.
$$c_1=\frac{h}{k}c_2$$
$$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\left[\frac{\dot{Q}}{6\alpha}L^2+c_1 L+c_2\right]$$
$$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\left[\frac{\dot{Q}}{6\alpha}L^2+c_1 L+\frac{k}{h}c_1\right]$$
$$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\frac{\dot{Q}}{6\alpha}L^2-\frac{h}{k}c_1 L-c_1$$
$$\left(2+\frac{h}{k} L\right)c_1=\frac{\dot{Q}}{3\alpha}L\left(1-\frac{h}{k}\frac{L}{2}\right)$$
With $c_1$ and $c_2$ known and $u=T-T_O$ the core and boundary temperatutes $T(0)$ and $T(L)$ can be calculated if so wished.
Note that the $\alpha$ and $\dot{Q}$ are not the same as in the derivation above this one.
Best Answer
The equation you've been trying to derive is $$\dot{Q}=2\pi k\frac{(T_1-T_0)}{\ln{(r_0/r_1)}}$$where $\dot{Q}$ is the rate of heat loss per unit length of pipe and k is the thermal conductivity of the pipe. Note that there are two unknowns ($\dot{Q}$ and $T_1$) but only one equation. To provide closure on this, as @Gert has indicated, you need to characterize the rate of heat loss from the pipe to the surrounding air in the room: $$\dot{Q}=2\pi r_0 h(T_0-T_{surr})$$where h is the convective heat transfer coefficient on the outside of the pipe. You can then get both $\dot{Q}$ and $T_1$ by combining these equations (using an estimate of h).