Assigning \$v_1\$ to the node between \$R_1\$ and \$C_1\$ and assigning \$v_2\$ to the remaining node (with \$R_2\$ and \$C_2\$) the KCL says:
$$\begin{align*}\frac1{R_1}v_1+C_1\frac{\text{d}}{\text{d}t}v_1 &= \frac1{R_1}v_{in}+C_1\frac{\text{d}}{\text{d}t}v_2\\\\\frac1{R_2}v_2+C_2\frac{\text{d}}{\text{d}t}v_2+C_1\frac{\text{d}}{\text{d}t}v_2&=C_1\frac{\text{d}}{\text{d}t}v_1 \end{align*}$$
You can solve that with Cramer's Rule., easy enough. Either for \$v_1\$ or \$v_2\$ (or both.) If you need hand-holding on that process, say so.
Doing this by hand (no computer) and for \$v_2\$ I get:
$$\begin{align*}\left[\left(\frac{\text{d}}{\text{d}t}\right)^2+\left(\frac1{R_1\, C_1}+\frac1{R_1\, C_2}+\frac1{R_2\, C_2}\right)\frac{\text{d}}{\text{d}t}+\frac1{R_1\, R_2\, C_1\, C_2}\right]v_2&=\frac1{R_1\, C_2}\frac{\text{d}}{\text{d}t}15\:\text{V} \cdot u(t)\\\\&= 0\end{align*}$$
(Obviously, the derivative on the right side goes to zero with constant voltage.)
That's homogeneous. And no Laplace required!
For the following I will use a simple calculator for getting some of the numbers. Nothing more, though.
Now, I happen to know that if \$a=\frac1{R_1\, C_1}+\frac1{R_1\, C_2}+\frac1{R_2\, C_2}=2\zeta\,\omega_{_0}\$ and \$b=\frac1{R_1\, R_2\, C_1\, C_2}=\omega_{_0}^{\:2}\$, that I can find \$\zeta=\frac{a}{2\,\omega_{_0}}\$ to determine if this is critically damped, over-damped, or under-damped. (The only meaning to that is whether or not I choose \$\cos\$ and \$\sin\$ or else \$\cosh\$ and \$\sinh\$ for the solution.) Here, I get \$a=17.5\:\text{k}\$ and \$b=25\:\text{M}\$, so \$\zeta=1.75\$. That means this is over-damped. So I'd select \$\cosh\$ and \$\sinh\$ for the solution.
Put \$\frac{\text{d}^2}{\text{d}t^2}+a\frac{\text{d}}{\text{d}t}+b\$ into standard form of \$\left(\frac{\text{d}}{\text{d}t}-\alpha\right)^2+\beta^2\$.
So \$\alpha=-\frac12 a=-8750\$ and \$\beta=\frac12\sqrt{4b-a^2}= 7180.70330817254\:j\$.
Then from the standard solution form:
$$\begin{align*}v_2 &=\exp\left(\alpha\,t\right)\cdot\left[A_1\cdot \cosh\left(\frac{\beta}{j}\,t\right)+A_2\cdot \sinh\left(\frac{\beta}{j}\,t\right)\right]\end{align*}$$
From the above, we know that at \$t=0\$ it must be that \$A_1=0\$. That simplifies things a lot. Then from the rate of change (derivative) we know that \$A_2=\frac{\frac{15\:\text{V}}{R_1\,C_2}\,j}{\beta}=10.4446593573419\$.
So the final answer:
$$v_2\approx 10.4446593573419\cdot\exp\left(-8750\,t\right)\cdot\sinh\left(7180.70330817254\,t\right)$$
You can expand the \$\sinh\$ and combine a few things, if you want. But that should get you there.
Now, at this point, I'll finally invoke LTspice:
I've included a behavioral voltage source that uses the above equation so that the circuit simulation can be compared with the mathematical answer.
They are the same.