I have the following two SDEs
$$dN_1=(2a-1)pN_1dt+\alpha_1 N_1dW_1$$
$$dN_2=(2pN_1-\mu N_2)dt+\alpha_2 N_2dW_2$$
where $W$ is the standard Brownian motion/Wiener process. This isn't homework, I'm just curious. I can solve the first one but the second one is in terms of $N_1$ and $N_2$ so I don't know how to go about it. I'm new to SDEs so any help is appreciated!
$$N_1(t)=N_1(0)exp\left\{((2a-1)p-\frac{1}{2}\alpha_1^2)t+\alpha_1 W_1\right\}$$
As mentioned in the comments the SDE
$$dN_1=(2a-1)pN_1dt+\alpha_1 N_1dW_1$$
is decoupled and so one can solve it
$$N_{1,t}=N_{1,0}\exp \left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_{1,t}\right).$$
for $\mu:=(2a-1)p$ and $\sigma:=\alpha_1$. Then since $W_{1}$ is independent of $W_{2}$ we can condition by it and so $N_{1,t}$ is simply a continuous process.
Then for the second equation,
$$dN_2=(2pN_1-\mu N_2)dt+\alpha_2 N_2dW_2$$
we simply use the Solution to General Linear SDE
\begin{align} dX_t = \big( a(t) X_t + b(t) \big) dt + \big( g(t) X_t \big) dB_t. \end{align}
\begin{align*} N_{2,t} =& \exp\left( \int_0^t\left( a(s) - \frac{1}{2}g^2(s) \right) \mathrm{d}s + \int_0^t g(s)\mathrm{d}B_s\right) \\ &\cdot \left(N_{2,0}+ \int_0^{t} b(s)\exp\left( \int_0^s\left( \frac{1}{2}g^2(r) - a(r)\right) \mathrm{d}r - \int_0^s g(r)\mathrm{d}B_r\right)\mathrm{d}s\right). \end{align*}