= Solution
The <Laplace transform> of a function on $t\geq0$ is
$$
\widehat f(s)=\mathcal L\{f\}(s)
=\int_0^\infty e^{-st}f(t)\,dt,
$$
for values of $s$ for which the <integral> converges. <Integration by parts> gives the <Laplace transform of a derivative>
$$
\boxed{\mathcal L\{f'\}(s)=s\widehat f(s)-f(0)}.
$$
Write $\widehat N_i(s)=\mathcal L\{N_i\}(s)$. Transforming the first two equations and using the initial data gives
$$
\widehat N_1(s)=\frac{N}{s+\lambda_1},
\qquad
\widehat N_2(s)
=\frac{N\lambda_1}{(s+\lambda_1)(s+\lambda_2)}.
$$
The third equation then gives
$$
(s+\lambda_3)\widehat N_3(s)-n
=\lambda_2\widehat N_2(s),
$$
and hence
$$
\widehat N_3(s)
=\frac n{s+\lambda_3}
+\frac{N\lambda_1\lambda_2}
{(s+\lambda_1)(s+\lambda_2)(s+\lambda_3)}.
$$
A <partial fraction decomposition> therefore yields the <sequential radioactive decay> formula
$$
\boxed{
\begin{aligned}
N_3(t)={}&ne^{-\lambda_3t}\\
&+N\lambda_1\lambda_2\left[
\frac{e^{-\lambda_1t}}{(\lambda_2-\lambda_1)(\lambda_3-\lambda_1)}
+\frac{e^{-\lambda_2t}}{(\lambda_1-\lambda_2)(\lambda_3-\lambda_2)}
+\frac{e^{-\lambda_3t}}{(\lambda_1-\lambda_3)(\lambda_2-\lambda_3)}
\right].
\end{aligned}
}
$$
Now let $\lambda_1=\lambda_2=\lambda$ and put $\Delta=\lambda_3-\lambda>0$. The transformed contribution originating from the initial $N_1$ population becomes
$$
\frac{N\lambda^2}{(s+\lambda)^2(s+\lambda_3)}.
$$
Equivalently, solve the middle equation first to obtain
$$
N_2(t)=N\lambda t e^{-\lambda t},
$$
and use an <integrating factor> in the final equation:
$$
N_3(t)=ne^{-\lambda_3t}
+N\lambda^2e^{-\lambda_3t}\int_0^t u e^{\Delta u}\,du.
$$
Evaluating the elementary integral gives
$$
\boxed{
N_3(t)
=ne^{-\lambda_3t}
+\frac{N\lambda^2}{(\lambda_3-\lambda)^2}
\left[
e^{-\lambda t}\bigl((\lambda_3-\lambda)t-1\bigr)
+e^{-\lambda_3t}
\right]
}.
$$
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