According to an economic model, the budget $b(t)$ at time $t\geq 0$ in a household is chosen to maximise the lifetime utility $U[b]=\int_{0}^{\infty}e^{-\beta t}u(c(t))dt$, where $u(c)\geq 0$ is the household utility function, $\beta>0$ is the discount rate, a constant, and $c(t)$ is the household consumption satisfying the budgetary equation $b'(t)=\gamma b(t)+\omega-c(t), b(0)=b_{0}$. In this equation, $\gamma>0$ is the bank interest rate, $\omega>0$ is the household wage (both assumed constant in this model), and $b_{0}>0$ is the initial household budget $b(0)$. Note that $b(t)$ may be negative if the household is in debt, but $c(t)>0$. The initial household consumption is $c(0)=c_{0}>0$. The budget $b(t)$ is subject to a No-Ponzi condition $\lim_{t \to \infty}e^{-\gamma t}b(t)\geq 0$ (which prevents the household financing current consumption through indefinitely borrowing and rolling over debt). Throughout this question, you may assume that the usual theory of the calculus of variations is valid for this model on the infinite time interval $t\in [0, \infty)$. By using the budgetary equation to express the functional $U[b]$ in terms of $b$ and $b'$, then calculating the Euler-Lagrange equation, we know that it may be written as a differential equation in $c(t)$:
$u''(c(t))c'(t)=(\beta-\gamma)u'(c(t))$.
Let the utility function $u(c)$ for $c>0$, be $u(c)=\frac{c^{1-\theta}}{1-\theta}$, where $0<\theta<1$ is a constant and $\beta\neq(1-\theta)\gamma$.
a) Show that the differential equation $u''(c(t))c'(t)=(\beta-\gamma)u'(c(t))$ may be written as $c'(t)=\kappa c(t)$, where $\kappa=(\gamma-\beta)/\theta$, and solve this equation for the stationary path $c(t)$.
b) Hence show that the budget on the stationary path is given by $b(t)=(b_{0}+\frac{\omega}{\gamma}+\frac{c_{0}}{\kappa-\gamma})e^{\gamma t}-\frac{\omega}{\gamma}-\frac{c_{0}}{\kappa-\gamma}e^{\kappa t}$.
c) Find conditions on the constants $b_{0}, c_{0}, \omega, \gamma$ and $\kappa$ that are required for the No-Ponzi condition to hold.
Here's my work:
a) Let $u(c)=\frac{c^{1-\theta}}{1-\theta}$, where $0<\theta<1$ is a constant and $\beta\neq(1-\theta)\gamma$.
Then $u'(c)=\frac{(1-\theta)(1-\theta)c^{-\theta}-c^{1-\theta}(0)}{(1-\theta)^2}=c^{-\theta}$ and $u''(c)=-\theta c^{-\theta-1}$.
Observe that $u''(c(t))c'(t)=(\beta-\gamma)u'(c(t))\implies (-\theta c^{-\theta-1})c'(t)=(\beta-\gamma)c^{-\theta}\implies c'(t)=\frac{(\beta-\gamma)c^{-\theta}}{-\theta c^{-\theta-1}}$.
Since $\kappa=(\gamma-\beta)/\theta$, it follows that $c'(t)=\kappa c(t)$.
Now we consider the equation $c'(t)=\kappa c(t)$.
Applying the method using separation of variables produces:
$\int \frac{dc}{c(t)}=\int \kappa dt$
$ln\lvert{c(t)}\rvert=\kappa t+K$
$c(t)=Ke^{\kappa t}$ where $K=constant$.
Note that the initial condition $c(0)=c_{0}>0$ implies $K=c_{0}$.
Hence $c(t)=c_{0}e^{\kappa t}$.
Therefore, the differential equation $u''(c(t))c'(t)=(\beta-\gamma)u'(c(t))$ may be written as $c'(t)=\kappa c(t)$, where $\kappa=(\gamma-\beta)\theta$, and the stationary path $c(t)$ is $c(t)=c_{0}e^{\kappa t}$.
b) Consider the budgetary equation $b'(t)=\gamma b(t)+\omega-c(t), b(0)=b_{0}$.
Note that $c(t)=c_{0}e^{\kappa t}$.
This means $b'(t)=\gamma b(t)+\omega-c_{0}e^{\kappa t}, b(0)=b_{0}$.
Multiplying both sides of this differential equation by the integrating factor $e^{\gamma t}$ gives: $e^{\gamma t}b'(t)-\gamma e^{\gamma t}b(t)=\omega e^{\gamma t}-c_{0}e^{(\kappa+\gamma)t}$.
Now we have $(e^{\gamma t}b(t))'=\omega e^{\gamma t}-c_{0}e^{(\kappa+\gamma)t}$.
Observe that $\int (e^{\gamma t}b(t))'dt=\int (\omega e^{\gamma t}-c_{0}e^{(\kappa+\gamma)t})dt\implies e^{\gamma t}b(t)=\frac{\omega}{\gamma}e^{\gamma t}-\frac{c_{0}}{\kappa+\gamma}e^{(\kappa+\gamma)t}+K$ where $K=constant$.
Thus $b(t)=\frac{\omega}{\gamma}-\frac{c_{0}}{e^{\gamma t}(\kappa+\gamma)}e^{(\kappa+\gamma)t}+\frac{K}{e^{\gamma t}}$.
Since $b(0)=b_{0}$, it follows that $\frac{\omega}{\gamma}-\frac{c_{0}}{\kappa+\gamma}+K=b_{0}\implies K=b_{0}-\frac{\omega}{\gamma}+\frac{c_{0}}{\kappa+\gamma}$.
From here, I'm stuck. Because I cannot see how the above work leads to showing that the budget on the stationary path is given by $b(t)=(b_{0}+\frac{\omega}{\gamma}+\frac{c_{0}}{\kappa-\gamma})e^{\gamma t}-\frac{\omega}{\gamma}-\frac{c_{0}}{\kappa-\omega}e^{\kappa t}$. I don't know what's wrong in here but something must be incorrect in my above work. For part c) of this problem, how should I find those conditions that are asked above?
Alright so we have the differential equation
$$u''(c(t))c'(t)=(\beta-\gamma)u'(c(t)),$$
where
$$u(c)=\frac{c^{1-\theta}}{1-\theta}$$
with $\theta\in(0,1)$. Observe first that then
$$u'(c)=c^{-\theta},\quad u''(c)=-\theta c^{-\theta-1},$$
so that the differential equation can be written
$$-\theta c(t)^{-\theta-1}c'(t)=(\beta-\gamma)c(t)^{-\theta}.$$
Dividing both sides by $-\theta c(t)^{-\theta-1}$ we get that
$$c'(t)=\underbrace{\frac{\gamma-\beta}{\theta}}_{=\kappa}c(t)=\kappa c(t).$$
Let us now solve this, and for this we will use an integrating factor. Since $\frac{\mathrm{d}}{\mathrm{d}t}e^{-\kappa t}=-\kappa e^{-\kappa t}$ we observe have that
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{-\kappa t}c(t)\right)=-\kappa e^{-\kappa t}c(t)+e^{-\kappa t}c'(t)=e^{-\kappa t}\left(c'(t)-\kappa c(t)\right)=0,$$
which implies that
$$e^{-\kappa t}c(t)=c_0$$
for some $c_0\in\mathbb{R}$, i.e.
$$c(t)=c_0e^{\kappa t}.$$
We now want to solve the differential equation
$$b'(t)=\gamma b(t)+\omega-c(t)$$
with this $c$. In particular we thus have the differential equation
$$b'(t)=\gamma b(t)+\omega-c_0e^{\kappa t}.$$
Once again using an integrating factor we have that
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{-\gamma t}b(t)\right)=e^{-\gamma t}\left(b'(t)-\gamma b(t)\right)=e^{-\gamma t}\left(\omega-c_0e^{\kappa t}\right).$$
Integrating from $0$ to $t$ yields that
\begin{align*} e^{-\gamma t}b(t)-b_0 &=\int_0^te^{-\gamma\tau}\left(\omega-c_0e^{\kappa\tau}\right)\,\mathrm{d}\tau \\ &=\int_0^t\left(\omega e^{-\gamma\tau}-c_0e^{(\kappa-\gamma)\tau}\right)\,\mathrm{d}\tau \\ &=\biggl[-\frac{\omega}{\gamma}e^{-\gamma\tau}-\frac{c_0}{\kappa-\gamma}e^{(\kappa-\gamma)\tau}\biggr]_0^t \\ &=-\frac{\omega}{\gamma}\left(e^{-\gamma t}-1\right)-\frac{c_0}{\kappa-\gamma}\left(e^{(\kappa-\gamma)t}-1\right) \end{align*}
with $b_0=b(0)$. Adding $b_0$ then multiplying by $e^{\gamma t}$ on both sides this means that
\begin{align*} b(t) &=\left[b_0-\frac{\omega}{\gamma}\left(e^{-\gamma t}-1\right)-\frac{c_0}{\kappa-\gamma}\left(e^{(\kappa-\gamma)t}-1\right)\right]e^{\gamma t}\\ &=b_0e^{\gamma t}-\frac{\omega}{\gamma}+\frac{\omega}{\gamma}e^{\gamma t}-\frac{c_0}{\kappa-\gamma}e^{\kappa t}+\frac{c_0}{\kappa-\gamma}e^{\gamma t} \\ &=\left(b_0+\frac{\omega}{\gamma}+\frac{c_0}{\kappa-\gamma}\right)e^{\gamma t}-\frac{\omega}{\gamma}-\frac{c_0}{\kappa-\gamma}e^{\kappa t}. \end{align*}
We finally investigate under what conditions on $b_0,c_0,\omega,\gamma,\kappa$ we have that
$$\lim_{t\to\infty}e^{-\gamma t}b(t)\geq0.$$
In the problem we are given that $b_0>0$, $c_0>0$, $\omega>0$, $\gamma>0$ and $\beta>0$ (recall the definition of $\kappa$ for this). Now
$$e^{-\gamma t}b(t)=b_0+\frac{\omega}{\gamma}+\frac{c_0}{\kappa-\gamma}-\frac{\omega}{\gamma}e^{-\gamma t}-\frac{c_0}{\kappa-\gamma}e^{(\kappa-\gamma)t}.$$
Observe that if $\kappa-\gamma>0$, i.e. $\kappa>\gamma$, then the last term goes to $-\infty$, so $e^{-\gamma t}b(t)\to-\infty$, meaning we must have that $\kappa\leq\gamma$. As $\kappa\neq\gamma$ (this is given in the problem) we have that $\kappa<\gamma$. In this case we can compute the limit as
$$\lim_{t\to\infty}e^{-\gamma t}b(t)=b_0+\frac{\omega}{\gamma}-\frac{c_0}{\gamma-\kappa}.$$
Consequently our condition is that $\gamma>\kappa$ and that
$$b_0+\frac{\omega}{\gamma}-\frac{c_0}{\gamma-\kappa}\geq0.$$
Perhaps this can be simplified further in a nice way but this should hopefully be enough for you to be satisfied with the solution.