In one of the Differential equation book says the following:
The ODE population equation is $dp/dt = p$....(1)
and the general solution of the above cited equations is: $p(t) = ce^t$. ...(2)
Solving of (1) is nothing but integration. But, after integrating the (1), I dint get (2). How (2) is solution of (1). I am wrong or text book wrong?
Also, the PDE of population equation look like $u_t$= $u$ ....(3), where $u$ is in $x$ and $t$. Now as per the same text book, the general solution of (3) is $u_(x,t)$ $= F(x) e^t$. ...(4). In (4), where we got F(x) and prove it (4) is solution of (3) as well as (2) is solution of (1).
Regards.
You can do a little deduction to get (2) from (1).
\begin{equation} \begin{array}{l} \frac{dp}{dt}=p\\ \frac{dp}{p}=dt\\ \mbox{By integrating for each sides:}\\ \int\frac{dp}{p}=\int dt\\ \ln p=t+C\quad \mbox{where $C$ is a constant to be determined}\\ p=e^Ce^t\\ p=C_0e^t\quad\mbox{where $C_0=e^C$ is also a constant} \end{array} \end{equation} You can get (4) from (3) by the same method but you should view $x$ as a constant when you do the analysis (say, integral) in $t$.
For the sake of simplicity, I use a alternative method, say integral factor, to solve (3). Note the integral factor of (3) is $e^{-t}$. By multiplying the factor on each sides of (3), we have \begin{equation} \frac{\partial u}{\partial t}e^{-t}-u e^{-t}=0\\ \frac{\partial}{\partial t}(e^{-t}u)=0\\ \end{equation} Integrate each side of above equation in $t$, and note u is function of both $x$ and $t$, we have: \begin{equation} \int \frac{\partial}{\partial t}(e^{-t}u) dt=\int 0 dt\\ e^{-t}u=F(x)\quad\mbox{(*)}\\ u=F(x)e^t \end{equation} where $F(x)$ is a function to be determined. I should point out that $F(x)$ appear in (*) because $u$ is a function of both $x$ and $t$. So when we integrate the both side of equation in $t$, there must have a function of $x$ instead of a constant. (In the case of ODE, such as the previous example, there is a constant $C$ after indefinite integral.) In other word, if we compute derivative of (*) on each sides in $t$, we can get the original equation. That is, \begin{equation} \frac{\partial}{\partial t} e^{-t} u=\frac{\partial}{\partial t} F(x)\\ \frac{\partial u}{\partial t}e^{-t}-u e^{-t}=0\\ \frac{\partial u}{\partial t}=u\\ \end{equation} The derivative of $F(x)$ in $t$ is zero because it is a function of $x$ and independent of $t$.