About dy/dx in application

Question

We know that the integral of $\frac{dy}{dx}$ is $y+c$.

However, I've encountered a problem modeling the growth of population, which can be expressed as

$\frac{dP}{dt}=kP$ . where $P$ is the population and $t$ is time

Integrating both sides with regard to dt gives $P(t)= kPt$.

I know this is wrong but can someone clarify for me what exactly did I do wrong? What is the specific thing in this case preventing me from integrating on both sides?

Edit:

Thanks I know it's separation of variables. But I don't really get when to separate variables and when not... Say for example $dy/dt=2r^3$, in this case you can directly integrate both sides with respect to $t$ and get $y(t)=(t^4)/2+C$. Why can't we view ($t^3$) as a certain $P(t)$??

Answer 1

If you integrate directly you get $$\int dp=\int kp dt$$ $$\implies p(t)=k\int p dt$$ you can't evaluate the integral you are stuck

Thats why we need to separate variables here $$\frac {dp}p=kdt$$ After integration $$\int \frac {dp}p=k\int dt \implies \ln p=kt+C$$ $$p=Ce^{kt}$$

With your example with y ,note that you have no y on the right side of the equation like the equation with p. These two differential equations are different.You have y on one side and t on the other side already. Thats what we tried to have with the equation with p(t). $$\frac {dy}{dt}=2t^3$$ $$\implies \int dy=2\int t^3dt$$ $$y(t)=\frac {t^4}2+C$$

Answer 2

Say for example $dy/dt=2\cdot t^3$, in this case you can directly integrate both sides with regard to t and get $y(t)=(t^4)/2+C$

Technically you have to separate the variables as well. Multiplying both sides by $dt$

$dy=2\cdot t^3 \, dt$

Integrating both sides

$\int \, dy=\int 2\cdot t^3 \, dt$

$y+c_1=\frac{t^4}{2}+c_2$

$y=\frac{t^4}{2}+c_2-c_1$

$C=c_2-c_1$

$y=\frac{t^4}{2}+C$

Answer 3

You can solve this first order differential equation, like the other posters have said, by separation of variables.

\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} &= k P(t) \\ \frac{\mathrm{d}P}{P(t)} &= k \,\mathrm{d}t \\ \int \frac{1}{P(t)} \mathrm{d}P &= k\int\mathrm{d}t \\ \ln{P(t)} + c_1 &= k\cdot t + c_2 \\ \ln{P(t)} &= k\cdot t + c \\ P(t) &= e^{kt + c} \\ P(t) &= e^{kt} + e^c \\ P(t) &= e^{kt} + C \end{align*}

Hence, we have an implicit solution to this ODE. You can see more about this method here.