What are differential equations and how do you solve ${dy \over dx}=y$ and find $y$ in terms $x$?

Question

I had been wondering about how to solve the equation $${dy \over dx}=y$$. My progress was to use the chain rule, like setting $$z=2x\;{dy\over dx}={dy\over dz}*{dz\over dx}={dy\over d(2x)}*2=y.$$ Now I’m officially stuck. What am I supposed to do?
An answer that do work(just for checking): $e^{x+1}$

Answer 1

This is a separable differential equation. Assuming $y\ne0$ which is a condition we will check separately, divide both sides of the equation by $y$:

$$ \frac{dy}{dx}=y\\ \frac{1}{y}\frac{dy}{dx}=1\\ \int\frac{1}{y}\frac{dy}{dx}\,dx=\int\,dx\\ \int\frac{1}{y}\,dy=x+C_1\\ \ln{|y|}+C_2=x+C_1\\ \ln{|y|}=x+C_1-C_2\\ \ln{|y|}=x+C_3\\ |y|=e^{x+C_3}\\ y=\pm e^{C_3}e^{x} $$

$\pm e^{C_3}$ covers all real numbers except $0$. But $y=0$ is also a solution because $0'=0$. Therefore, we can write the following as the general solution to this equation:

$$y=Ce^x,\ C\in\mathbb{R}.$$

Answer 2

$\frac{dy}{dx}=y\Rightarrow\frac{dy}{y}=dx\Rightarrow\int\frac{dy}{y}=\int dx$

$\Rightarrow \ln\left(y\right)=x+C\Rightarrow y=e^{x+C}=C'e^x$

$C$ and $C'=e^{C}$ are constants; in order to put integrals, you should quote the theorem of separation of variables for differential equations.

Answer 3

Here comes a solution without "magical" manipulation of differentials.

Assuming that $y \neq 0$ on some interval, divide the equation with $y$: $$\frac{y'(x)}{y(x)} = 1.$$

Now the left hand side can be written as the derivative of $\ln |y(x)|$ and the right hand side as the derivative of $x$: $$\frac{d}{dx} \ln |y(x)| = \frac{d}{dx} x.$$

Therefore, $\ln |y(x)| = x + C,$ where $C$ is some constant, and on the interval we thus have $$y(x) = \pm e^{x+C} = \pm e^C e^x = C' e^x,$$ where $C' = \pm e^C$ is a nonzero constant.

We can see though that also when $C' = 0,$ we get a solution: $y(x) \equiv 0.$ We therefore have a general solution $y(x) = C' e^x,$ where $C'$ is a constant.

Answer 4

The first question in your title is

What are differential equations?

The informal answer is that often what we know about a quantity we are interested in is its rate of change. For example, if you know the velocity as a function $v(t)$ of time you can find the position $p$ by solving the differential equation $$ p'(t) = v(t) . $$ That's just integration. You use any known position to find the "constant of integration".

The differential equation you ask about, $$ y' = y, $$ says that the rate of change of $y$ is $y$ itself. In slightly more generality $$ y' = ky $$ says that the rate of change of $y$ is proportional to $y$. That classic equation models population growth over time or compound interest: the more stuff there is the faster it's growing. The solution is exponential growth: $$ y = Ce^{kt}. $$

The second order differential equation $$ y" = -ky $$ models a spring: the second derivative of position is (essentially) a force (that's Newton's law $F=ma$). The equation says that the farther $y$ is from $0$ the greater the force returning it to $0$. The solutions are the oscillating functions $y = \sin(t+ \phi)$.