The given differential equation is:

$y' = y$

This is a first-order linear differential equation, and we can solve it using separation of variables or by recognizing it as a standard equation with the solution $y = Ce^x$, where $C$ is a constant.

Steps for solving:

1. Rewrite the equation: $\frac{dy}{dx} = y$

2. Separate variables: $\frac{1}{y} \, dy = dx$

3. Integrate both sides: $\int \frac{1}{y} \, dy = \int dx$ The integral gives: $\ln|y| = x + C_1$ where $C_1$ is a constant of integration.

4. Solve for $y$: Exponentiate both sides to eliminate the logarithm: $|y| = e^{x + C_1} = e^{x} \cdot e^{C_1}$ Let $C = e^{C_1}$, so: $y = Ce^x$ where $C$ is a constant.

Thus, the general solution to the differential equation $y' = y$ is:

$y = Ce^x$

Would you like more details or further clarification?

Related Questions:

1. What is the solution to the equation $y' = ky$, where $k$ is a constant?
2. How do you solve the differential equation $y' = y^2$?
3. What is the role of the constant $C$ in the solution of a differential equation?
4. How do you solve second-order differential equations like $y'' = y$?
5. Can you use Laplace transforms to solve differential equations like $y' = y$?

Tip:

When solving separable differential equations, always make sure to correctly apply the initial conditions to find the specific solution.