Worked exercise — y' = 3y with y(0)=2

Problem

Solve $y'=3y$ with the initial condition $y(0)=2$.

Solution

It is separable: $\dfrac{dy}{y} = 3\,dx$. Integrating: $\ln|y| = 3x + C$, from which $y = Ae^{3x}$. The condition $y(0)=2$ gives $A=2$, hence: $\boxed{y(x) = 2e^{3x}}$

Links

Topics: Differential equations
Concepts: Initial condition · Exponential growth · Separable variables
Functions: Exponential function
Skills: Integrating · Solving equations
Exercise type: Equation solving