Introduction to Differential Equations

What is a Differential Equation?

A differential equation is simply an equation that contains an unknown function and one or more of its derivatives. While standard algebraic equations are solved for a number (like $x = 5$), differential equations are solved to find a function (like $y = x^2 + 3$).

For example, $\frac{dy}{dx} = 3y$ or $y' = 3y$ is a differential equation. It asks: "What function $y$ has a derivative that is exactly 3 times the original function?"

Verifying a Solution

To verify if a given function is a solution to a differential equation, just plug the function and its derivative into the equation.

Example: Verify that $y = Ce^{3x}$ is a solution of $y' = 3y$.

1. Find the derivative: $y' = \frac{d}{dx}(Ce^{3x}) = 3Ce^{3x}$.
2. Substitute into the right side: $3y = 3(Ce^{3x})$.
3. Since $y' = 3y$, the function is indeed a valid solution.

Solving Separable Differential Equations

A separable differential equation is a specific type of equation where you can completely separate the variables. You can move all the $y$ terms (including $dy$) to one side of the equation and all the $x$ terms (including $dx$) to the other side.

Here are the steps to solve them:

1. Separate the variables: Manipulate the equation so that it looks like $g(y)dy = f(x)dx$.
2. Integrate both sides: Set up the integrals: $\int g(y)dy = \int f(x)dx$.
3. Solve for $y$: Evaluate the integrals (don't forget to add a constant of integration, $C$, to one side) and isolate $y$ if possible.

General Solutions vs. Initial Conditions

When you integrate, you get a $+C$ in your equation. This creates a general solution, which represents an infinite family of functions.

To find a particular solution (a specific function), you need an initial condition. This is a given point, like $y(0) = 1$, which tells you that when $x = 0$, $y = 1$. You can plug these values into your general solution to solve for the exact value of $C$.

Example Problem

Solve the differential equation $\frac{dy}{dx} = 2xy$ with the initial condition $y(0) = 1$.

Step 1: Separate the variables. Divide both sides by $y$ and multiply by $dx$: $\frac{1}{y} dy = 2x dx$

Step 2: Integrate both sides. $\int \frac{1}{y} dy = \int 2x dx$ $\ln|y| = x^2 + C$ (Note: We only need to put the constant $C$ on the $x$ side.)

Step 3: Solve for $y$ (General Solution). Exponentiate both sides to get rid of the natural log: $|y| = e^{x^2 + C}$ Using exponent rules, $e^{x^2 + C} = e^{x^2} \cdot e^C$. Since $e^C$ is just another constant, we can rename it $A$: $y = A e^{x^2}$

Step 4: Apply the initial condition. We know that $y = 1$ when $x = 0$. Plug these in to find $A$: $1 = A e^{(0)^2}$ $1 = A \cdot 1 \implies A = 1$

Step 5: Write the final particular solution. Substitute $A = 1$ back into the general solution: $y = e^{x^2}$