Solving Differential Equations of the Form dy/dx = f(x)

Understanding the Problem

When you are asked to solve an expression of the form $\frac{dy}{dx} = f(x)$, you are being asked to find what $y$ is when expressed in terms of $x$. Specifically, you are looking for a function $y$ that, when differentiated, results in $f(x)$. In simpler terms, you must determine what function you would differentiate to arrive at $f(x)$.

Solving Without Additional Conditions

To find $y$, we integrate both sides of the equation with respect to $x$. This process relies on the Fundamental Theorem of Calculus, which establishes that integration and differentiation are inverse operations. If you differentiate $y$ and then integrate the result, you return to $y$ (plus a constant).

Example

Solve $\frac{dy}{dx} = 3x^2 + 4x - 3$

To solve this, we integrate both sides with respect to $x$:

1. Integrating the left-hand side, $\int \frac{dy}{dx} dx$, gives us $y$.
2. Integrating the right-hand side, $\int (3x^2 + 4x - 3) dx$, gives us $x^3 + 2x^2 - 3x + c$.

Therefore, the general solution is: $y = x^3 + 2x^2 - 3x + c$

In this solution, we assume that all constants generated during the integration process are combined into a single constant, written as $c$ on the side of the independent variable $x$.

Solving With Additional Conditions

In many exam problems, you will be given additional information, such as specific values for $y$ at a given $x$. These are often referred to as boundary conditions or initial conditions. These values allow you to find the specific value of the constant $c$. There are two equivalent methods for doing this.

Example

Given $y = 5$ when $x = 1$ and $\frac{dy}{dx} = 3x^2 + 4x - 3$, find $y$ in terms of $x$.

Method 1: Using the Constant of Integration

In this method, we first find the general solution as we did above and then substitute the given values to find $c$.

1. From our previous work, we know the general solution is $y = x^3 + 2x^2 - 3x + c$.
2. Substitute $x = 1$ and $y = 5$ into this equation: $5 = (1)^3 + 2(1)^2 - 3(1) + c$ $5 = 1 + 2 - 3 + c$ $5 = 0 + c$ $c = 5$
3. Substitute the value of $c$ back into the general solution: $y = x^3 + 2x^2 - 3x + 5$

Method 2: Definite Integration with Corresponding Limits

Alternatively, we can use the given conditions as limits in a definite integral. We set the bottom limits as the known pair $(x=1, y=5)$ and the top limits as the variables $(x=x, y=y)$.

$\int_{y=5}^{y=y} \frac{dy}{dx} dx = \int_{x=1}^{x=x} (3x^2 + 4x - 3) dx$

While the notation on the left-hand side is slightly informal (as we are integrating with respect to $x$ but using $y$ limits), it is a practical way to reach the solution. This leads to:

$[y]_5^y = [x^3 + 2x^2 - 3x]_1^x$

Evaluating the limits gives: $y - 5 = (x^3 + 2x^2 - 3x) - (1^3 + 2(1)^2 - 3(1))$ $y - 5 = (x^3 + 2x^2 - 3x) - (0)$ $y = x^3 + 2x^2 - 3x + 5$

Both methods are entirely equivalent and will lead to the same particular solution.