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

What is a Differential Equation?

Solving an equation of the form $\frac{dy}{dx} = f(x)$ is essentially asking: what function $y$, when differentiated with respect to $x$, results in the expression $f(x)$? This is the simplest type of differential equation. In the TMUA, you are expected to understand that this process is the inverse of differentiation, meaning it is solved using integration.

Finding the General Solution

To find the expression for $y$, we integrate both sides of the equation with respect to $x$. Based on the Fundamental Theorem of Calculus, integrating the derivative $\frac{dy}{dx}$ returns the function $y$. However, because the derivative of any constant value is zero, we must always add a constant of integration, denoted as $+c$, to our result.

If we integrate $\frac{dy}{dx}$ with respect to $x$, we find $\int \frac{dy}{dx} \, dx$. While the integration of the derivative returns $y$, we combine all constants generated during the process on the $x$ side of the equation as a single constant $c$. This gives us the general solution.

Worked Example

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

To find $y$, we integrate the function on the right hand side with respect to $x$:

$y = \int (3x^2 + 4x - 3) \, dx$

$y = x^3 + 2x^2 - 3x + c$

This expression represents a family of curves that all have the same gradient function but are shifted vertically relative to one another.

Particular Solutions and Boundary Conditions

In many exam questions, additional information is provided, such as a specific point $(x, y)$ that lies on the curve. These values are known as boundary conditions or initial conditions. They allow us to calculate the specific value of $c$, leading to a particular solution.

Worked Example

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

We can solve this using two equivalent methods.

Method 1: Algebraic Substitution

First, we find the general solution as shown in the previous example:

$y = x^3 + 2x^2 - 3x + c$

Next, we use the boundary condition $y = 5$ when $x = 1$ to find the value of $c$ by substitution:

$5 = (1)^3 + 2(1)^2 - 3(1) + c$

$5 = 1 + 2 - 3 + c$

$5 = 0 + c$, which means $c = 5$

Substituting this back into our general solution gives the particular solution:

$y = x^3 + 2x^2 - 3x + 5$

Method 2: Definite Integration

Alternatively, we can treat $y = 5$ when $x = 1$ as the lower limits and $y$ and $x$ as the upper limits in a definite integral. This ensures that the variables correspond to each other exactly:

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

It is worth noting a slight technicality here: because we are integrating with respect to $x$, the limits on the left hand side should technically be $x$ values. However, in practice, it is often easier to write the corresponding $y$ values as limits directly, as these are the values we substitute after the integration is performed. Following this through:

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

$y - 5 = (x^3 + 2x^2 - 3x) - (1^3 + 2(1)^2 - 3(1))$

$y - 5 = x^3 + 2x^2 - 3x - 0$

$y = x^3 + 2x^2 - 3x + 5$

Both methods provide the same unique particular solution.