Did you know that we can solve a differential equation by integration?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
And the simplest of all differential equations to solve are the first-order equations with separable variables.
Why Separable Differential Equations?
If a function can be separated or factored into a function of \(\mathrm{x}\) times \(\mathrm{y}\), then we can detach and regroup the variables so that all the xs are together and all the ys are together. Thus, allowing us to integrate both separately and without too much difficulty.
So, a first-order differential equation of the form \(\frac{d y}{d x}=f(x) g(y)\) is said to be separable or to have separable variables such that:
\begin{align*}
\begin{aligned}
& \frac{d y}{g(y)}=f(x) d x \\\\
& \int \frac{d y}{g(y)}=\int f(x) d x \\\\
& G(y)=F(x)+C
\end{aligned}
\end{align*}
Where \(G(y)\) and \(F(x)\) are antiderivatives of \(\frac{d y}{g(y)}\) and \(f(x) d x\) respectively.
Separation of Variables Technique
The technique of separation of variables is relatively straightforward and one we often employ in calculus 1 and 2.
The only difference now is that we will make our separable differential equations a bit harder.
Integration Strategies
In Calculus 2, you only needed to use a limited amount of integration knowledge and never encountered anything more difficult than u-substitution. Now, you’re going to explore a wide range of integration strategies, including:
- Integration by parts
- Partial fractions decomposition
- And more!
Are you up for the challenge? Of course, you are! Let’s look at an example before diving into more challenging integration problems.
Example: Solving a Separable Differential Equation
Let’s solve the differential equation \(\left(1+x^{2}\right) y^{\prime}-2 x y=0\).
- Separate the variables:
- Integrate both sides separately:
- Solve for y by exponentiating both sides:
\begin{align*}
\begin{aligned}
& \left(1+x^{2}\right) \frac{dy}{dx}-2 x y=0 \\\
& \left(1+x^{2}\right) \frac{dy}{dx}=2 x y \\\
& \frac{dy}{dx}=\frac{2 x y}{\left(1+x^{2}\right)} \\\
& \frac{dy}{y}=\frac{2 x}{\left(1+x^{2}\right)} dx
\end{aligned}
\end{align*}
\begin{equation}
\int \frac{1}{y} d y=\int \frac{2 x}{\left(1+x^2\right)} d x
\end{equation}
\begin{equation}
\int \frac{1}{y} d y=\int \frac{2 x}{\left(1+x^2\right)} d x \quad u=1+x^2, d u=2 x
\end{equation}
\begin{equation}
\int \frac{1}{y} d y=\int \frac{1}{u} d u
\end{equation}
\begin{equation}
\ln |y|=\ln |u|+C
\end{equation}
\begin{equation}
\ln |y|=\ln \left(1+x^2\right)+C
\end{equation}
\begin{align*}
\begin{aligned}
& e^{\ln |y|}=e^{\ln \left(1+x^{2}\right)+C} \\\
& e^{\ln |y|}=e^{\ln \left(1+x^{2}\right)} \cdot e^{c} \\\
& y=C\left(1+x^{2}\right)
\end{aligned}
\end{align*}
And that’s it.
What’s Next?
As you progress, don’t worry about feeling overwhelmed. You will:
- Review exponential and logarithmic properties
- Practice various integration techniques
- Learn to find explicit solutions for given initial value problems using separation of variables
Get ready for an exciting journey as you master separable differential equations!
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