Today we'll learn about our first type of differential equation, a [b]separable differential equation[/b], and an algebraic method for solving it. When you're done with today you will know how to identify a separable first order differential equation, how to use the process to solve a separable differential equation, how to use GeoGebra to assist with the solution, and how to check that your solution is correct visually and algebraically. [br][br][b]Definition:[/b][br][br]A separable differential equation is one that can be manipulated algebraically so that it is of the form:[br][br][math]N\left(y\right)\frac{dy}{dx}=M\left(x\right)[/math][br][br][b]Solution Process[/b][br][br]To solve a separable differential equation, multiply both sides by [i]dx[/i] to clear the [i]dx[/i] in the denominator of [i]dy[/i]/[i]dx[/i], and then integrate both sides.[br][br][math]N\left(y\right)\frac{dy}{dx}\cdot dx=M\left(x\right)\cdot dx[/math][br][br][math]N\left(y\right)dy=M\left(x\right)dx[/math][br][br][math]\int N\left(y\right)dy=\int M\left(x\right)dx[/math][br][br]After completing the integral (which might be quite difficult--use GeoGebra to assist!), it's typical to solve the resulting (non differential) equation for [i]y [/i]to obtain the [b]general solution[/b] of the differential equation[i].[br][br][/i]The general solution of a differential equation is the solution that involves unspecified constants of integration. If a differential equation includes an [b]initial condition[/b] then we can use the initial condition to obtain a [b]specific solution. [/b]The difference between general and specific solutions will become clearer as we work through examples, so don't stress if you don't quite understand it yet. [br][br]A good reference on separable differential equations can be found here: [br][br][url=https://tutorial.math.lamar.edu/classes/de/separable.aspx]https://tutorial.math.lamar.edu/classes/de/separable.aspx[/url][br][br]

These are examples of separable differential equations. [br][br][b]Example 1 [/b]This differential equation from day 1 is separable.[br][br][math]\frac{dy}{dx}=\frac{x}{y}[/math][br][br]However, it is not of the form [math]N\left(y\right)\frac{dy}{dx}=M\left(x\right)[/math] at present. First you need to multiply both sides by [i]y[/i]:[br][br][math]y\cdot\frac{dy}{dx}=y\cdot\frac{x}{y}[/math][br][br][math]y\cdot\frac{dy}{dx}=x[/math][br][br]Now [math]N\left(y\right)=y[/math] and [math]M\left(x\right)=x[/math], and so we can confirm that this is in fact a separable differential equation. [br]The next step is to multiply both sides by [i]dx[/i], and then we will integrate.[br][br][math]y\cdot\frac{dy}{dx}\cdot dx=x\cdot dx[/math][br][br][math]y\cdot dy=x\cdot dx[/math][br][br][math]\int ydy=\int xdx[/math][br][br][math]\frac{1}{2}y^2+c_1=\frac{1}{2}x^2+c_2[/math][br][br]Notice two constants of integration. Since [math]c_2-c_1[/math] is equivalent to a single arbitrary constant of integration [math]c_3[/math], we can subtract [math]c_1[/math] from both sides, and replace [math]c_2-c_1[/math] with [math]c_3[/math].[br][br][math]\frac{1}{2}y^2=\frac{1}{2}x^2+c_2-c_1[/math][br][br][math]\frac{1}{2}y^2=\frac{1}{2}x^2+c_3[/math][br][br]Now, as stated above, solve for y, which means multiplying both sides by 2. Notice that [math]2c_3[/math] is still an arbitrary constant, so we can replace [math]2c_3[/math] with [math]c_4[/math].[br][math]2\cdot\frac{1}{2}y^2=2\cdot\frac{1}{2}x^2+2\cdot c_3[/math][br][br][math]y^2=x^2+c_4[/math][br][br]Lastly, take the square root of both sides to obtain the general solution[br][br][math]y=\pm\sqrt{x^2+c_4}[/math][br][br]We can also just refer to [math]c_4[/math] as just [i]c[/i]. In fact, whenever multiple constants of integration are absorbed into a single constant of integration, it's normal to simply refer to them as [i]c[/i]. With this final tweak, we obtain the general solution. Noe that this is the same solution we saw on day 1; now we know how it was gotten![br][br][math]y=\pm\sqrt{x^2+c}[/math][br][br]The differential equation, in its original "slope field form" -- [math]\frac{dy}{dx}=\frac{x}{y}[/math] -- is shown in the GeoGebra applet below. The solution is also shown. Note that the + and the - parts of the square root of the solution are shown as two different functions [math]f_1(x)[/math] and [math]f_2\left(x\right)[/math].[br][br]

[b]Example 2:[/b] The following is also a separable differential equation with "initial value". [br][br][math]\frac{dy}{dx}=x^2y[/math], [math]y\left(0\right)=1[/math][br][br]We'll solve it together in class. See the notes for the worked solution. [br][br]The GeoGebra applet below shows the slope field of the differential equation. If you want, after we find the solution (or solutions), plot it (or them) in the applet.

[b]Example 3:[br][br][math]y'=\frac{3x^2+4x-4}{2y-4};y\left(1\right)=3[/math][br][/b][br]We may not get to this in class. If so, then I will update this page with the solution. A slope field is visible below however.