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Integration II

Solving differential equations

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Summary

Solving differential equations

In a nutshell

Differential equations are equations that involve a function and its derivative. It is possible to solve a very specific type of differential equation (called "separable differential equations") using a method called "separation of variables".

Separable differential equations

A separable differential equation is an equation of the form:

$\dfrac{dy}{dx}=f(x)g(y)$

where $f(x)$ and $g(y)$ are given functions.

Separation of variables

Separation of variables is the technique used to solve separable differential equations. The process is as follows.

procedure

1.	Make sure the equation is in the form $\dfrac{dy}{dx}=f(x)g(y)$ .
2.	Bring all the functions of $y$ and $dy$ to one side and all the functions of $x$ and $dx$ to the other side. This will give $\dfrac{1}{g(y)}\,dy=f(x)\,dx$ .
3.	Integrate both sides with respect to their respective variables. Don't forget to include a constant on one side. Note: There is no need to include one on both sides, since if you do that and then subtract one constant from the other, the result is that one side has the addition of a constant while the other does not.
4.	Solve for the constant using extra given information (if applicable).

Note: Here you are treating $dy$ and $dx$ as if they are themselves variables. While this is not quite true, you can manipulate them as such.

Example 1

Solve the differential equation $\dfrac{dy}{dx}=xy$ .

"Separate the variables", so divide both sides by $y$ and multiply both sides by $dx$ :

$\begin{aligned} \dfrac{dy}{dx}&=xy\\ \Rightarrow \dfrac1y \,\dfrac{dy}{dx}&=x\\ \Rightarrow \dfrac1y \, dy &= x \, dx \end{aligned}$

Integrate both sides:

$\begin{aligned}\dfrac1y \, dy &= x \, dx \\ \int \dfrac1y \, dy &= \int x \, dx \\ \ln\vert y \vert &= \dfrac12 x^2+C\end{aligned}$

Rearrange for $y$ :

$\begin{aligned} \ln\vert y \vert &= \dfrac12 x^2+C\\y&=e^{\frac12 x^2+C}\\&=e^Ce^{\frac12 x^2}\\&=Ae^{\frac12 x^2}, \ where \ A=e^c \end{aligned}$

$\underline{y=Ae^{\frac12 x^2}}$

Note: It is perfectly fine to replace a constant (in this case $e^c$ ) with a simpler constant (such as $A$ ). The are both just constants.

General and particular solutions

The solution to the above example is a general solution to the differential equation. This is because there are an infinite number of unique functions for $y$ , as any value of $A$ can be chosen.

A particular solution is a solution that has no arbitrary constants - it is the only solution satisfying the differential equation and an added boundary condition $(a,b)$ . This means that when $x=a$ , $y=b$ and hence the constant of integration can be found.

Example 2

Find the particular solution of the differential equation $\dfrac{dy}{dx}=\dfrac{\tan(y)}{x}$ that passes through the point $(1,\dfrac{\pi}{2})$ .

Separate the variables by multiplying both sides by $dx$ and dividing both sides by $\tan(y)$ :

$\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\tan(y)}{x}\\\dfrac{1}{\tan(y)}\,dy&= \dfrac{1}{x}\,dx\\ \cot(y)\,dy&=\dfrac1x \,dx \end{aligned}$

Integrate both sides:

$\begin{aligned} \cot(y)\,dy&=\dfrac1x \,dx \\ \int \cot(y)\,dy&=\int \dfrac1x \,dx \\\ln \vert\sin(y)\vert &=\ln \vert x \vert+C \end{aligned}$

Use the boundary condition:

$\begin{aligned}\ln \vert\sin(y)\vert &=\ln \vert x \vert+C \\x=1&, y=\dfrac{\pi}{2}\\ \ln\left\vert \sin\left(\dfrac{\pi}{2}\right) \right\vert&=\ln \vert1 \vert+C\\ \ln(1)&=\ln(1)+C\\C&=0 \end{aligned}$

Rewrite the solution with the solved value of $C$ and rearrange:

$\begin{aligned}\ln \vert \sin(y)\vert &= \ln(\vert x\vert)\\ \vert \sin(y)\vert&= x \end{aligned}$

$\underline{x=\vert\sin(y)\vert}$

Note: This example also showcases that, unless otherwise instructed, it is not necessary for the solution to be written in the form $y=f(x)$ .

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