We recognize many types of differential equation. Such recognizing is the key for solving, because then we can apply the proper method, which is able to bring the solution of DE.

We know already how to solve simple DE in the form

$$ \frac{dy}{dx} = g(x). $$Just integrating both sides gives us the solution:

$$ \begin{align} dy &= g(x)\ dx, \label{ref:separ1} \\ y &= \int g(x)\ dx, \nonumber \\ y &= G(x)+c, \nonumber \end{align} $$where $G(x)$ is an antiderivative of $g(x)$. Now what if in $(\ref{ref:separ1})$ is on the left side not only $dy$, but a function depending on $y$? The solution would be that simple as well. Just to integrate both left and right side.

That means **if we can separate
members depending on $\boldsymbol x$ together with $\boldsymbol{dx}$ and at the
same time do the same for $\boldsymbol y$, then it is easy to solve DE by
integrating both sides.** So

would be the general form of DE, which can be solved by separating variables. For example

$y' = xy$ | $y' = xy \color{red}{+ \sin(x+y)}$ |

or | or |

$y' = x/y$ | $y' = x/y \color{red}{+ \log(x+y)}$ |

are separable | are not separable |

In many practical problems a rough geometrical approximation to a solution might be all that is needed. Sometimes we may even be not able to solve the DE symbolically.

We can construct a **direction field:** a rectangular area with
grid points and compute/draw tangent $dy/dx$ **for each grid point.** See the
graphs within this chapter, e.g. the last example for illustration.

Producing direction field by hand might be time consuming. It is handy if
one can find so called **isoclines,** i.e. lines which have the
same value of the slope.

Solve DE $x\ dy -y\ dx = 0$. Find particular solution for $y(2)=4$.

There is not much effort needed to reshuffle DE into

$$ \frac 1 y dy - \frac 1 x dx = 0 $$The variables $x$ are separated from $y$. It is simple task to integrate both sides now.

$$ \begin{align} \log |y| &= \log |x| + c \nonumber \\ y &= e^{\log |x| + c} \nonumber \\ y &= e^{\log|x|} \cdot e^c \nonumber \\ y &= xC \nonumber \end{align} $$To find the particular solution we have to substitute initial value $y(x=2) = 4$ into solution. Then value $C=2$ is found.

$$ \underline{\underline{y = 2x}} $$Solve DE $yx^2dy -y^3dx = 2x^2dy$.

The DE is somehow more complicated than the previous one. We have to put some effort into separating variables. If we succeed to separate variables, then after integrating we have the solution.

Let us divide by $x^2y^3$ and see what happens

$$ \frac 1 {y^2}dy - \frac{1}{x^2}dx = 2\frac{1}{y^3}dy $$We separated the variables quite easily. Now all members can be integrated.

$$ \begin{align} -x^{-2}dx &= (2y^{-3}-y^{-2})\ dy \nonumber \\ x^{-1} &= -y^{-2}+y^{-1} + C \nonumber \\ \frac 1 x &= -\frac 1 {y^2} + \frac 1 y + C \hskip2em x\neq 0,\ y\neq 0 \nonumber \\ y^{2} &= -x + xy+xy^2C\nonumber \end{align} $$We can use quadratic formula or leave the solution in implicit form.

$$ \underline{\underline{ y^2(1-xC) = x(y-1)}} $$**Note: we have lost a solution.** The above family of solutions is not the general solution.

During the operations we ruled out $y=0$. If we substitute $y=0$ (and $dy=0$, since $y$ is constant) into given DE, then it is quite clear that the equality holds for any $x$. Thus $y=0$ is also a solution.

$$ \underline{\underline{y=0}} \hskip2em \text{(singular solution)} $$Solve DE $y' = e^{\sqrt x}/y$ for initial value $y(1) = 4$.

$$ y\ dy = \color{red}{e^{\sqrt x} dx} $$When the variables are separated, the left side is easy to integrate, but the right side is more complicated.

We have to solve $\color{red}{\int e^{\sqrt x} dx}$. It contains internal function so we have to use the chain rule backwards. First let us differentiate $e^{\sqrt x}$ and then integrate the product by integration by parts (the product rule backwards): $$ (e^{\sqrt{x}})' = \frac 1 2 e^{\sqrt{x}}x^{-1/2} $$

Now to integrate $1/2\ e^{\sqrt{x}}x^{-1/2}$ we use integration by parts. From $(uv)' = uv' + u'v \implies uv' = (uv)' - u'v$ we have $\color{blue}{\int uv' = uv - \int u'v}$. One member is easy to differentiate ($e^{\sqrt{x}}$) and the second one is easy to integrate ($1/2\ x^{-1/2}$).

$$ \begin{align} e^{\sqrt{x}} &= \int \frac 1 2 e^{\sqrt{x}}x^{-1/2}\ dx = \nonumber \\ &= \int uv'\ dx = \begin{vmatrix} u = e^{\sqrt{x}} & \implies & u' = \frac 1 2 e^{\sqrt{x}}x^{- 1/2} \\ v' = \frac 1 2 x^{- 1/2} & \implies & v = x^{ 1/2 } \end{vmatrix} = \nonumber \\ &= e^{\sqrt{x}} x^{1/2} -\frac 1 2 \int e^{\sqrt x}dx \nonumber\\ \frac 1 2 \int e^{\sqrt x}dx &= e^{\sqrt{x}} x^{1/2} - e^{\sqrt{x}}\nonumber\\ \color{red}{\int e^{\sqrt x}dx} &= 2e^{\sqrt{x}} x^{1/2} - 2e^{\sqrt{x}}\nonumber \end{align} $$Now we have figured out how to integrate both left and right side, so we can integrate, therefore complete the solution of DE:

$$ \begin{align} \frac 1 2 y^2 = 2e^{\sqrt{x}} x^{1/2} - 2e^{\sqrt{x}} + c. \nonumber \\ y = 2\sqrt{e^{\sqrt{x}} \sqrt x - e^{\sqrt{x}} + C}. \nonumber \end{align} $$If the initial value $y(x=1) = 4$ is substituted into the general solution, the value of constant is determined as $C = 4$. The particular solution is then

$$ \underline{\underline{y = 2\sqrt{e^{\sqrt{x}} \sqrt{x} - e^{\sqrt{x}} + 4}}}. $$Solve DE $y' +y = 0$ for initial value $y(1) = 1$.

$$ \begin{align} \frac 1 y dy + dx &= 0 \nonumber\\ \log|y|+x &= c \nonumber\\ \log|y| &= c-x \nonumber\\ y &= e^{c-x} \nonumber \end{align} $$Substituting initial value $y(x=1) =1$ into solution gives $c=1$

$$ \underline{\underline{y=e^{1-x}}} $$Solve DE $\sin x \cos 2y\ dx + \cos x \sin 2y\ dy = 0$ for initial value $y(0) = \pi/2$.

$$ \begin{align} \frac{\sin x}{\cos x}\ dx + \frac{\sin 2y}{\cos 2y}\ dy &= 0 \nonumber \\ \tan x\ dx + \tan 2y\ dy &= 0 \nonumber \\ -\log|\cos x|-\frac 1 2 \log|\cos 2y| &= C \nonumber \\ \log|\cos x|+\frac 1 2 \log|\cos 2y| &= D \nonumber \\ \log \sqrt{|\cos 2y|} &= D - \log|\cos x| \nonumber \\ \sqrt {|\cos 2y|} &= e^{D - \log|\cos x|} \nonumber \\ \sqrt {|\cos 2y|} &= \frac{e^D}{\cos x} \nonumber \\ \cos 2y \cdot \cos^2x &= E \nonumber \\ \end{align} $$If we use initial value within the solution we get $E=-1$, so the particular solution is

$$ \underline{\underline{\cos 2y \cdot \cos^2x = -1}} \\ $$Solve DE $dr/d \theta = -\sin \theta$.

$$ dr = -\sin \theta\ d \theta $$After integrating both sides:

$$ \underline{\underline{r = \cos \theta + C}} $$Solve DE $dr/ d \theta \cot \theta - r = 2$. (Note that $\cot \theta = \cos \theta / \sin \theta$)

$$ \begin{align} \frac {1}{2+r} dr &= \tan \theta \ d \theta \nonumber \\ \log |2+r| &= \log |\sec \theta| + C \nonumber \\ e^{\log|2+r|} &= e^{\log|\sec \theta |} + C \nonumber \\ \end{align} $$Note: function $\sec \theta = 1/\cos \theta = \text{hypotenuse}/\text{adjacent}$.

Assuming $\theta \neq \dots,\ \pi/2,\ 3\pi/2,\ \dots$ and $r\neq -2$, the general solution is

$$ \begin{equation} \underline{\underline{2+r = D\sec \theta}} \label{ref:first_order_sol1} \end{equation} $$Now we have reached a solution but when solving DE, esp. difficult ones, it is quite
possible we could make a mistake. For that reason, we might be interested **whether
there is a method to check the solution for the correctness.**

Well, there is a **simple numerical way of checking by using a calculator.** Let us choose the constant $D$ of any value, e.
g. $D=2$ and take any value for $\theta$, e.g. $\theta=1$ and evaluate $r$
from the solution. Then let us choose any *small* $d \theta$, e.g. $d \theta =
0.001$.

So, from solution, we have

- $D=2,\ \theta = 1 \implies r = 1.7016314$
- $D=2,\ \theta = (1+d\theta) \implies r = 1.7074072 \implies dr = 0.0057758$

**We got two points (from the solution), which are close to each
other.** Not only we can draw the graph of solution, but **we can
numerically evaluate approximate tangent** at a point and note that
the tangent is used also within the context of DE **(it is
$\boldsymbol{y'}$ within DE indeed)**. If there is no mistake within the
solution, we can expect that after substituting prepared numerical values into
DE, left and right side have to hold equality (roughly):

Despite all the simplifications we have used, the last line shows that the tangent we got from our solution fits into the DE and that suggests that the solution $(\ref{ref:first_order_sol1})$ is correct.

Solve DE with initial values. Draw direction field and solution on $x = \{-3, -2,-1,0,1,2\} \times y = \{-2,-1,0,1,2\}$

$$ y'=\frac{2x+1}{2y},\hskip2em y(-2) = -1,\hskip2em y\neq 0 $$After some shuffling and integrating both sides:

$$ y^2 = x^2 + x + c $$For given initial value then $c=-1$

Now we have implicit solution which can be splitted into two solutions

$$ \begin{align} y & = +\sqrt{x^2 + x -1} \hskip2em \text{and} \nonumber \\ y & = -\sqrt{x^2 + x -1} \nonumber \end{align} $$The initial value fits only to the second one. For drawing the direction field, let us form a table, things go fast and easy then.

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | |
---|---|---|---|---|---|---|---|---|

y | ||||||||

-2 | 5/4 | 3/4 | 1/4 | -1/4 | -3/4 | -5/4 | -7/4 | |

-1 | 5/2 | 3/2 | 1/2 | -1/2 | -3/2 | -5/2 | -7/2 | |

0 | — | — | — | — | — | — | — | |

1 | -5/2 | -3/2 | -1/2 | 1/2 | 3/2 | 5/2 | 7/2 | |

2 | -5/4 | -3/4 | -1/4 | 1/4 | 3/4 | 5/4 | 7/4 |