The linear DE of first order can be described as

$$ a_1(x)\ y' + a_0(x)\ y = g(x). $$That means both $y,\ y'$ are involved with coefficients, which are either constant or are a function of $x$.

Note: DE e.g.

$$ \begin{align} y' + \sin y &= g(x) \hskip1em \text{or} \nonumber \\ (y')^2 + y &= g(x) \nonumber \end{align} $$are **not** linear DE.

We will rather work with **linear DE in the standard form**

In some cases the DE above, which is linear, can be solved also by
separating variables. General method to solve linear DE involves a special function $\mu(x)$. As for now, do
not think where this function comes from. If we use such **special function
$\boldsymbol{\mu(x)}$ to multiply DE,** it will help us to bring the solution. So let
us multiply $(\ref{ref:linea1_1})$ by $\mu(x)$ to derive the principle:

Now it is important to pay attention to the left side of
$(\ref{ref:linea1_2})$. Because it will be shown that ** after
multiplying both sides of DE by $\boldsymbol{\mu(x)}$, on the left side of DE is being left
$\boldsymbol{\mu(x) y}$ differentiated with respect to
$\boldsymbol{x}$.** Let us differentiate $\mu(x)y$:

By $(\ref{ref:linea1_2a})$ it was shown that $\frac{d}{dx}\mu(x) y$ is on the left side of $(\ref{ref:linea1_2})$ supposing

$$ \begin{equation} \frac{d\mu}{dx} = \mu(x) P(x) \label{ref:linea1_3} \end{equation} $$The DE $(\ref{ref:linea1_3})$ above can be solved by separating variables and will answer the question where does the function $\mu(x)$ come from.

$$ \begin{align} \frac{1}{\mu(x)}d \mu &= P(x)\ dx \nonumber \\ \log |\mu(x)| &= \int P(x)\ dx + c_1 \nonumber \\ \mu(x) &= c_2 e^{\int P(x)\ dx} \nonumber \end{align} $$It can be shown that the value of the constant has no effect on solution of linear DE. So, we may put $c_2 = 1$:

$$
\begin{equation}
\color{red}{\mu(x) = e^{\int P(x)\ dx}}. \label{ref:linea1_4}
\end{equation}
$$

$$
\begin{align}
\frac{dy}{dx}+P(x)\cdot y &= f(x) \hskip4em / \cdot e^{\int P(x)\ dx} \nonumber \\
\color{blue}{\left[e^{\int P(x)\ dx}\right]\frac{dy}{dx}+ \left[e^{\int P(x)\ dx}\right]P(x)\cdot y} &= \left[e^{\int P(x)\ dx}\right]f(x) \nonumber \\
\color{blue}{\frac{d}{dx}\left(e^{\int P(x)\ dx} y\right)} &= e^{\int P(x)\ dx} f(x) \nonumber
\end{align}
$$
Since we have identified $\mu(x)$ as $e^{\int P(x)\ dx}$, let us use $\mu(x)$ again to multiply the DE $(\ref{ref:linea1_1})$. Because it will be shown, that after multiplying, the DE is readily solved. Let us bring back DE in standard form and multiply by integrating factor:

The most important observation is that **we have $\boldsymbol{d/dx
(\mu(x)y)}$ (as a product of the product rule) on the left side.** Now
we have to integrate both sides of DE. Especially the left side is simple and
the remaining tasks are not so difficult, which will be seen on examples.

**The steps to solve linear DE can be summarized as**

- Bring DE into standard form $\frac{dy}{dx}+P(x)\cdot y = f(x)$.
- Identify $P(x)$ and form the integrating factor $\mu(x)$ from $(\ref{ref:linea1_4})$.
- Multiply both sides of DE by integrating factor $\mu(x)$.
- Integrating both sides of DE is simple then considering the fact on the left side is $d/dx (\mu(x)y)$ to be found.

Solve DE $dy/dx + 3x^2y = 6x^2$.

Given DE is linear, already in standard form, so we can locate $P(x) = 3x^2$ to evaluate integrating factor $\mu(x)$:

$$ \mu(x) = e^{\int 3x^2\ dx} = e^{x^3}. $$The integrating factor will be used to multiply both sides of DE in the standard form. Then:

$$ e^{x^3}\frac{dy}{dx} + e^{x^3}\ 3x^2y = e^{x^3}\ 6x^2 $$We know in advance that what will be found on the left side is a product of product rule of $d/dx(\mu(x)\ y) = d/dx(e^{x^3} y)$:

$$ \frac{d}{dx}e^{x^3}y = e^{x^3}\ 6x^2 $$Now all we need is to integrate both sides and especially the left side is trivial.

$$ \underline{\underline{e^{x^3}y = 2e^{x^3}+ C}} $$Solve DE $x^2y' + xy = 1,\ x \gt 0,\ y(1)=2$.

The first step is to bring the DE into standard form, then locate $P(x)$ in order to evaluate integrating factor $\mu(x)$:

$$ y' = \color{red}{\frac{1}{x}}y = \frac{1}{x^2} \implies P(x)= \frac 1 x\hskip2em x \neq 0 \\ \mu(x) = e^{\int\frac 1 x \ dx} = e^{\log x} = x $$Integrating factor has been found. Now let us use it to multiply DE:

$$ y'x +y = \frac 1 x $$On the left side is product of the product rule applied on $d/dx(\mu(x)y) = d/dx(xy) = (xy)'$ as anticipated. So it is easy to integrate both sides then.

$$ \begin{align} (xy)' &= \frac 1 x \nonumber \\ xy &= \log |x| + C \nonumber \\ y &= \frac{\log |x| + C}{x} \nonumber \end{align} $$That is general solution, constant $C$ has to be found for supplied IV $y(x=1) = 2$:

$$ 2 = \frac{\log |1| + C}{1} \implies C=2 \\ $$The particular solution is then

$$ \underline{\underline{y = \frac{\log x + 2}{x}}}. $$Solve DE $y' + \color{red}{2x}y = 1$.

The given DE is already in the standard form, so we can find the integrating factor $\mu(x)$ for $P(x) = 2x$:

$$ \mu(x) = e^{\int P(x)\ dx} = e^{\int 2x\ dx} = e^{x^2}. $$Let us use $\mu(x)$ to multiply each part of DE:

$$ y'e^{x^2} + 2xy e^{x^2} = e^{x^2} $$Again, what is on the left side is $(\mu(x)y)'$, thus trivial to integrate.

$$ \begin{align} (e^{x^2} y)' &= e^{x^2} \nonumber \\ e^{x^2}y &= \int e^{x^2}\ dx + C \nonumber \end{align} $$However this time on the right side is a function which antiderative can not be expressed in terms of elementary functions. So it is OK to leave the member as $\int e^{x^2}\ dx$.

$$ \underline{\underline{y = e^{-x^2}\int e^{x^2}\ dx + Ce^{-x^2}}} $$Solve DE $xy' + y =x^3$.

If we compare this DE with linear DE $(\ref{ref:linea1_1})$ in standard form, we observe that given DE fits into linear DE if is divided by $x$:

$$ y' + \color{red}{\frac 1 x} y = x^2. $$Now it is clear that $P(x) = 1/x$ and we can assemble integrating factor $\mu(x)$ which will be used to multiply both sides of the DE.

$$ \begin{align} \mu(x) &= e^{\int P(x)\ dx} = e^{\int 1/x\ dx} = e^{\log x} \implies \nonumber \\ \implies \mu(x) &= x \nonumber \\ \end{align} $$As reminded, now the integrating factor $\mu(x) = x$ is good to multiply DE and then $(\mu(x)y)'$ appears on the left side.

$$ \begin{align} xy' + y &= x^3 \nonumber \\ (\mu(x)\ y)' = (xy)' &= x^3 \nonumber \\ xy &= \int x^3 \ dx \nonumber \\ \end{align} $$Not much can be done when the right side is finally integrated. The solution will be left in implicit form.

$$ \underline{\underline{xy = \frac 1 4 x^4 + C}} $$Solve DE $y'+\color{red}{a}y=b$.

DE is in the standard form, now $P(x) = a$, integrating factor

$$ \begin{align} \mu(x) = e^{\int P(x)\ dx} &= e^{\int a\ dx} = e^{ax} \nonumber \\ a^{ax}\frac{dy}{dx} + e^{ax}ay &= a^{ax}b \nonumber \\ \frac{d}{dx} e^{ax}y &= e^{ax}b \nonumber \\ e^{ax}y &= \int e^{ax}b\ dx \nonumber \\ e^{ax}y &= b \frac 1 a e^{ax} + C\nonumber \\ \end{align} $$And after we divide by $e^{ax}$: $$ \underline{\underline{y = \frac b a + ce^{-ax}}}. $$

Solve DE $dy/dx=2x-3y,\ y(0)=1/3$.

First we have to shuffle the members to bring DE into standard form $(\ref{ref:linea1_2})$ of linear DE.

$$ \begin{align} \frac{dy}{dx} + \color{red}{\color{red}{3}}y = &2x \nonumber \\ \mu(x) &= e^{\int P(x)\ dx} = e^{\int 3\ dx} = e^{3x} \nonumber \\ e^{3x}\frac{dy}{dx} + e^{3x}\cdot 3y &= e^{3x}\cdot 2x \nonumber \\ \end{align} $$Again, after multiplying by $\mu(x)$, on the left side is $(\mu(x)y)'$ to be found.

$$ \begin{align} \frac{d}{dx}e^{3x}y &= e^{3x}\cdot 2x \nonumber \\ e^{3x}y &= 2\int e^{3x}x\ dx \nonumber \end{align} $$To integrate $e^{3x}x$ we have to use integration by parts. Integration by parts is product rule used backwards.

$$ \color{blue}{(uv)'} = u'v + uv' \implies uv' = \color{blue}{(uv)'} - u'v \implies \int uv' = \color{blue}{uv}-\int u'v $$For integration by parts there have to be one member which is easy to integrate and the other one which is easy to differentiate. In this case we differentiate $x$ and integrate $e^{3x}$.

$$ \int e^{3x}x\ dx = \begin{vmatrix} u = x & u'=1\\ v'=e^{3x} & v = \frac 1 3 e^{3x} \end{vmatrix} = x\frac 1 3 e^{3x} - \int\frac 1 3 e^{3x}\ dx = \frac 1 3 e^{3x}x-\frac 1 9e^{3x} $$Considering given IV $y(x=0) = 1/3$:

$$ \frac 1 3 = \frac 2 3 (0-\frac 1 3 )+ Ce^{-3\cdot 0} \implies C = \frac 5 9 $$Applying evaluated constant into the general solution we get particular solution for IV:

$$ \underline{\underline{y = \frac 2 3 (x-\frac 1 3) + \frac 5 9 e^{-3x}}}. $$Solve DE $xy'+y=e^x,\ y(1)=2$.

DE is not in the standard form yet.

$$ y' + \color{red}{\frac 1 x} y = \frac 1 x e^x \implies P(x) = \color{red}{1/x} \nonumber \\ \mu(x) =e^{\int P(x)\ dx} = e^{\int 1/x\ dx} = e^{\log x} = x $$The integrating factor $\mu(x)$ was needed to multiply the DE, because that is the way how to solve it:

$$ \begin{align} xy' + x\frac 1 x y &= x\frac 1 x e^x \nonumber \\ xy' + y &= e^x \nonumber \\ \frac d {dx}xy &= e^x \nonumber \end{align} $$As expected, on the left side is $d/dx\ \mu(x)y = d/dx\ xy$, which when integrated become $xy$:

$$ \begin{align} (xy)' &= e^x \nonumber\\ xy &= \int e^x\ dx \nonumber \\ xy&= e^x + C \nonumber\\ y &= \frac 1 x(e^x+C) \nonumber \end{align} $$Applying IV $y(x=1) = 2$ brings value of the constant $C=(2-e)$ and the particular solution is

$$ \underline{\underline{y = \frac 1 x(e^x+2-e)}}. $$