*Notes to a video lecture on http://www.unizor.com*

__Linear Ordinary Differential Equations__

Standard form of linear ordinary differential equations is

**f(x)·y' + g(x)·y + h(x) = 0**As the first step, we can divide all members of this equation by

*(assuming it's not identically equal to 0), getting a simpler equation*

**f(x)**

**y'+u(x)·y+v(x) = 0**The suggested solution lies in the substitution

*, where*

**y(x)=p(x)·q(x)***and*

**p(x)***are unknown (for now) functions.*

**q(x)**Express

*in terms of*

**y'(x)***and*

**p(x)***:*

**q(x)**

**y' = p·q'+q·p'**Substitute this into our equation:

**p·q'+q·p'+u·p·q+v = 0**Let's simplify this

**p(q'+u·q)+q·p'+v = 0**If there are such functions

*and*

**p(x)***that satisfy conditions*

**q(x)**(1)

*and*

**q'+u·q = 0**(2)

**q·p'+v = 0**our job would be finished.

Let's try to find such functions.

From the equation (1) in our pair of equations we derive

*,*

**q'/q = −u**which can be converted into

*d*

**q(x)/q(x) = −u(x)·**d**x**that can be solved by integrating:

**ln(q(x)) = −∫u(x)·**d**x**

**q(x) = e**^{−∫u(x)·dx}Once

*is found, we solve the equation (2) for*

**q(x)***:*

**p(x)***,*

**p'(x) = −v(x)/q(x)**which can be integrated to find

**p(x) = −∫v(x)/q(x)**d**x**and, consequently,

*can be fully determined.*

**y(x)=p(x)·q(x)**Let's consider a few examples.

*Example 1*

Solve the following linear differential equation

**y' + y + x = 0**Let's look for a solution in a form

**y(x)=p(x)·q(x)**Then

**y'(x) = p'(x)·q(x)+p(x)·q'(x)**Our equation looks like this now

**p'·q+p·q' + p·q + x = 0**Factor out

*, getting*

**p**

**p·(q'+q) + (p'·q+x) = 0**We will try to find

*and*

**p(x)***to separately bring to zero*

**q(x)***and*

**q'+q***.*

**p'·q+x**Let's look for a function

*that brings expression*

**q(x)***to zero:*

**q'+q**

**q'+q = 0***d*

**q/q = −**d**x**

**∫**d**q/q = −∫**d**x**

**ln(q) = −x + A**(where

*is any constant)*

**A**

**q(x) = B·e**^{−x}(where

*, so it represents any positive number)*

**B=e**^{A}Next, let's find a solution to

**p'·q+x = 0**

**p'·B·e**^{−x}+x = 0

**p(x) = −(1/B)·∫x·e**d^{x}**x**This integral can be found using the "by parts" technique:

**p(x) = −(1/B)·(x·e**d^{x}−∫e^{x}**x) =**

= −(1/B)·(x·e

= −(1/B)·(x−1)·e= −(1/B)·(x·e

^{x}−e^{x}−C) == −(1/B)·(x−1)·e

^{x}+C/B(where

*is any positive constant and*

**B***is any constants)*

**C**Now let's find

*:*

**y(x)=p(x)·q(x)***[*

**y(x) =***]·[*

**−(1/B)·(x−1)·e**^{x}+C/B*] =*

**B·e**^{−x}=

**1−x+C·e**^{-x}Checking:

**y'(x) = −1 − C·e**^{-x}

**y'+y+x = −1 − C·e**^{-x}+1−x+C·e^{-x}+x = 0Solution was correct.

*Example 2*

Solve the following linear differential equation

**y'·cos(x) + y·sin(x) − 1 = 0**First of all, let's normalize it by dividing by

*, noticing that*

**cos(x)***and*

**sin(x)/cos(x)=tan(x)**

**1/cos(x)=sec(x)**:

**y' + y·tan(x) − sec(x) = 0**Let's look for a solution to this equation in a form

**y(x)=p(x)·q(x)**Then

**y'(x) = p'(x)·q(x)+p(x)·q'(x)**Our equation looks like this now:

**p'·q+p·q'+p·q·tan(x)−sec(x) = 0**Factor out

*, getting*

**p**

**p·(q'+q·tan(x))+p'·q−sec(x) = 0**First, let's find function

*such that*

**q(x)**

**q' + q·tan(x) = 0**It can be solved using the technique of separation:

*d*

**q/q = −tan(x)**d**x**Since [

*]' =*

**ln(x)***and [*

**1/x***]' =*

**cos(x)***, the last equation can be transformed into*

**−sin(x)***d*

**(ln(q)) =**d**(cos(x))/cos(x)***d*

**(ln(q)) =**d**(ln(cos(x)))**Now it's easy to integrate, the result is

**ln(q) = ln(cos(x))+C**where

*- any real number, from which, raising number*

**C***to both left and right sides, follows that*

**e**

**q = D·cos(x)**(new constant

*represents any positive number)*

**D=e**^{C}Now let's find function

*such that*

**p(x)**

**p'·q − sec(x) = 0**Substitute already found

*getting*

**q(x)**

**p'·D·cos(x) − sec(x) = 0**

**p'(x) = (1/D)·(1/cos²(x))***[*

**p(x) = (1/D)·∫**d**x/cos²(x) =**

= (1/D)·= (1/D)·

*]*

**tan(x) + E**where

*and*

**D***are constants (*

**E***- any positive,*

**D***- any real)*

**E**Let's determine

*now.*

**y=p·q***[*

**y(x) = p(x)·q(x) =**

= (1/D)·= (1/D)·

*]*

**tan(x) + E**

**·D·cos(x) =**

= sin(x) + E·cos(x)= sin(x) + E·cos(x)

Let's check this result.

**y'(x) = cos(x)−E·sin(x)**

**y'·cos(x) + y·sin(x) − 1 =**

= cos²(x)−E·sin(x)·cos(x) + sin²(x)+E·cos(x)·sin(x)−1 =

= sin²(x) + cos²(x) −1 = 0= cos²(x)−E·sin(x)·cos(x) + sin²(x)+E·cos(x)·sin(x)−1 =

= sin²(x) + cos²(x) −1 = 0

which proves the correctness of our answer.

*Example 3*

Solve the following differential equation

**ln(x·y'+y) = ln(2x)+x²**It's not linear, but can be made linear if we raise

*to a power defined by its left and right sides, getting*

**e**

**x·y'+y = 2x·e**^{x²}Let's normalize it by dividing by

*:*

**x**

**y' + y/x = 2e**^{x²}Now it's a linear equation that we know how to solve.

Let's look for a solution in a form

**y(x)=p(x)·q(x)**Then

**y'(x) = p'(x)·q(x)+p(x)·q'(x)**Our equation looks like this now

**p'·q+p·q' + p·q/x = 2e**^{x²}Factor out

*, getting*

**p**

**p·(q'+q/x) + p'·q = 2e**^{x²}We will try to find

*and*

**p(x)***to separately*

**q(x)**(a) bring to zero

*and*

**q'+q/x**(b) equalize

*with*

**p'·q***.*

**2e**^{x²}Let's solve equation (a) and look for a function

*that brings expression*

**q(x)***to zero:*

**q'+q/x**

**q'+q/x = 0***d*

**q/q = −**d**x/x**

**∫**d**q/q = −∫**d**x/x**

**ln(|q|) = −ln(|x|) + A**where

*- any constant.*

**A**Raising

*to both sides of this equation, we get*

**e**

**|q| = B/|x|**where

*- any positive number.*

**B=e**^{A}Let's get rid of absolute values in the above equation by allowing

*to be any non-zero real number, so*

**B**

**q = B/x**Substitute it to equation (b):

**p'·B/x = 2e**^{x²}

**p(x) = (1/B)∫2x·e**d^{x²}·**x**Since derivative of

*is*

**x²***,*

**2x**

**p(x) = (1/B)∫e**d^{x²}·**(x²)**Now we can integrate directly:

**p(x) = (1/B)e**^{x²}+ Cwhere

*is any real number.*

**C**This allows to express the solution to our differential equation in the form

**y(x) = p(x)·q(x)**where

*and*

**p(x) = (1/B)e**^{x²}+ C

**q(x) = B/x**That produces

**y = e**^{x²}/x + C/x = (e^{x²}+C)/xwhere

*- any real number.*

**C**Checking:

**y' = −(e**

= e^{x²}+C)/x² + e^{x²}·2x/x == e

^{x²}·(2−1/x²) −C/x²

**y/x = e**^{x²}/x² + C/x²*,*

**y' + y/x = 2e**^{x²}which corresponds to the original equation after multiplying both sides by

*and taking logarithm.*

**x**The end.

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