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

__Homogeneous Ordinary Differential Equations__

We have defined homogeneous ordinary differential equations of the first order as an equation

**F(x, y, y')=0**which does not change if we replace

*with*

**x***and*

**λ·x***with*

**y***, where*

**λ·y***- any real number not equal to zero.*

**λ**In other words,

**F(x, y, y') = F(λ·x, λ·y, y')**Examples:

**F(x, y, y') = y'+y/x**

**F(x, y, y') = 3y'+x·y/(x²+y²)**etc.

The recommended technique to solve these equations is to substitute function

*with*

**y(x)***and solve the equation for*

**x·z(x)***, after which determine*

**z(x)***.*

**y(x)=x·z(x)**Let's solve a few equations of this kind.

*Example 1*

Check for homogeneousness and solve the following equation:

**x·y' = x·sin(y/x) + y**Checking for homogeneousness.

Substitute

*with*

**x***and*

**λ·x***with*

**y***:*

**λ·y**

**λ·x·y' = λ·x·sin(λ·y/(λ·x)) + λ·y**Obviously,

*cancels out completely, which proves homogeneous character of the equation.*

**λ**Now let's solve this equation using the substitution

*, which results in*

**z(x)=y(x)/x***, and express the initial equation in terms of*

**y(x)=z(x)·x***,*

**x***and*

**z***.*

**z'**

**x·(z'·x + z) = x·sin(z) + z·x**Simplifying:

**z'·x + z = sin(z) + z**

**z'·x = sin(z)***d*

**z/sin(z) =**d**x/x**

**∫**d**z/sin(z) = ∫**d**x/x**The right side is easy, the integral equals to

*.*

**ln(|x|)+C**The left side is more involved.

**∫**d**z/sin(z) =**

= ∫d= ∫

**z/(2sin(z/2)·cos(z/2)) =**

= ∫d= ∫

**(z/2)/(sin(z/2)·cos(z/2))**Substitute

*, getting*

**u=z/2***=*

**∫**d**u/(sin(u)·cos(u))**=

**∫cos(u)**d**(u)/(sin(u)·cos²(u)) =**

= ∫d= ∫

**(sin(u))/(sin(u)·cos²(u))**Substitute

*, getting*

**t=sin(u)***[*

**∫**d**t/***]*

**t·(1−t²)**The polynomial in the denominator is

**t·(1−t²) = t·(1−t)·(1+t)**Its inverse can be represented as

*[*

**1/t − 1/***]*

**2(1+t)***[*

**+ 1/***]*

**2(1−t)**which makes our integral equal to

*{*

**∫***[*

*]*

**2/t−1/(1+t)+1/(1−t)***}*

**/2***d*

**t**The last expression can be represented as a sum of three integrals, the result of integration is:

**ln(|t|) − ln(|1+t|)/2 − ln(|1−t|)/2**where

**t=sin(z/2)**This leads us to a final solution of our differential equation.

**ln(|sin(z/2)|) − ln(1+sin(z/2))/2 − ln(1−sin(z/2))/2 = ln(|x|)+C**and then we should substitute

*to get the final expression*

**z=y/x**

**ln(|sin(y/2x)|) − ln(1+sin(y/2x))/2 − ln(1−sin(y/2x))/2 = ln(|x|)+C**Using this as an exponent, we come up with an expression without logarithms

*[*

**|sin(y/2x)| /***]*

**(1+sin(y/2x))·(1−sin(y/2x))**

**= C·|x|**A simplification in the denominator results in

**|sin(y/2x)| / cos²(y/2x) = C·|x|**We leave it "as is" without resolving for

*.*

**y(x)***Example 2*

Check for homogeneousness and solve the following equation:

*[*

*]*

**(y − x·y')/x**

^{x}= e^{y}Checking for homogeneousness.

Substitute

*with*

**x***and*

**λ·x***with*

**y***:*

**λ·y***[*

*]*

**(λy − λx·y')/λx**

^{λx}= e^{λy}Cancel

*in the ratio, getting:*

**λ***[*

*]*

**(y − x·y')/x**

^{λx}= e^{λy}This can be written as

*{*

*[*

*]*

**(y − x·y')/x***}*

^{x}*[*

^{λ}=*]*

**e**^{y}

^{λ}Raising both sides to power

*(or, which is the same, extracting a root of power*

**1/λ***) we come to the original equation, which proves homogeneous character of the equation.*

**λ**Now we will solve it using the recommended technique.

Substitute

*, which results in*

**z(x)=y(x)/x***and express the initial equation in terms of*

**y(x)=z(x)·x***,*

**x***and*

**z***.*

**z'**The expression for a derivative

*is:*

**y'**

**y' = (z·x)' = z'·x+z**New equation is, therefore,

*[*

*]*

**(z·x − x·(z'·x+z))/x**

^{x}= e^{z·x}Simplifying it by raising to power

*both sides (or, equivalently, extracting a root of power*

**1/x***):*

**x***[*

*]*

**(z·x − x·(z'·x+z))/x**

^{}= e^{z}Cancel

*:*

**x**

**z − (z'·x+z) = e**^{z}Cancel

*:*

**z**

**−z'·x = e**^{z}This equation is separable, let's separate

*from*

**x***, getting*

**x**

**−e**d^{−z}·**z =**d**x/x**Ready to integrate:

**∫−e**d^{−z}·**z = ∫**d**x/x**

**e**^{-z}= ln(x)+C(assuming for simplicity positive only sign for

*, so integral on the right is*

**x***instead of*

**ln(x)***)*

**ln(|x|)**From the last equation we derive:

**−z = ln(ln(x)) + C**

**z = −ln(ln(x)) + C**Now we can use it to find an expression for

*:*

**y**

**y = −x·ln(ln(x)) + C**Solution must be checked.

It's easier, instead of checking the original equation

*[*

*]*

**(y − x·y')/x**

^{x}= e^{y}to check the equality of logarithms from both sides:

*[*

**x·ln***]*

**(y − x·y')/x**

**= y**or, simpler,

**ln(y/x − y') = y/x**where we should substitute

**y = −x·ln(ln(x)) + C**and

**y' = −ln(ln(x))−x·(1/ln(x))·(1/x)**or, simpler,

**y' = −ln(ln(x)) − 1/ln(x)**Let's disregard constant

*in this checking to make manipulations simpler.*

**C**Then, since

**y/x = −ln(ln(x))**we will have to check that

**ln(−ln(ln(x)) + ln(ln(x)) + 1/ln(x)) = −ln(ln(x))**Canceling opposite positive and negative members under logarithm on the left, we come to an obvious equality

**ln(1/ln(x)) = −ln(ln(x))**which proves the correctness of our solution.

*Example 3*

Check for homogeneousness and solve the following equation:

**x·y·y' = (x+y)²**Checking for homogeneousness.

Substitute

*with*

**x***and*

**λ·x***with*

**y***:*

**λ·y**

**λx·λy·y' = (λx+λy)²**

**λ²x·y·y' = λ²(x+y)²**Obviously,

*cancels out, and we get the same original equation.*

**λ**Now let's solve it by substituting

*, which results in*

**z(x)=y(x)/x***and express the initial equation in terms of*

**y(x)=z(x)·x***,*

**x***and*

**z***.*

**z'**The expression for a derivative

*is:*

**y'**

**y' = (z·x)' = z'·x+z**So, our equation looks like

**x·(z·x)·(z'·x+z) = (x+z·x)²**Simplifying by opening all parenthesis, we get

**x²·(x·z·z'+z²) = x²·(1+z)²**

**x·z·z'+z² = (1+z)²**

**x·z·z' = 1+2z**This equation can be solved using the method of

*separation*.

**z·**d**z/(1+2z) =**d**x/x**Integrating the left side of this equation:

*[*

**∫z·**d**z/(1+2z) =**

= (1/2)∫(1+2z−1)·d= (1/2)∫(1+2z−1)·

**z/(1+2z) =**

= (1/2)= (1/2)

*]*

**∫**d**z − ∫**d**z/(1+2z)***[*

**=**

= (1/2)= (1/2)

*]*

**z−(1/2)ln(1+2z)**

**+ C**Integrating the right side of the equation:

**∫**d**x/x = ln(x) + C**Since integral of both sides are equal,

*[*

**(1/2)***]*

**z−(1/2)ln(1+2z)**

**=**

= ln(x)+ C= ln(x)+ C

which can be simplified

**2z − ln(1+2z) = 4ln(x) + C**Though this equation for

*cannot be easily solve for*

**z(x)***, it allows to replace the original differential equation for*

**z***with purely algebraic one, replacing*

**y***with*

**z***:*

**y/x**(A)

**2y/x − ln(1+2y/x) =**

= 4ln(x) + C= 4ln(x) + C

This is the final algebraic answer to our differential equation. Though it's not resolved for

*, it's still the best solution we can come up with.*

**y(x)**Solution must be checked.

If this equality that includes function

*is correct, derivatives of both parts are also equal. Let's differentiate them both.*

**y(x)**

**−2y/x² + 2y'/x − (1/(1+2y/x))·(−2y/x²+2y'/x) = 4/x**Simplifying by multiplying by

*:*

**x²**

**−2y + 2xy' − x·(−2y+2xy')/(x+2y) = 4x**Multiplying by

*:*

**x+2y**

**−2xy−4y²+2x²y'+4xyy'+2xy−2x²y' = 4x²+8xy**After cancellation of mutually opposing by sign members and dividing by

*we get:*

**4**

**−y²+xyy' = x²+2xy**which easily transforms into

**xyy' = (x+y)²**that corresponds to original differential equation.

This proves the correctness of the answer (A) as an equation that includes

*and*

**x***without derivatives that we obtained above.*

**y(x)**
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