mathematics-class-12-unit-09-chapter-05-differential-equations-5-8-ql4rwwa7zoi

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Homogeneous subsets of the plane </primary>
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-  Definition :-  A subset $D$ of $\mathbb{R}^2$ is said to be  homogeneous  if whenever $(x, y) \in D$ the point ( $b x, t y$ ) also belongs to $D$ for any $t \in R$ and $t \neq 0$. Thus a subset of the plane is homogeneous if and only if it is a union of lines through the origin (possibly with origin removed).

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<footer style = "font-size: 18px"> $\rightarrow$ Differential equation lecture-5 $\rightarrow$ <span style = "color:blue"> Homogeneous subsets of the plane </span> $\rightarrow$ Homogeneous subsets of the plane $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Homogeneous subsets of the plane </primary>
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- Often we are interested not in homogeneity with respect to $t \in R$ but  positive homogeneity  namely a subset $D$ of the plane is said to be  positively homogeneous  if whenever $(x, y) \in D$ the point $(t x, t y)$ also belongs to $D$ for any $t>0$.

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<footer style = "font-size: 18px">Differential equation lecture-5 $\rightarrow$ Homogeneous subsets of the plane $\rightarrow$ <span style = "color:blue"> Homogeneous subsets of the plane </span> $\rightarrow$ Examples $\rightarrow$ Homogeneous functions </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Examples </primary>
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* The entire plane is obviously a homogeneous set.

* The subset $\mathbb{R}^2-\{(0.0)\}$ is homogeneous.
* The open first quadrant $\{(x, y) \in \mathbb{R}^2 ; x>0, y>0\}$ is only positively homogeneous. The union of the open first and third quadrants is a homogeneous subset of the plane.
* The function $f(x, y)=\log \left|\frac{x-y}{x+y}\right|$ is defined on the plane minus the two lines $x \neq y-0$ and the domain of $f$ is homogeneous.
* Examine whether the set $\{(x, y) \in \mathbb{R}^2: x>y\}$ is positively homogeneous? Is it homogeneous?

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<footer style = "font-size: 18px">Homogeneous subsets of the plane $\rightarrow$ Homogeneous subsets of the plane $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ Homogeneous functions $\rightarrow$ Examples of homogeneous functions </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Homogeneous functions </primary>
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-  Definition:-  Given a (positively) homogeneous subset $D$ of the plane, a function $f_k D \longrightarrow \mathrm{R}$ is said to be  homogeneous of degree $k$ (resp. positively homogeneous of degree $k$ )  if

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<footer style = "font-size: 18px">Homogeneous subsets of the plane $\rightarrow$ Examples $\rightarrow$ <span style = "color:blue"> Homogeneous functions </span> $\rightarrow$ Examples of homogeneous functions $\rightarrow$ Some important subsets of the plane </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Examples of homogeneous functions </primary>
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- $ f(t x, t y)=t^k f(x, y)$
- for all $(x, y) \in D$ and $t \in \mathbb{R}-[0]$ (resp $t>0)$   
- (1) The function $f(x, y)=x^2+y^2-7 x y$ is homogeneous of degree two.

- (2) The function $\tan ^{-1}(y / x)$ defined on $R^2$ minus the $y$-axis is homogeneous of degree zero.

- (3) Note that $f(x, y)=\sqrt{x^3-y^3}$ defined on $\{(x, y) \in \mathbb{R}^2 ; x>y\}$ is positively homogeneous of degree $3 / 2$ and $f(x, y)=\sqrt{x^3-y^3 /(x^2+y^2)^{3 / 2}}$ is positively homogeneous of degree zero.

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<footer style = "font-size: 18px">Examples $\rightarrow$ Homogeneous functions $\rightarrow$ <span style = "color:blue"> Examples of homogeneous functions </span> $\rightarrow$ Some important subsets of the plane $\rightarrow$ Homogeneous differential equation </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Some important subsets of the plane </primary>
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- We shall be working with the following types of subsets of the plane all of which are positively homogeneous. The letter $D$ will always denote one of these sets.
- The entire plane
- Open half planes
- $\{(x, y) \in \mathbb{R}^2: a ^ x+b y>0\}$
- where $a$ and $b$ are constants not both zero. Note the four special cases $a=1 . b=0 ; a=-1, b=0 ; a=0 . b=-1$ and $a=0 . b=1$.
- The open first quadrant
- Finite intersections of open half planes describe in item (2) above.

- $ R^2-\{(0,0)\}$
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<footer style = "font-size: 18px">Homogeneous functions $\rightarrow$ Examples of homogeneous functions $\rightarrow$ <span style = "color:blue"> Some important subsets of the plane </span> $\rightarrow$ Homogeneous differential equation $\rightarrow$ Not homogeneous </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Homogeneous differential equation </primary>
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- Assume that $M(x, y)$ and $N(x, y)$ are both defined on a homogeneous subset $D$ of the plane. The differential equation
- $- M(x, y) d x+N(x, y) d y=0$
- is said to be homogeneous if $M(x, y)$ and $N(x, y)$ are homogeneous of the  same degree  $k$.

-  Examples:-  The following differential equations are homogeneous.

- $\begin{array}{r}\left(x^2-y^2\right) d x+2 x y d y=0 \\\  \left(\log x^2-\log y^2\right) d x+\left(\tan ^{-1}(y / x)\right) d y=0 .\end{array}$

- The first is defined on the whole plane, the second is defined on the plane minus the coordinate axes.

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<footer style = "font-size: 18px">Examples of homogeneous functions $\rightarrow$ Some important subsets of the plane $\rightarrow$ <span style = "color:blue"> Homogeneous differential equation </span> $\rightarrow$ Not homogeneous $\rightarrow$ Positively homogeneous differential equations </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Not homogeneous </primary>
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- As per this definition,  the following equation would not be homogeneous. 
- $ (x+y) d x-\left(\sqrt{x^2+y^2}\right) d y=0 .$  $\quad (2.1)$

- The reason being that both the coefficients are  only  positively homogeneous. While $x+y$ is homogeneous, the second coefficient is positively homogeneous but not homogeneous.
- $\sqrt{(t x)^2+(t y)^2}=\sqrt{t^2} \sqrt{x^2+y^2}=|t| \sqrt{x^2+y^2} .$

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<footer style = "font-size: 18px">Some important subsets of the plane $\rightarrow$ Homogeneous differential equation $\rightarrow$ <span style = "color:blue"> Not homogeneous </span> $\rightarrow$ Positively homogeneous differential equations $\rightarrow$ Theorem-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Positively homogeneous differential equations </primary>
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- Again look at the following differential equation
- $\left(\sqrt{x^2+y^2}\right) d x+\left(\sqrt{x^2-y^2}\right) d y=0 . $.....$(2.2)$

- Defined on $x>y>0$. The domain on which $\sqrt{x^2-y^2}$ is defined is only positively homogeneous! If $(x, y)$ belongs to the domain and $t<0$ then $(t x, t y)$ does not belong to the domain. It is appropriate to call $(2.1)$ and $(2.2)$ positively homogeneous differential equations  and we shall do so  However books would call equations (2.1) (2.2) homogeneous  since the technique for solving homogeneous equations applies ALSO to (2.1) and (2.2) as long as we restrict to the part of the domain in the open first quadrant. In short the method that follows works for any positively homogeneous differential equation  in the first quadrant .

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<footer style = "font-size: 18px">Homogeneous differential equation $\rightarrow$ Not homogeneous $\rightarrow$ <span style = "color:blue"> Positively homogeneous differential equations </span> $\rightarrow$ Theorem-1 $\rightarrow$ Proof of the theorem contd... </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Theorem-1 </primary>
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- <ins>**The $ y = v x$ substitution**</ins>

- As indicated in the very first lecture, this idea goes back to Leibnitz (circa 1693)

- The substitution $y=v x$ transforms a homogeneous differential equation into a variable separable equation. The same holds for a positively homogeneous differential equation defined in the first quadrant.

- Proof is very easy. Well,
- $\frac{d y}{d x}=v+x \frac{d v}{d x} .$.....$(2.3)$

- Substituting into the differential equation
- $ M(x, y)+N(x, y) \frac{d y}{d x}=0$
- we get

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<footer style = "font-size: 18px">Not homogeneous $\rightarrow$ Positively homogeneous differential equations $\rightarrow$ <span style = "color:blue"> Theorem-1 </span> $\rightarrow$ Proof of the theorem contd... $\rightarrow$ Exercise </footer>

### <Success style="font-size:25px"> Differential Equations L-5 </Success>
### <primary style="font-size:24px"> Orthogonal trajectories </primary>
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- (2)Find the orthogonal trajectories of the system of curves

- $ x^3-3 x y^2=c .$

- The astute student will recognize that $x^3-3 x y^2$ is the real part of $z^3$ where $z=x+iy$   and will guess the orthogonal trajectories!  Differentiating, the $c$ disappears and we get at once the differential equation for the family of curves
- $\left(x^2-y^2\right) d x-2 x y d y=0 .$

- The differential equations for the orthogonal trajectories are
- $\left(x^2-y^2\right) d y+2 x y d x=0 .$

- This is a homogeneous differential equation.  Apply the method you have learned to solve this and find the orthogonal trajectories. 
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<footer style = "font-size: 18px">Exercise $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Orthogonal trajectories </span> $\rightarrow$ Examples contd $\rightarrow$ Examples contd </footer>

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- Example: Consider the following equation in the first quadrant
- $ x>0, y>0 \text {. }$
- $ y d x+x(\log y-\log x) d y=0 .$.....$(2.7)$

- We shall find the solution of the differential equation with intial conditions $x=1$ when $y=1$. Think of $x$ as a function of $y$ and we denote it as $x(y)$. Comment: Since $\frac{d x}{d y}=0$ when $y=1$ and $x=1$, we see that $\frac{d y}{d x}$ would be meaningless

- The differential equation is easily seen to be POSITIVELY homogeneous. Substituting $y=v x$ and using (2.3) we get
- $ v d x+(\log v)(v d x+x d v)=0 .$

- Rearranging this, we get the variable separable equation:
- $ v(1+\log v) d x+x(\log v) d v=0 $.....$(2.8)$

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<footer style = "font-size: 18px">Exercise $\rightarrow$ Orthogonal trajectories $\rightarrow$ <span style = "color:blue"> Examples contd </span> $\rightarrow$ Examples contd $\rightarrow$ Examples contd </footer>

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-  (3a)  Is there any value $y>1$ at which the derivative of $x^{\prime}(y)$ vanishes?

-  (3b)  Can you say $x(y)$ is a monotone decreasing function of $y$ for $y>1$ ? Sketch the graph of the function $x(y)$ near $(1,1)$.

-  (3c)  Do you think $x(y) \longrightarrow 0$ when $y \longrightarrow a$ for some finite $a>1$ ?

-  (3d)  What happens as $y \rightarrow \infty$ ? Can you think of $y$ as a function of $x$ on the interval $(0,1)$ ? Hint: Where does the graph $(2.10)$ cut the line $y=x$ ? Now if $y<1$ and $y$ is close to 1 then $x^{\prime}(y)>0$ and so $y<x$. Does this persist for all $0<y<1$ ? Can $y>x$ at some point on the curve $(2.10)$ with $0<y<1$ ?

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<footer style = "font-size: 18px">Examples contd $\rightarrow$ Examples contd $\rightarrow$ <span style = "color:blue"> Examples contd </span> $\rightarrow$ Using the differential equation to graph the solution $\rightarrow$ Using the differential equation to graph the solution </footer>

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### <primary style="font-size:24px"> Using the differential equation to graph the solution </primary>
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- It is important to get as much information about the solutions WITHOUT recourse to the explicit solution obtained.

- The differential equation $(2.7)$ gives:
- $ x^{\prime}=\frac{x(\log x-\log y)}{y} $

- Now using quotient rule and the fact that $x^{\prime}(1)=0$,
- $ x^{\prime \prime}(1)=y x \frac{d}{d y}(\log x-\log y)-x(\log x-\log y), \quad x=1, y=1 .$

- The denominator is 1 and the term $x^{\prime}(1)$ obtained drops out and so these have not been displayed. We see that $x^{\prime \prime}(1)=-1$. The function $x(y)$ has a local maximum at $y=1$.

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### <primary style="font-size:24px"> Using the differential equation to graph the solution </primary>
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- $y dx + x(\log y - \log x ) d y = 0 $

- Curve passing through $(1,1)$

- As $x \rightarrow 0+ ,$

- Necessarily $ y \rightarrow 0+ $

- Regard $x$ =$y$

- $ y - \log y = 1 - \log x $

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<footer style = "font-size: 18px">Examples contd $\rightarrow$ Using the differential equation to graph the solution $\rightarrow$ <span style = "color:blue"> Using the differential equation to graph the solution </span> $\rightarrow$ Graphing the solution contd $\rightarrow$ Maxima or minima at a point </footer>

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- At $y=1$, we know that $x=1$ .  $y=1$ is a local maximum.  If $y$ increases beyond 1 , then $x$ must decrease. So
- $ x<1<y .$

- Then,
- $ y / x>1$

- And so.
- $\log (y / x)>0 \text {. }$

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<footer style = "font-size: 18px">Using the differential equation to graph the solution $\rightarrow$ Using the differential equation to graph the solution $\rightarrow$ <span style = "color:blue"> Graphing the solution contd </span> $\rightarrow$ Maxima or minima at a point $\rightarrow$ Example contd </footer>

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- Note: Directly from the diffrential equation we get $x(y)$ has a local max at $y=1$

- Also $x\prime(y)= \frac {\log(y/x)}{-(y/x)}  $   (from diffrential equation)

- So $x\prime$ has no other roots. That is $x\prime (y) \neq 0  $

- $ if, y \neq 1 ,(y>0) $

- $x\prime(y)$ must remain negative throughout $(1 , \infty)$

- and $x\prime(y)$ must be positive on $(0,1)$

- So $x(y)$ is strictly monotone on $(1, \infty) $ as well as on $(0,1)$

- Note: All this information from the diffrential equation directly !

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<footer style = "font-size: 18px">Using the differential equation to graph the solution $\rightarrow$ Graphing the solution contd $\rightarrow$ <span style = "color:blue"> Maxima or minima at a point </span> $\rightarrow$ Example contd $\rightarrow$ Example contd </footer>

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- Note that since the value of $y$ is 1 at $x=1$, we get $v=1$ when $x=1$. This equation can be readily integrated and the solution is
- $\int_1^v \frac{d u}{u}+\int_1^v \frac{\log t d t}{t(1+\log t)}=0 .$

- An easy integration gives
- $\log y-\log (1+\log y-\log x)=0 $.....$(2.9)$   
- or equivalently.
- $ y=1+\log y-\log x $.....$(2.10)$

- <ins>**Behaviour of solution as $y \longrightarrow \infty$**</ins>

- We have
- $= y-\log y-1=-\log x $.....$(2.10)$

- So when $y \longrightarrow+\infty$, can we say $y-\log y-1 \longrightarrow+\infty$ ? How? If so.
- $\log x \longrightarrow+\infty$ and consequently $x \longrightarrow 0+$.

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<footer style = "font-size: 18px">Maxima or minima at a point $\rightarrow$ Example contd $\rightarrow$ <span style = "color:blue"> Example contd </span> $\rightarrow$ Comparison of infinities $\rightarrow$ Exercise </footer>

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### <primary style="font-size:24px"> Comparison of infinities </primary>
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- As $y \rightarrow \infty$,

-  (i)  $\log y$ goes to infinity faster than $\log (\log y)$    
-  (ii)  $y$ goes to infinity faster than $\log y$    
-  (iii)  $y^2$ goes to infinity faster than $y$    
-  (iv)  expy goes to infinity faster than $y$

- <ins>**Comparison of infinities**</ins>

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<footer style = "font-size: 18px">Example contd $\rightarrow$ Example contd $\rightarrow$ <span style = "color:blue"> Comparison of infinities </span> $\rightarrow$ Exercise $\rightarrow$ Comparison of infinities contd </footer>

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-  (4)   .

- What does it mean to say that a function $f(t)$ tends to infinity slower than $g(t)$ as $t \longrightarrow a$ (where $f$ and $g$ are defined on the same interval containing a). In the present context can you think of applying the L'Hospital's rule?

- For instance, show that $(\log t)^{1000}$ goes to infinity slower than $t^{1 / 10000}$ as $t \rightarrow \infty$. No matter how large $N$ is and no matter how small $a>0$ is, $(\log t)^N$ goes to infinity slower than $t^3$ as $t \rightarrow \infty$.

-  Remark:  Comparing the rate of growth to infinity of two functions is an extremely important idea in mathematics. An important point which is not usually stressed in calculus courses:

-  Calculus in essence, is the comparison of the growths of functions relative to one another (or comparison of rates at which two functions tend to zero).     
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<footer style = "font-size: 18px">Example contd $\rightarrow$ Comparison of infinities $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Comparison of infinities contd $\rightarrow$ Comparison of infinities contd </footer>

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- Gauss and Legendre independently conjectured that $f(t)$ and $g(t) g o$ to infinity at the same rate but this was notoriously difficult to establish and remained open for a hundred years.

- A development of the idea of comparison of infinities is available in   
-  G. H. Hardy, Orders of infinity, Cambridge university press, 1910.

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<footer style = "font-size: 18px">Exercise $\rightarrow$ Comparison of infinities contd $\rightarrow$ <span style = "color:blue"> Comparison of infinities contd </span> $\rightarrow$ Orders of infinity and differential equations $\rightarrow$ Thank you </footer>

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- As an amusing exercise consider the variable separable differential equation
- $\frac{d y}{d x}=\frac{y}{(1+y) x}, \quad(x>0, y>0) $.....$(2.11)$   .

-  (5)  Find the solutions $y(x)$ of the differential equation (2.11).

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<footer style = "font-size: 18px">Comparison of infinities contd $\rightarrow$ Comparison of infinities contd $\rightarrow$ <span style = "color:blue"> Orders of infinity and differential equations </span> $\rightarrow$ Thank you $\rightarrow$  </footer>