mathematics-class-12-unit-09-chapter-06-differential-equations-6-8-2phmut5ykdc

### <Success style="font-size:25px"> Differential Equations L-6 </Success>
### <primary style="font-size:24px"> Orders of infinity and differential equations </primary>
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- As an amusing exercise consider the variable separable differential equation
- $\frac{d y}{d x}=\frac{y}{(1+y) x}, \quad(x>0, y>0)$

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

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<footer style = "font-size: 18px"> $\rightarrow$ Differential equations lecture-6 $\rightarrow$ <span style = "color:blue"> Orders of infinity and differential equations </span> $\rightarrow$ Orders of infinity and differential equations examples $\rightarrow$ Discussion of exercise </footer>

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- (i) Show that $y(x)$ cannot escape to infinity in finite time.

- (ii) Show that $y(x)$ tends to infinity at the same rate as $\log x$.

- (iii) Use L'Hospital's rule to compute the limit
- $\lim _{x \rightarrow \infty} \frac{y(x)-\log x}{\log (\log x)}$

- Thus for large values of $x$, the solution $y(x)$ "behaves like" $\log x-\log (\log x)$. Try out a similar exercise for the solution $(2.10)$ to equation (2.7). calculate $\lim y(x)$ as $x \rightarrow 0+$.

- Discussion of exercise (5) contd...

- The differential equation gives $y(x)$ is monotone increasing! On solving it,
- $ y+\log y=c+\log x . $
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- Discussion of exercise 5 contd..

- Well.
- $ y+\log y=c+\log x$

- Suppose $y(x) \rightarrow+\infty$ as $x \rightarrow a$ (where $a$ is finite and positive). If $y(x) \longrightarrow+\infty$ then $\log y \longrightarrow+\infty$.

- Thus in the displayed equation, LHS $\rightarrow \infty$ but RHS $\rightarrow c+\log a$. Contradiction. So $y(x)$ cannot excape to infinity in finite time: To give a slightly different argument, write the above equation as
- $ y\left(1+\frac{\log y}{y}\right)=c+\log x$

- Now $(\log y) / y \rightarrow 0$ and therefore $y\left(1+\frac{\log y}{y}\right) \rightarrow+\infty$.
- Hence LHS tends to infinity but the RHS tends to the finite limit $c+\log a$ which is a contradiction

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- Discussion of exercise 5 contd...

- $\frac{d y}{d x}=\frac{y}{x(1+y)}, \quad x>0, y>0 .$

- The diff. eqn. tells us the derivative is positive. Look at the equation
- $ y+\log y=c+\log x$
- and see what happens as $x \longrightarrow+\infty$.
- If $y(x)$ has a finite limit then so would $\log y$. Thus
- $\text { LHS }=y+\log y \longrightarrow \text { Finite Limit }$
- whereas
- $\text { RHS }=c+\log x \longrightarrow+\infty$

- Again we get a contradiction.

- Orders of infinity and differential equations

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- Discussion of exercise (5) contd...

- Apply L'Hospital's rule again and the limit is -1 . We thus see that for large values of $x$,
- $ y(x) \sim \log x-\log (\log x), \quad x \rightarrow \infty,$
- where the symbol $\sim$ simply means the ratio tends to one as $x \rightarrow \infty$. Loosely put, the solution $y(x)$ behaves like $\log x-\log (\log x)$ for large values of $x$.

- Note that we have obtained VERY PRECISE information about the growth of the solution $y(x)$ !

- It is HOPELESS trying to solve the equation $y+\log y=c+\log x$ to obtain $y(x)$ explicitly in terms of $x$.
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<footer style = "font-size: 18px">Discussion of exercise $\rightarrow$ Discussion of exercise $\rightarrow$ <span style = "color:blue"> Discussion of exercise </span> $\rightarrow$ Reference for the last example $\rightarrow$ Two exercises from Rainville's book </footer>

### <Success style="font-size:25px"> Differential Equations L-6 </Success>
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- The example in the last slide is taken from the following paper: J Shackell, Growth orders occuring in expansions of Hardy-field solutions of algebraic differential equations, Annals de L.'nst. Fouriet, 45 (1995) 183-221. This equation appears on page 219

- Clearly, understanding the behaviour of solutions of a differential equation for large times is extremely important as it fumishes information about the ultimate state towards which a physical system evolves. However in the toy problem 5(iii) we got this by solving the differential equation which is rarely possible :

- So one needs to develop theorems that provide such information without solving the differential equations.

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### <Success style="font-size:25px"> Differential Equations L-6 </Success>
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- Here are two exercises appearing on page 30 from Rainville. Elementary Diflerential Equations, lifth edition, Macmillan, 1974 If the minus signs in the second term are tampered, the integration would get lengthy
- (6) Solve the following differential equation on the first quadrant $x>0$, $y>0:$
- $\left(y-\sqrt{y^2+x^2}\right) d x-x d y=0$
- and sketch the family of curves obtained Specifially find the solution satisfyes the condition $y(\sqrt{3} / 2)=1 / 2$.
- (7) Solve the following differential equation on the first quadrant $x>0$. $y>0$ :
- $\left(y+\sqrt{y^2+x^2}\right) d x-x d y=0$
- and sketch the family of curves obtained. Specifially find the solution satisfying the condition $y(\sqrt{3} / 2)=1 / 2$.

- This is WRONG! because if you differentiate (2.14)

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- Rainville asks for the solution only in the first quadrant by giving the initial conditions in there. However let us ask ourselves how to solve (2.12) in the second quadrant? We shall see that if we proceed blindly setting $y=v x$ we would get
- $ x d v+\left(\sqrt{1+v^2}\right) d x=0$
- which upon solving gives
- $\log |x|+\log \left(v+\sqrt{1+v^2}\right)=c . $

- Now, $v+\sqrt{1+v^2}$ is always positive but in the second quadrant, $|x|=-x$ and so
- $y+\sqrt{x^2+y^2}=e^c .$

- then we do not get back the differential equation (2.12) but get instead the equation
- $\left(y+\sqrt{x^2+y^2}\right) d x-x d y=0 .$

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- The differential equation is only positively homogeneous and so the solution procedure as such is valid ONLY in the first quadrant. Let us work afresh and set $y=v x$ and $d y=v d x+x d v$ which is perfectly fine even in the second quadrant. However if $M(x, y)$ is positively homogeneous and $N(x, y)$ is homogeneous and $x<0$,
- $\begin{aligned} M(x, v x) d x+N(x, v x) d y & =M(-(-x),-(-x) v) d x+x^k N(1, v) d y \\\\& =(-x)^k M(-1,-v) d x+x^k N(1, v) d y \\\\& =x^k\left((-1)^k M(-1,-v) d x+N(1, v) d y\right)\end{aligned} $

- The differential equation then transforms into
- $ (-1)^k M(-1,-v) d x+N(1, v)(v d x+x d v)=0 .$
- which is again variable separable.

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- o The $y=v x$ method completely fails along the $y$-axis since the variable $v$ would have no meaning. At points on the $y$-axis except the origin, one employs the substitution $x=v y$.
- Let us examine whether the differential equation makes sense along points on the $y$-axis. At a point $(0, t)$ equation $(2,12)$ reads
- $\frac{d y}{d x}=\frac{y-\sqrt{x^2+y^2}}{x}=\frac{-x}{y+\sqrt{x^2+y^2}} .$
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- The first expression is indeterminate while the second gives the slope of the curve as zero for points $(0, t)$ with $t>0$. Hovever if $t<0$. neither expression makes sense.

- Write the differential equation as
- $\frac{d x}{d y}=\frac{x}{y-\sqrt{x^2+y^2}}$

- Does it make sense along the points $(0, t)$ with $t<0$ ?

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- (8) Find the solution curve of $(2.12)$ passing through the point $(0, c)$ for various values of $c \neq 0$. Sketch the curves. Do you see any exceptional features worthy of comment?

- (9) Solve the differential equation
- $\frac{d y}{d x}=\frac{x+y}{x-y}, \quad x \neq y .$

- Ans: $\log \left(x^2+y^2\right)-2 \tan ^{-1}(y / x)=C$

- (10) Solve the equation
- $\frac{d y}{d x}-\frac{y}{x}=\exp (y / x)$

- Ans: $\log x+\exp (-y / x)=C$.

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- We now make a few comments on the geometry of the homogeneous differential equation:
- $\frac{d y}{d x}=f(x, y)$
- where $f(x, y)$ is homogeneous (or positively homogeneous in say the first quadrant) of degree zero (which would be so since $M(x, y)$ and $N(x, y)$ are homogeneous of the SAME degree).

- Equation (2.19) states that the tangent to each solution curve at $(x, m x)$ has the same slope
- $\tan ^{-1}(f(1, m))$
- so that the solutions curves all cut the line $y=m x$ at the same angle.
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- Similarity: Principle of symmetry in homogeneous differential equations

- In classical geometry, a family of curves with the property described in the previous slide is called a family of similarly placed similar curves - with the origin as the center of similitude and if we know ONE generic member of such a family
- $\phi(x, y)=0 .$
- then all other members can be obtained as
- $\phi\left(\frac{x}{c}, \frac{y}{c}\right)=0 .$

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- Usually, if one solution is known then one can obtain ALL solutions This however may fail in certain degenerate cases (such as the case of a trivial solution $y(x) \equiv 0$ ) which is why we say "Usually" and have not elevated the above statement to a theorem.
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- Similarity and the principle of symmetry contd...
- Given a non-zero real number $t$ the substitution
- $ X=t x, \quad Y=t y$
- is called a homothety or a similarity transformation. Let us now take a homogeneous differential equation and write it in the form
- $\frac{d y}{d x}=f(x, y)$
- where $f(x, y)=-M(x, y) / N(x, y)$ is homogeneous of degree zero. Let us introduce the change of variables $(2.23)$ into $(2.24)$ :
- $\frac{d Y}{d X}=\frac{d Y}{d y} \frac{d y}{d x} \frac{d x}{d X}=t \frac{d y}{d x} \frac{1}{t}=\frac{d y}{d x}$

- So the LHS of $(2.24)$ retains its form. The RHS also retains its form since $f(x, y)$ is homogeneous of degree zero and so $(2.24)$ transforms into
- $\frac{d Y}{d X}=f(X, Y) \text {. }$

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- Similarity and the principle of symmetry contd...

- Note that $(2.24)$ and $(2.25)$ are the same equations. We say the equation $(2.24)$ is invariant under homotheties or similarity transformations.

- Now the principle of symmetry states that if $\phi(x, y)=0$ is a solution then $\phi(\alpha, c y)=0$ is ALSO a solution of the differential equation and all solutions can be obtained thus from a single (non-degenerate) solution $\phi(x, y)=0$.

- Stated differently, the homothety takes a solution to another solution. Or the set of solutions as a whole is invariant under homotheties
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- Discussion of Example 1
- $\phi\left(\frac{x}{c}, \frac{y}{c}\right)=\left(\frac{x}{c}\right)^2+\left(\frac{y}{c}\right)^2-\frac{y}{c}$

- So $\phi\left(\frac{x}{c}, \frac{y}{c}\right)=0$ gives,
- $ x^2+y^2-c y=0 .$

- We see that we get ALL the solutions! out of the specific solution
- $\phi(x, y)=x^2+y^2-y=0 .$
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<footer style = "font-size: 18px">Principle of symmetry contd $\rightarrow$ Principle of symmetry contd $\rightarrow$ <span style = "color:blue"> Principle of symmetry contd </span> $\rightarrow$ Examples of homogeneous equations $\rightarrow$ y dx + x(log y - log x) dy =0 </footer>

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- Let us examine this in the light of the many examples of homogeneous equations that we have solved. The numbers of the equations have been indicated for reference.
- For the equation $2 x y d x-\left(x^2-y^2\right) d y=0$ which is equation (2.4) we have the solution $x^2+y^2-c y=0$. Taking $c=1$. look at
- $\phi(x, y)=x^2+y^2-y$
- and we see that the $\phi(x / c, y / c)=0$ gives all the solutions of the equation as $c$ varies. So all the circles can be obtained from the specific one $\phi(x, y)=0$ through scaling the variables $x$ and $y$ by a the same numerical factor $c$.

- $2 x y d x+\left(x^2-y^2\right) d y=0$, the general solution (which was an exercise for you) is $3 x^2 y-y^3=c$. Taking $c=1$ again we get $\phi(x, y)=3 x^2 y-y^3-1$ and all solutions are obtained as $\phi(x / c, y / c)=0$.

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<footer style = "font-size: 18px">Principle of symmetry contd $\rightarrow$ Principle of symmetry contd $\rightarrow$ <span style = "color:blue"> Examples of homogeneous equations </span> $\rightarrow$ y dx + x(log y - log x) dy =0 $\rightarrow$ Discussion of example </footer>

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- Next. $y d x+x(\log y-\log x) d y=0$ for which we got the solution $y-1-\log y+\log x=0$ which is (2.10). Taking
- $\phi(x, y)=y-1-\log y+\log x$
- we see that $\phi(x / c, y / c)=0$ is the general solution.
- (11) Exercise: Check the claim made in the last line !
- Let us look at the example from Rainville's book (equation (2.12))
- $\left(y-\sqrt{x^2+y^2}\right) d x-x d y=0$

- Then $\phi(x, y)=y+\sqrt{x^2+y^2}-1$ does the job but ONLY in the first quadrant since the differential equation is ONLY positively homogeneous. Thus $\phi(x / c, y / c)=0$ gives all the solutions as $c$ varies over $(0, \infty)$.
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- Discussion of example 3

- In this case
- $\phi(x, y)=y-1-\log y+\log x .$

- We restrict to the first quadrant only (the symmetry principle works).
- $\phi\left(\frac{x}{c}, \frac{y}{c}\right)=\frac{y}{c}-1-\log (y / c)+\log (x / c) .$

- Or.
- $\phi\left(\frac{k x}{c}, \frac{y}{c}\right)=\frac{1}{c}(y-c-c \log y+c \log x) .$

- So the equation $\phi(x / c, y / c)=0$ reads
- $ y-c-c(\log y-\log x)=0$

- Check that this is the general solution!
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- Two more examples from Rainville's book
- Consider the homogeneous equation
- $ x y d x-\left(x^2+4 y^2+4 x y\right) d y=0$

- Observe that if we take $\phi(x, y)=y+x$ then we see that $\phi(x, y)=0$ is a solution of (2.26). However if we look at $\phi(x / c, y / c)=0$ then we get the same solution! Does this contradict the principle of symmetry?

- (12) Find the general solution of (2.26) and identify this special solution.

- Ans: $y^3 x+y^4=\operatorname{cexp}(x / y)$.

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- Again for the differential equation
- $ x^2 d y-\left(4 x^2+2 y^2+7 x y\right) d x=0$

- we see that $\phi(x, y)=0$ furnishes a solution for each of the choices $\phi(x, y)=x+y$ and $\phi(x, y)=2 x+y$. However $\phi(x / c, y / c)=0$ as $c$ varies would not yield any new solutions!
- (13) Solve the differential equation completely and try to figure out what is going on. Ans: $x^2 y+2 x^3-c x-c y=0$.

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