mathematics-class-12-unit-09-chapter-02-differential-equations-2-8-l6wjntlr-9e

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Variable Separable Equations </primary>
<hr>
<main>

- Topic

- Definite integrals are superior beings !

- When feasible, it is better to use definite integrals since the integral
- $\int_a^b f(x) d x$
- is defined for ANY continuous function on $[a, b]$. The symbol has a precise mathematical meaning (say for continuous functions) regardless of whether or not one is able to find an explicit anti-derivative. Another advantage is that definite integrals incorporate the given initial conditions and we do not have to worry about finding the value of the integration constant by plugging in the initial conditions.

- Use of definite integrals

- Let us now rework the same problem with definite integrals. Integrating
- $\frac{1}{y^2} \frac{d y}{d t}=1$

</div>

</div>

</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Differential equation lecture-2 $\rightarrow$ <span style = "color:blue"> Variable Separable Equations </span> $\rightarrow$ Variable Separable Equations $\rightarrow$ Solution escapes to infinity in finite time </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Solution escapes to infinity in finite time </primary>
<hr>
<main>

- Let us look at the innocent differential equation
- $\frac{d y}{d t}=y^2 $
- and prescribe initial condition $y(0)=c$ where $c>0$. Then proceeding as above,
- $\int \frac{d y}{y^2}=\int d t$

- Integrate (and remember the const. of integration I) $y(t)=-1 /(t+b)$. Put $t=0$ and the corresponding $y=c$. We get $b=-1 / c$ and so the solution is:
- $ y(t)=\frac{c}{1-c t}$

- As $t \longrightarrow 1 / c$ from the fift, the solution $y(t)$ escapes to infinity ! Thus, the interval I on which the solution would make sense is usually $\left(t_0-a, t_0+a\right)$ and rarely the whole real line !

</div>
<div style='float: right; width:1%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Variable Separable Equations $\rightarrow$ Variable Separable Equations $\rightarrow$ <span style = "color:blue"> Solution escapes to infinity in finite time </span> $\rightarrow$ Exercises $\rightarrow$ Discussion of solution exercise-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Exercises </primary>
<hr>
<main>

- (1) Consider the differential equation
- $\frac{d y}{d t}=1+y^2$
- with initial condition $y(0)=1$. Does the solution escape to infinity in finite time? What if the right hand side is replaced by $1+y^{10}$ ?

- (2) Consider the differential equation
- $\frac{d y}{d t}=\frac{1}{1+y^2}$
- with initial condition $y(0)=1$. Does the solution escape to infinity in finite time? What if the right hand side is replaced by $\left(1+y^{10}\right)^{-1}$ ?

- Note that in these examples we are not asking for the exact solution.
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Variable Separable Equations $\rightarrow$ Solution escapes to infinity in finite time $\rightarrow$ <span style = "color:blue"> Exercises </span> $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-1 </primary>
<hr>
<main>

- So our differential equation is
- $\frac{d y}{d t}=1+y^2, \quad y(0)=1$

- Then surely
- $\frac{d y}{d t}>y^2, \quad y(0)=1$
- which means, since the derivative is always positive, $y(t)$ would increase for $t>0$ and so would be positive.
- $\frac{1}{y^2} \frac{d y}{d t}>1$

- Now integrate over say $[0, s]$.   With $1+y^{10}$ in place of $1+y^2$
- $\frac{d y}{d t}=1+y^{10}, \quad y(0)=1,$

- Knock off the 1 in $1+y^{10}$ and $\quad \frac{d y}{d t}>y^{10}, \quad y(0)=1,$

- Divide by $y^{10}$ and proceed along the same lines: $\quad \int_1^{y(s)} \frac{d y}{y^{10}}>\int_0^s d t=s$

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution escapes to infinity in finite time $\rightarrow$ Exercises $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-1 </span> $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-1 </primary>
<hr>
<main>

- We immediately see from here that as $s \rightarrow 1$ - the right hand side of the inequality
- $ y(s)>\frac{1}{1-s}$
- goes to infinity as $s \rightarrow 1-$. Thus the solution $y(s)$ certainly cannot live for time beyond $s=1$.
</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Exercises $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-1 </span> $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-1 </primary>
<hr>
<main>

- Replace the Differential Equation by a differential Inequality
- $\frac{d y}{d t}>y^2, \quad y(0)=1 .$

- Or in the other case
- $\frac{d y}{d t}>y^{10}, \quad y(0)=1$

- An Important Comment

- Note that we passed from the differential equation
- $\frac{d y}{d t}=1+y^2, \quad y(0)=1$
- to the differential inequality
- $\frac{d y}{d t}>y^2, \quad y(0)=1$
- and concluded that $y(s)>1 /(1-s)$ which means $y(s)$ would ALREADY have gone to infinity when $s \rightarrow 1-$.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-1 </span> $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-3 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-1 </primary>
<hr>
<main>

- That is to say, the time of existence of $y(s)$ CANNOT exceed 1 but can actually be strictly smaller than 1 . We saw this by obtaining the explicit solution $y(t)=\tan \left(t+\frac{\pi}{4}\right)$ for the solution of the original differential equation.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-1 </span> $\rightarrow$ Discussion of solution exercise-3 $\rightarrow$ Discussion of solution exercise-3 </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-3 </primary>
<hr>
<main>

- $\log \left(\sec ^2 y+\sec y \tan y\right)=\log \left(\sec ^2 x+\sec x \tan x\right)+c$

- Exponentiate
- $\sec ^2 y+\sec y \tan y=e^c\left(\sec ^2 x+\sec x \tan x\right) .$

- Here we could stop and declare the last displayed line as the solution! Well, we have obtained the solution in IMPLICIT FORM
</div>

</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-1 $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-3 </span> $\rightarrow$ Discussion of solution exercise-3 $\rightarrow$ Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Discussion of solution exercise-3 </primary>
<hr>
<main>

- (3) Consider $\cos x(1+\sin y) \frac{d y}{d x}=(1+\sin x) \cos y$ Here $x$ and $y$ lie in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Prove that the solution $y(x)$ is defined on the entire interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and as $x \longrightarrow \frac{\pi}{2}-$,

- $y(x) \longrightarrow \frac{\pi}{2}$. Discuss the special case $y(0)=0$.
- Solution: Separate variables and integrate:
- $\int\left(\frac{1+\sin y}{\cos y}\right) \frac{d y}{d x} d x=\int \frac{(1+\sin x)}{\cos x} d x .$

- The solution in implicit form is ( $c$ is the const. of integration):

- $ \sec ^2 y+\sec y \tan y=e^c\left(\sec ^2 x+\sec x \tan x\right) . $

- $ \sec ^2 y+\sec y \tan y=\left(\sec ^2 x+\sec x \tan x\right) .$
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-1 $\rightarrow$ Discussion of solution exercise-3 $\rightarrow$ <span style = "color:blue"> Discussion of solution exercise-3 </span> $\rightarrow$ Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) $\rightarrow$ Example from geometry </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) </primary>
<hr>
<main>

- Suppose that $h\left(y_0\right)=0$ and we are looking for a solution of the equation
- $\frac{d y}{d x}=g(x) h(y)$
- with initial conditions $y\left(x_0\right)=y_0$. Note that the constant function $y(x)=y_0$ satisfies the differential equation as well as the initial conditions.
- On the other hand if $h\left(y_0\right) \neq 0$ and $g\left(x_0\right)=0$. The only point to be addressed is whether we are justified in using the substitution rule in our method. We shall assume that $g(x) \neq 0$ if $x \neq x_0$ in a neighbohood of $x_0$. Then $y^{\prime}$ is not zero for $x \neq x_0$ in a neighborhood of $x_0$ and we have a right to invoke the substitution theorem on the half intervals $\left(x_0, x_0+a\right)$ and $\left(x_0-a, x_0\right)$.
- 7
- II
</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-3 $\rightarrow$ Discussion of solution exercise-3 $\rightarrow$ <span style = "color:blue"> Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) </span> $\rightarrow$ Example from geometry $\rightarrow$ Example from geometry contd.. </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Example from geometry </primary>
<hr>
<main>

- A popular example from geometry

- Find a plane curve $y=f(x)$ with the property that all its normals pass through the same point.

- Let us assume that the point through which all the normals pass is the origin. Let $\left(x_0, y_0\right)$ be a point on the curve whose equation is $y=f(x)$. The slope of the normal at $\left(x_0, y_0\right)$ is
- $\left(f^{\prime}\left(x_0\right)\right)^{-1}$

- The equation of the normal is
- $\left(y-y_0\right) f^{\prime}\left(x_0\right)+\left(x-x_0\right)=0 .$

- Since this passes through the origin,
- $ x_0+y_0 f^{\prime}\left(x_0\right)=0 .$

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Discussion of solution exercise-3 $\rightarrow$ Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) $\rightarrow$ <span style = "color:blue"> Example from geometry </span> $\rightarrow$ Example from geometry contd.. $\rightarrow$ Example from Rainville's elem. diff eqns. </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Example from geometry contd.. </primary>
<hr>
<main>

- This holds for all the points $\left(x_0, y_0\right)$ on the curve and so we rewrite (1.13) by omitting the subscript zero and using the notation $\frac{d y}{d}$ in place of $f^{\prime}(x)$
- $ y \frac{d y}{d x}=-x$

- Integrating with respect to $x$ we get
- $ y^2+x^2=2 c$
- where $c$ is the constant of integration. The curve is thus a circle.
</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Cases when h(y)=0 or g(x)=0 in y'=h(y) g(x) $\rightarrow$ Example from geometry $\rightarrow$ <span style = "color:blue"> Example from geometry contd.. </span> $\rightarrow$ Example from Rainville's elem. diff eqns. $\rightarrow$ Example from Rainville's elem. diff eqns. contd </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Example from Rainville's elem. diff eqns. </primary>
<hr>
<main>

- An interesting example from Rainville's Elem. Diff. Eqns.

- Consider the differential equation with initial condition:
- $\sqrt{1-x^2} \frac{d y}{d x}+\sqrt{1-y^2}=0, \quad y(0)=\frac{\sqrt{3}}{2}$

- Proceeding as usual we get
- $\int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=0 .$

- Integrating we get
- $\sin ^{-1} y+\sin ^{-1} x=C$

- To determine the constant of integration we put in the given initial conditions and see that $C=\frac{\pi}{3}$. while this may sound pretty easy let us see the solution in some detail.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Example from geometry $\rightarrow$ Example from geometry contd.. $\rightarrow$ <span style = "color:blue"> Example from Rainville's elem. diff eqns. </span> $\rightarrow$ Example from Rainville's elem. diff eqns. contd $\rightarrow$ Example from Rainville's elem. diff eqns. contd </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Example from Rainville's elem. diff eqns. contd </primary>
<hr>
<main>

- which gives
- $\sin ^{-1} y=\frac{\pi}{3}-\sin ^{-1} x$
- $ y=\sin \left(\frac{\pi}{3}-\sin ^{-1} x\right)$

- Recalling the addition formula for sine we get
- $ y=\frac{\sqrt{3}}{2} \cos \left(\sin ^{-1} x\right)-\frac{1}{2} x$
- which after squaring gives
- $\left(y+\frac{x}{2}\right)^2=\frac{3}{4}\left(1-x^2\right)$

- Or elegantly
- $- x^2+y^2+x y=\frac{3}{4}$
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Example from geometry contd.. $\rightarrow$ Example from Rainville's elem. diff eqns. $\rightarrow$ <span style = "color:blue"> Example from Rainville's elem. diff eqns. contd </span> $\rightarrow$ Example from Rainville's elem. diff eqns. contd $\rightarrow$ Birth of elliptic functions </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Example from Rainville's elem. diff eqns. contd </primary>
<hr>
<main>

- To examine the nature of the curve
- $ x^2+y^2+x y=\frac{3}{4}$

- Let us put
- $ y=\frac{1}{\sqrt{2}}(x+Y) . \quad x=\frac{1}{\sqrt{2}}(x-Y) .$

- We get an ellipse
- $ 3 x^2+y^2=\frac{3}{2}$

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Example from Rainville's elem. diff eqns. $\rightarrow$ Example from Rainville's elem. diff eqns. contd $\rightarrow$ <span style = "color:blue"> Example from Rainville's elem. diff eqns. contd </span> $\rightarrow$ Birth of elliptic functions $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Differential Equations L-2 </Success>
### <primary style="font-size:24px"> Birth of elliptic functions </primary>
<hr>
<main>

- The example above goes back to L. Euler and the differential equation expresses the addition formula for the sine function. The technique employed in essence can be used to prove
- $\int_0^a \frac{d t}{\sqrt{1-t^2}}+\int_0^e \frac{d t}{\sqrt{1-t^2}}=\int_0^{(u, v)} \frac{d t}{\sqrt{1-t^2}} $
- where $o(u, v)=u \sqrt{1-v^2}+v \sqrt{1-u^2}$. Euler went further in this direction and derived a similar formula for the integrals:
- $\int_0^e \frac{d t}{\sqrt{1-t^4}}$

- Three decades later Gauss (1796 approx) studied the inverse of these functions and obtained an addition formula for them analogous to the addition formula for the sine function.
</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Example from Rainville's elem. diff eqns. contd $\rightarrow$ Example from Rainville's elem. diff eqns. contd $\rightarrow$ <span style = "color:blue"> Birth of elliptic functions </span> $\rightarrow$ Thank you $\rightarrow$  </footer>