mathematics-class-12-unit-09-chapter-01-differential-equations-1-8-yxxvrbrbixw

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Brief history-2 </primary>
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* Issac Barrow
* Issac Newton (1687 - Linear Diff. Eqn)
* Leibnitz (1693 - The $y= tx$ substitution for homogeneous equations)
* Jean Bernoulli (1697)
* Jacopo Riccati $y^{\prime}=A(x) y^2+B(x) y+C(x)$

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<footer style = "font-size: 18px">Differential equation lecture-1 $\rightarrow$ Brief history-1 $\rightarrow$ <span style = "color:blue"> Brief history-2 </span> $\rightarrow$ Brief history-3 $\rightarrow$ Brief history-4 </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Brief history-4 </primary>
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- **Origin and scope of differential equations**
- Laws of physics when expressed in mathematical terms give rise to differential equations !
- The law of mass action in chemistry - Differential equations of chemical kinetics and enzyme kinetics.
- Models arising in Ecology and demography. Striking similiarities with equations modelling chemical kinetics.
- Mathematical Biology: Spread of diseases, Models of heart-beat, Mathematical Physiology...
- (i) E. C. Zeeman: Modelling of heart beat

- (ii) Hodgkin and Huxley: Nerve impulses (1952) Received Nobel Prize for their work.
- Problems arising in geometry (applications within the field of mathematics)

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<footer style = "font-size: 18px">Brief history-2 $\rightarrow$ Brief history-3 $\rightarrow$ <span style = "color:blue"> Brief history-4 </span> $\rightarrow$ The simple pendulum-1 $\rightarrow$ The simple pendulum-1 </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> The simple pendulum-1 </primary>
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- $m l^2 \frac{d^2 y}{d t^2}=-m \lg \sin (y(t))$

- which simplifies to the differential equation

- $\frac{d^2 y}{d t^2}+\frac{g}{l} \sin (y)=0$

- We have derived the differential equation governing the motion of a simple pendulum. This is a second order differential equation.
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<footer style = "font-size: 18px">Brief history-4 $\rightarrow$ The simple pendulum-1 $\rightarrow$ <span style = "color:blue"> The simple pendulum-1 </span> $\rightarrow$ More examples from physics - SHM $\rightarrow$ Solutions of a differential equation </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> More examples from physics - SHM </primary>
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- A particle moving along a straight line executes simple harmonic oscillations. The force acting on it is proportional to the displacement and in the opposite direction. Thus if $y(t)$ is the displacement form the origin at time $t$ and $m$ is the mass of the particle then Newton's second law of motion gives

- m $\frac{d^2 y}{d t^2}=-k y$

- where $k$ is a positive constant (namely the constant of proportionality).
- Denoting $k / m$ by $\omega^2$ we get the differential equation for harmonic oscillations as:

- $\frac{d^2 y}{d t^2}+\omega^2 y=0$     $\quad$ (1.2)

- which is again a second order differential equation

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<footer style = "font-size: 18px">The simple pendulum-1 $\rightarrow$ The simple pendulum-1 $\rightarrow$ <span style = "color:blue"> More examples from physics - SHM </span> $\rightarrow$ Solutions of a differential equation $\rightarrow$ Solutions of a differential equation </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Solutions of a differential equation </primary>
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- Let us take up the differential equation of SHM:

- $\frac{d^2 y}{d t^2}+\omega^2 y=0$        $\quad$ (1.2), $\frac{d^2y} {dt^2} + \frac{g} {l} \sin y$

- A direct substitution confirms that if we set $y(t)=\cos \omega t$ then equation $(1.2)$ is satisfied and also when we substitute $y(t)=\sin \omega t$. We say that $\cos \omega t$ and $\sin \omega t$ are two solutions of the differential equation (1.2). The student can verify that if $A$ and $B$ are any two constants then

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<footer style = "font-size: 18px">The simple pendulum-1 $\rightarrow$ More examples from physics - SHM $\rightarrow$ <span style = "color:blue"> Solutions of a differential equation </span> $\rightarrow$ Solutions of a differential equation $\rightarrow$ Examples from electrical circuits </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Solutions of a differential equation </primary>
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- $y(t)=A \cos \omega t+B \sin \omega t$      $\quad$ (1.3)

- also satisfies the equation (1.2). Thus we have enlisted many solutions of (1.2).

- Question: Have we enlisted all the solutions? How do we know that if $z(t)$ is a solution of (1.2) then $z(t)=A \cos \omega t+B \sin \omega t$ for certain constants $A$ and $B$ ?We see that we are led to the problem of describing the class of ALL the solutions.
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<footer style = "font-size: 18px">More examples from physics - SHM $\rightarrow$ Solutions of a differential equation $\rightarrow$ <span style = "color:blue"> Solutions of a differential equation </span> $\rightarrow$ Examples from electrical circuits $\rightarrow$ Examples arising out of ecology </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Examples from electrical circuits </primary>
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- Analogue of equation (1.2) also appears in the theory of electric circuits the LC circuits. For a discussion on this see p. 845 (equation 38.5) in D. Halliday and R. Resnick, Physics, Vol 2, Third Edition, John Wiley and Sons, 1978. On p. 848 of this important book one finds the following differential equation for an LCR circuit:

- $\frac{d^2 q}{d t^2}+\frac{R}{L} \frac{d q}{d t}+\frac{q}{L C}=0$

- which is the electrical analogue of the damped harmonic motion exhibited in springs (see equation 15.37 of Halliday-Resnick).

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<footer style = "font-size: 18px">Solutions of a differential equation $\rightarrow$ Solutions of a differential equation $\rightarrow$ <span style = "color:blue"> Examples from electrical circuits </span> $\rightarrow$ Examples arising out of ecology $\rightarrow$ Verhulst's model </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Examples arising out of ecology </primary>
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- Malthus (in 1798 but earlier and independently L. Euler) had proposed the following model for population growth in an ecology consisting of a single species. If $y(t)$ is the population at time $t$ then the rate of change of population is proportional to $y(t)$ which means

- $\frac{d y}{d t}=k y$     $\quad$ (1.4)

- where $k$ is the constant of proportionality. Observe that (1.4) implies $y(t)=A \exp (k t)$ showing that the population $y(t)$ must grow exponentially fast !

- Is this practical?

- Would nature allow such a exponential growth of population??

- Would there not be some inhibiting factors caused by limitations of natural resources preventing this exponential growth?
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<footer style = "font-size: 18px">Solutions of a differential equation $\rightarrow$ Examples from electrical circuits $\rightarrow$ <span style = "color:blue"> Examples arising out of ecology </span> $\rightarrow$ Verhulst's model $\rightarrow$ Two species ecology predator-prey model </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Verhulst's model </primary>
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- Verhulst (1836) proposed a modification of the Malthusian/Eulerian model taking into account the limitations of natural resources. Paucity of resources would produce social friction which would check the rate of growth of population by introducting the term $-k y^2 / R$ in the differential equation:

- $\text { of } \frac{d y}{d t}=k y\left(1-\frac{y}{R}\right)$     $\quad$(1.5)

- where the constant $R$ is called the carrying capacity of the environment.
- J. D. Murray: Mathematical Biology Volume - I. Springer Verlag 2002.

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<footer style = "font-size: 18px">Examples from electrical circuits $\rightarrow$ Examples arising out of ecology $\rightarrow$ <span style = "color:blue"> Verhulst's model </span> $\rightarrow$ Two species ecology predator-prey model $\rightarrow$ Predator-prey model contd </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Two species ecology predator-prey model </primary>
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- Let us consider a two species ecology consisting of a predator with population $x(t)$ and a prey with population $y(t)$. For the predators, the source of food is the availability of prey and the latter are herbivorous living on say algae.

- In the absence of prey $y(t)$ the population of the predators $x(t)$ would decrease exponentially:

- $\frac{d x}{d t}=-a x, \quad a>0$    $\quad$(1.6)

- whereas the population of prey would increase:

- $\frac{d y}{d t}=k y, \quad k>0$     $\quad$(1.7)

- However when the two are put together, the rate of decrease indicated in the RHS of (1.6) would have to be modified as

- $-a x+b x y, \quad b>0 .$

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<footer style = "font-size: 18px">Examples arising out of ecology $\rightarrow$ Verhulst's model $\rightarrow$ <span style = "color:blue"> Two species ecology predator-prey model </span> $\rightarrow$ Predator-prey model contd $\rightarrow$ Coming to specific differential equations </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Predator-prey model contd </primary>
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- and the rate of increase indicated on the right hand side of (1.7) would be checked by the addition of a term $-cxy$ namely ky $- cxy$.

- Thus we get the pair of differential equations:

- $\frac{d x}{d t}=-a x+b x y, \quad \frac{d y}{d t}=ky-cxy$     $\quad$(1.8) .

- where the constants $a, b, c$ and $k$ are all positive.

- This model is known as the Volterra-Lotka system of equations.

- It is a system of simultaneous differential equations and is a coupled

- systems since the unknown $y$ appears in the $x$ equation and vice-versa. We shall not undertake a systematic analysis of this system.

- This was mainly as an illustration of how differential equations arise in the study of Ecology

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<footer style = "font-size: 18px">Verhulst's model $\rightarrow$ Two species ecology predator-prey model $\rightarrow$ <span style = "color:blue"> Predator-prey model contd </span> $\rightarrow$ Coming to specific differential equations $\rightarrow$ The variable separable equation </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Coming to specific differential equations </primary>
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- From now on we shall only deal with differential equations of first order. that is involving only one derivative with respect to time. The differential equations of the pendulum, SHM and LCR circuits involve two time derivatives and so are second order equations. The general first order equations that we shall deal with are of the form

- $\frac{d y}{d x}=f(x, y)$     $\quad$(1.9)

- where $f(x, y)$ is a function of two variables. We shall look at three types of equations
* Variable separable equations
* Homogeneous equations
* Linear equations and its close cousin namely the Bernoulli equation.

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<footer style = "font-size: 18px">Two species ecology predator-prey model $\rightarrow$ Predator-prey model contd $\rightarrow$ <span style = "color:blue"> Coming to specific differential equations </span> $\rightarrow$ The variable separable equation $\rightarrow$ The variable separable equation </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> The variable separable equation </primary>
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- This is a case when the function $f(x, y)$ appearing in (1.9) is a product of $g(x) h(y)$ in which case we rewrite the equation in the form
- $\frac{1}{h(y)} \frac{d y}{d x}=g(x) .\quad (1.10)$

- This is permissible as long as $h(y)$ is different from zero. Infact we shall assume that neither $h(y)$ nor $g(x)$ is zero. Such cases as always would have to be dealt with separately working on intervals where they are non-zero to begin with. Integrating both sides of (1.10) with respect to $x$ we get
- $\int \frac{1}{h(y)}\left(\frac{d y}{d x}\right) d x=\int g(x) d x$
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<footer style = "font-size: 18px">Predator-prey model contd $\rightarrow$ Coming to specific differential equations $\rightarrow$ <span style = "color:blue"> The variable separable equation </span> $\rightarrow$ The variable separable equation $\rightarrow$ Initial conditions </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> The variable separable equation </primary>
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- The non-vanishing of $g(x) h(y)$ shows that $y^{\prime}(x)$ is non zero and so we may invoke the substitution theorem and rewrite this as
- $\int \frac{d y}{h(y)}=\int g(x) d x  \quad (1.11)$

- $\frac{d y}{d x}=f(x, y)=g(x) h(y)$   .

- Disclaimer: We shall assume that the functions $g(x)$ and $h(y)$ are continuous on the interval in question.

- $\frac{1}{h(y)} \frac{d y}{d x}=g(x), \quad h(y) \neq 0$

- Integrate both sides with respect to $x$ :

- $\int \frac{1}{h(y)} \frac{d y}{d x} d x=\int g(x) d x$  
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<footer style = "font-size: 18px">Coming to specific differential equations $\rightarrow$ The variable separable equation $\rightarrow$ <span style = "color:blue"> The variable separable equation </span> $\rightarrow$ Initial conditions $\rightarrow$ Solution escapes to infinity in finite time </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Initial conditions </primary>
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- Usually a differential equation comes equipped with some initial condition. We think of the independent variable as time and the state of the system is prescribed at a certain time instant $t=t_0$. For example if the differential equations describes chemical kinetics. the concentrations of the various reacting substances are prescribed at time $t=t_0$. In the case of an ecological system the populations of the species at time $t=t_0$ is specified We then seek the solution $y(t)$ (the values of the concentrations of the reacting substances or the population of the species) for times in a certain interval $1=\left(t_0-a, t_0+a\right)$.
- Warning: Even if the differential equation is defined everywhere the interval I on which the solution makes sense may be limited. Let us see this in a noteworthy special case.

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<footer style = "font-size: 18px">The variable separable equation $\rightarrow$ The variable separable equation $\rightarrow$ <span style = "color:blue"> Initial conditions </span> $\rightarrow$ Solution escapes to infinity in finite time $\rightarrow$ Solution escapes to infinity in finite time </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Solution escapes to infinity in finite time </primary>
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- 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 !) $y(t)=-1 /(t+b)$. 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 left, the solution $y(t)$ escapes to infinity !

- $\frac{d y}{d t}=y^2, \quad y(0)=c$   
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<footer style = "font-size: 18px">The variable separable equation $\rightarrow$ Initial conditions $\rightarrow$ <span style = "color:blue"> Solution escapes to infinity in finite time </span> $\rightarrow$ Solution escapes to infinity in finite time $\rightarrow$ Solution escapes to infinity in finite time </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Solution escapes to infinity in finite time </primary>
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- where, $b$ is the constant of integration.

- $y(t)=\frac{-1}{t+b}$

- Put in the initial conditions: When $t=0, y(0)=c$. We get

- $c=\frac{-1}{b}, \quad \text { or } b=\frac{-1}{c}$

- Feedback into the previous equation:

- $y(t)=\frac{-1}{t+b}=\frac{-1}{t-\frac{1}{c}}$

- Solution of the differential equation is
- $y(t)=\frac{c}{1-c t}$   
- Notice what happens when $ t \rightarrow \frac{1}{c}-$ ?

- The RHS $c /(1-c t)$ blows up to infinity!
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<footer style = "font-size: 18px">Solution escapes to infinity in finite time $\rightarrow$ Solution escapes to infinity in finite time $\rightarrow$ <span style = "color:blue"> Solution escapes to infinity in finite time </span> $\rightarrow$ Solution escapes to infinity in finite time $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Differential Equations L-1 </Success>
### <primary style="font-size:24px"> Solution escapes to infinity in finite time </primary>
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- 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 !) $y(t)=-1 /(t+b)$. 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 left, the solution $y(t)$ escapes to infinity !
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