Application Of Linear Differential Equation Final

Differential Equation

ASLAM U ALIKUM

Application of linear Differential Equation By Shahid bashir 08105012 Umair munir 08105037 Umar maqsood 08105002 CH.Usman ali 08105022 Nouman dilawar 08105035

After presentation Should Be Clear   

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Linear differential equation General form of linear equation Application of linear differential equation Growth and decay Half life Mixture problem Series circuit Cooling Carbon dating Falling Object

What is linear differential equation? An equation involving derivatives in which the dependent variable an all derivatives appearing in the equation are raised to the first power. A linear differential equation of the form dy/dx +p(x)y=f(x) Is said to be linear differential equation

OR

Linear Differential Equations 

A first-order differential equation is said to be linear if, in it, the unknown function y and its derivative y' appear with non-negative integral index not greater than one and not as product yy' either. Hence, The most general first order linear differential equation is given by:dx /dt + p(t)y = q(t)

(1)

where p and q are given real-valued function in 1.

General form of linear equation

General form of linear diffrential equation of order n to be An(x)dny/dxn+ an-1(x)dn-1 /dxn-1+………+a1(x)dy/dx+a0(x)y= g(x)

Applications of linear Differential Equations

Growth And Decay The initial value problem dx/dt=kx, x(to)=xo, There k is a constant of proportionality occur in many physical theories involving either growth and decay. For example: In biology it is often observed that the rate at which certain bacteria grow is proportional to the number of bacteria present in any time.

Find Exponential GrowthPopulation Let P(t) be a quantity that increases with time t and the rate of increase is proportional to the same quantity P as follows dP/dt=kP where d p/d t is the first derivative of P, k>0 and t is the time. The solution to the above first order differential equation is given by P(t) = A ekt where A is a constant not equal to 0. If P = P0 at t = 0, then P 0 = A e0 which gives A = P0 The final form of the solution is given by P(t) = P0 ekt

Half life 

In physics the half life is a measure of stability of a radio activate substance. The half life is simply the time it takes for one half of the atoms in initial amount A0 is disintegrate or transmute into the atoms of another element, the longer the half life of a substance the more stable it is,

For example:The half life is highly radioactive radium Ra 266 is about 1700 years.

Half life The most commonly occurring uranium isotopes U-238 has a half life of approximately 4,500,000,000 years. in about 4.5billion years one half of a quantity of U -238 is transmuted in to lead, pb-206..

Find Exponential Decay Radioactive Material 

Let M(t) be the amount of a product that decreases with time t and the rate of decrease is proportional to the amount M as follows dM/dt=-kM where d M / d t is the first derivative of M, k > 0 and t is the time. Solve the above first order differential equation to obtain M(t) = Ae -kt where A is non zero constant. It we assume that M = M0 at t = 0, then M0 = A e0 which gives A = M0 The solution may be written as follows M(t) = M0 e-

kt

Find Exponential Decay Radioactive Material Assuming M0 is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M is also called an exponential decay model.

Carbon dating The theory of carbon is based on the fact that the isotopes carbon 14 is produced in the atmosphere by the actions of cosmic radiations on nitrogen. The ratio of the amount of C-14 to ordinary carbon in the atmosphere appears to be a constant, and as a consequence the proportionate amount of isotope present in a all living organisms is the same as that in the atmosphere. when an organism dies, the absorption of C- 14 , by either breathing or eating ceases. Thus by comparing the proportionate amount of C-14 present, say, in a fossil with a constant ratio found In the atmosphere, it is possible to obtain a reasonable estimation of it’s age.

Carbon dating

The method is based on the knowledge that the half life of the radioactive C- 14 is approximately 5600 years.

COOLING 

Newton’s law of cooling states that the rate at which the temperature T (t) changes in a cooling body is proportional to the difference between the temperature in the body and the constant temperature Tm of the surrounding medium that is dT/dt=k(T-Tm) Where k is constant of proportionality.

Newton's Law of Cooling It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object. d T / d t = - k (T - Te) Let x = T - Te so that dx / dt = dT / dt Using the above change of variable, the above differential equation becomes dx/dt=-kx The solution to the above differential equation is given by x=Ae-kt

Newton's Law of Cooling substitute x by T - Te T - Te = A e - k t Assume that at t = 0 the temperature T = To T0 - Te = A e

0

which gives A = To - Te The final expression for T(t) i given by T(t) = Te + (T0 - Te)e - k t This last expression shows how the temperature T of the object changes with time.

SERIES CIRCUITS 

In a series circuit containing only a resister and an inductor, Kirchof’s second law states that the sum of the voltage drop across the inductor (L(di/dt)) and the voltage drop across the resister (iR) is the same as the impressed voltage (E(t)) on the circuit. Thus we obtain the linear equation for the current i(t) L di/dt + Ri = E (t)

Where L and R are constants known as the inductance and the resistance respectively, the current i(t) is sometimes called the response of the system.

RL Circuit

RL circuit Let us consider the RL (resistor R and inductor L) circuit shown above. At t = 0 the switch is closed and current passes through the circuit. Electricty laws state that the voltage across a resistor of resistance R is equal to R i and the voltage across an inductor L is given by L di/dt (i is the current). Another law gives an equation relating all voltages in the above circuit as follows: L di/dt + Ri = E , where E is a constant voltage. Let us solve the above differential equation which may be written as follows L [ di / dt ] / [E - R i] = 1

RL Circuit Which may be written as - (L / R) [ - R d i ] / [E - Ri] = dt Integrate both sides - (L / R) ln(E - R i) = t + c , c constant of integration. Find constant c by setting i = 0 at t = 0 (when switch is closed) which gives c = (-L / R) ln(E) Substitute c in the solution - (L / R) ln(E - R i) = t + (-L/R) ln (E) which may be written (L/R) ln (E)- (L / R) ln(E - R i) = t ln[E/(E - Ri)] = t(R/L)

RL circuit Change into exponential form [E/(E - Ri)] = et(R/L) Solve for i to obtain i = (E/R) (1-e-Rt/L) The starting model for the circuit is a differential equation which when solved, gives an expression of the current in the circuit as a function of time.

Mixture problem 

In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t). Note as well that in many situations we can think of air as a liquid for the purposes of these kinds of discussions and so we don’t actually need to have an actual liquid, but could instead use air as the “liquid”.

Mixture problem 

The main assumption that we’ll be using here is that the concentration of the substance in the liquid is uniform throughout the tank. The main “equation” that we’ll be using to model this situation is

Mixture problem Rate at Rate of = which Q(t) change of enters the Q(t) tank

Rate at - which Q(t) exits the tank

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering) Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Mixture problem Consider a tank which initially holds V0 gal of brine that contains a lb of salt. Another solution, containing b lb of salt per gallon, is poured into the tank at the rate of e gal/min while simultaneously, the well-stirred solution leaves the tank at the rate of f gal/min (Figure 3-2). The problem is to find the amount of salt in the tank at any time t. Let Q denote the amount (in pounds) of salt in the tank at any time. The time rate of change of Q, dQ/ dt, equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume V0

Mixture problem 

plus the volume of brine added et minus the volume of brine removed ft. Thus, the volume of brine at any time is

Mixture problem e gal/min

f gal/min

Mixture problem V0+et−ft  The concentration of salt in the tank at any time is Q / (V0+et−ft), from which it follows that salt leaves the tank at the rate of F(Q/V0+et−ft) 1b/min Thus, dQ/dt=be-F(Q/V0+et−ft) OR dQ/dt+F(Q/(V0+(e−f)t)=be

Falling Object 

An object is dropped from a height at time t = 0. If h(t) is the height of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows: a(t) = dv / dt , v(t) = dh / dt. For a falling object, a(t) is constant and is equal to g = -9.8 m/s. Combining the above differential equations, we can easily deduce the following equation d 2h / dt 2 = g

Falling Object Integrate both sides of the above equation to obtain dh / dt = g t + v0

Falling Object 

Integrate one more time to obtain h(t) = (1/2) g t + v0 t + h0

The above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.

Find Falling Object An object is dropped from a height at time t = 0. If h(t) is the height of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows: a(t) = dv / dt , v(t) = dh / dt. For a falling object, a(t) is constant and is equal to g = -9.8 m/s. Combining the above differential equations, we can easily deduce the following equation d 2h / dt 2 = g Integrate both sides of the above equation to obtain dh / dt = g t + v0

Find Falling Object Integrate one more time to obtain h(t) = (1/2) g t + v0 t + h0 The above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.