Solve the second order differential equations, Mathematics

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Solve the subsequent IVP

Y'' - 9 y = 0, y(0) = 2, y'(0) = -1

Solution

First, the two functions

 y (t ) = e3t  and  y(t ) = e-3t

That is "nice enough" for us to by the general solution to the differential equation. At this point, please only believe this. You will be capable to verify this for yourself in a couple of sections.

The general solution to our differential equation is now,

y (t ) = c1 e-3t  + c2e3t

Here all we require to do is apply the initial conditions. It means that we require the derivative of the solution.

y' (t ) = -3 e-3t  + 3e3t

Plug in the initial conditions

2 = y (0) = c1 + c2

-1 = y'(0) = -3 c1 +3 c2

This provides us a system of two equations and two unknowns which can be solved.  Doing this yields

c1 = 7/6, c2 = 5/6

The solution to the IVP is so,

 y(t) = (7/6)e-3t + (5/6) e3t

Up to such point we've only looked at a particular differential equation and we found its solution by inspection. For rare little differential equations we can do this. Though, for the huge majority of the second order differential equations out there we will be not capable to do this.

Therefore, we would like a method for arriving at the two solutions we will require so as to form a general solution which will work for any linear, second order differential equation and constant coefficient equation. It is easier than it might initially look.

We will utilize the solutions we found in the first illustration as a guide. Each of the solutions in this illustration were in the form

y (t ) = ert

Remember, that we didn't include a constant in front of it as we can literally comprise any constant which we want and still get a solution. The significant idea here is to find the exponential function. One time we have which we can add on constants to our hearts content.

Thus, let's suppose that all solutions to,

ay′′ + by′ + cy = 0   ...................(4)

It will be of the form as

y (t ) = e rt        ........................(5)

 

To notice if we are correct all we require to do is plug this in the differential equation and notice what occurs.  Thus, let's get several derivatives and after that plug in.

y′ (t )= rert

 y′′ (t ) = r2ert

 a (r2ert )+ b (rert )+ c (ert ) = 0

ert (ar2 + br + c )= 0

Therefore, if (5) is to be a solution to (4) then the subsequent must be true

ert (ar2 + br + c ) = 0

It can be decreased further by noting as exponentials are never zero. Thus, (5) will be a solution to (4) given r is a solution to

 ar2 + br + c = 0    .......... (6)

 This equation is usually termed as the characteristic equation for (4).

Okay, then how do we use this to get solutions to a linear, constant coefficient, second order differential equation? First write down the feature equation, (6), for the differential equation, (4). It will be a quadratic equation and thus we must expect two roots, r1 and r2. One time we have these two roots we have two solutions to the differential equation.

1538_Solve the SECOND ORDER DIFFERENTIAL EQUATIONS.png...............(7)


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