Second order differential equation, Mathematics

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Write the subsequent 2nd order differential equation as a system of first order, linear differential equations.

2 y′′ - 5 y′ + y = 0

 y (3) = 6

 y′ (3) = -1

 We can write higher order differential equations like a system with a extremely simple change of variable. We'll begin with defining the following two new functions.

x1 (t )= y (t)

x2 (t ) = y′ (t)

Now see that if we differentiate both sides of these we determine,

x1' = y' = x2

x2' = y'' = -(1/2)y + (5/2) y' = -(1/2)x1 + (5/2)x2

remember the use of the differential equation in the second equation. We can also change the initial conditions in excess of to the new functions.

x1 (3)= y (3) = 6

x2 (3 ) = y′ (3) = -1

Putting all of this together provides the subsequent system of differential equations.

x1' =  x2                                                                        x1 (3)= 6

x2' = -(1/2)x1 + (5/2)x2             x2(3) = -1

We will call such system in the above illustration an Initial Value Problem just as we done for differential equations with initial conditions.


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