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.