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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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1+1=
9:59-10:45
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