Second Order first degree differential equations with constant coefficients: Solved Example Problems

Example 4.25

Solve (D ² − 3D − 4)y = 0

Solution:

Given          (D ² − 3D − 4)y = 0

The auxiliary equations is

m² − 3m − 4 = 0

( m − 4)( m + 1) = 0

m = −1, 4

Roots are real and different

∴ The complementary function is Ae^−x + Be^{4 x}

The general solution is y = Ae^−x + Be^{4 x}

Example 4.26

Solve 9 y ′′ − 12 y′ + 4 y = 0

Solution:

Given          ( 9D² − 12D + 4) y         = 0                                

The auxiliary equation is

 ( 3m – 2)²   = 0

(3m – 2) ( 3m – 2)          = 0    ⇒ m = 2/3,2/3

Roots are real and equal

The C.F. is ( Ax + B )e ^2x/3

The general solution is y = ( Ax + B )e ^2x/3

Type II : f ( x ) = e ^ax (i.e ) f( D ) y = e^ax

Example 4.32

Suppose that the quantity demanded Q_d =  and quantity supplied Q_s = 5 + 4 p where p is the price. Find the equilibrium price for market clearance.

Solution: