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

**Example 4.25**

Solve (*D* ^{2} −
3*D* − 4)*y* = 0

*Solution:*

Given (*D* ^{2}
− 3*D* − 4)*y* = 0

The auxiliary equations is

*m*^{2}* *−* *3*m *−* *4* *= 0

(* m *−* *4)(* m *+* *1)* *= 0

*m *=* *−1,
4

Roots are real and different

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

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

**Example 4.26**

Solve 9 *y* ′′ − 12 *y*′ + 4 *y* =
0

*Solution:*

Given ( 9D^{2}
− 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:*

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12th Business Maths and Statistics : Chapter 4 : Differential Equations : Second Order first degree differential equations with constant coefficients: Solved Example Problems | with Answer, Solution, Formula

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