Differential Equation: Eliminate $C_1$, $C_2$, and $C_3$ from $y=C_1e^x+C_2e^{2x}+C_3e^{3x}$

EngineerInTheMaking's picture

How to eliminate arbitrary constants
y=C1e^x+C2e^2x+C3e^3x

thank you :)

Jhun Vert's picture

$y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}$   ←   Equation (1)

$y' = C_1 e^x + 2C_2 e^{2x} + 3C_3 e^{3x}$   ←   Equation (2)

$y'' = C_1 e^x + 4C_2 e^{2x} + 9C_3 e^{3x}$   ←   Equation (3)

$y''' = C_1 e^x + 8C_2 e^{2x} + 27C_3 e^{3x}$   ←   Equation (4)
 

Equation (2) - 3 × Equation (1)
$y' - 3y = -2C_1 e^x - C_2 e^{2x}$   ←   Equation (5)
 

Equation (3) - 3 × Equation (2)
$y'' - 3y' = -2C_1 e^x - 2C_2 e^{2x}$   ←   Equation (6)
 

Equation (4) - 3 × Equation (3)
$y''' - 3y'' = -2C_1 e^x - 4C_2 e^{2x}$   ←   Equation (7)
 

Equation (6) - 2 × Equation (5)
$(y'' - 3y') - 2(y' - 3y) = 2C_1 e^x$   ←   Equation (8)
 

Equation (7) - 2 × Equation (6)
$(y''' - 3y'') - 2(y'' - 3y') = 2C_1 e^x$   ←   Equation (9)
 

Equation (9) - Equation (8)
$[ \, (y''' - 3y'') - 2(y'' - 3y') \, ] - [ \, (y'' - 3y') - 2(y' - 3y) \, ] = 0$

$(y''' - 5y'' + 3y') - (y'' - 5y' + 3y) = 0$

$y''' - 6y'' + 8y' - 3y = 0$

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