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- General Solution of $y' = x \, \ln x$
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- Special products and factoring
- Newton's Law of Cooling
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- Eliminate the Arbitrary Constants
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$y = C_1 e^x + C_2 e^{2x} + C
$y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}$ ← 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)
$[ \, (y''' - 3y'') - 2(y'' - 3y') \, ] - [ \, (y'' - 3y') - 2(y' - 3y) \, ] = 0$
$(y''' - 5y'' + 3y') - (y'' - 5y' + 3y) = 0$
$y''' - 6y'' + 8y' - 3y = 0$