kindly help me with my homework. eliminate the following arbitrary constants... y= x^2 + c1 e^3x + c2 e^3x.
Are you sure that the power of e is the same for the 2nd and 3rd terms? Both are e3x.
In case if you mean y = x^2 + c1 e^2x + c2 e^3x: $y = x^2 + c_1 e^{2x} + c_2 e^{3x}$ ← Equation (1)
$y' = 2x + 2c_1 e^{2x} + 3c_2 e^{3x}$ ← Equation (2)
$y'' = 2 + 4c_1 e^{2x} + 9c_2 e^{3x}$ ← Equation (3)
Equation (2) - 3 × Equation (1) $y' - y = (2x - 3x^2) - c_1 e^{2x}$ ← Equation (4)
Equation (3) - 3 × Equation (2) $y'' - y' = (2 - 6x) - 2c_1 e^{2x}$ ← Equation (5)
Equation (5) - 2 × Equation (4) $(y'' - y') - 2(y' - y) = (2 - 6x) - 2(2x - 3x^2)$ collect similar terms to get to your final answer.
thank you so much :)
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Are you sure that the power of e is the same for the 2nd and 3rd terms? Both are e3x.
In case if you mean y = x^2 + c1 e^2x + c2 e^3x:
$y = x^2 + c_1 e^{2x} + c_2 e^{3x}$ ← Equation (1)
$y' = 2x + 2c_1 e^{2x} + 3c_2 e^{3x}$ ← Equation (2)
$y'' = 2 + 4c_1 e^{2x} + 9c_2 e^{3x}$ ← Equation (3)
Equation (2) - 3 × Equation (1)
$y' - y = (2x - 3x^2) - c_1 e^{2x}$ ← Equation (4)
Equation (3) - 3 × Equation (2)
$y'' - y' = (2 - 6x) - 2c_1 e^{2x}$ ← Equation (5)
Equation (5) - 2 × Equation (4)
$(y'' - y') - 2(y' - y) = (2 - 6x) - 2(2x - 3x^2)$ collect similar terms to get to your final answer.
thank you so much :)
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