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 e^{3x}.

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 :)

MATHalino

Are you sure that the power of

eis the same for the 2nd and 3rd terms? Both aree^{3x}.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 :)