y=A e^x+B e^{2 x}+C e^{3 x} satisfies the dif | vedclass

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(A) Given that, $y = A e^x + B e^{2x} + C e^{3x}$ ... $(i)$ Differentiating with respect to $x$, we get: $y^{\prime} = A e^x + 2B e^{2x} + 3C e^{3x}$

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$y^{\prime \prime \prime}-6 y^{\prime \prime}+11 y^{\prime}-6 y=0$

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(A) Given that, $y = A e^x + B e^{2x} + C e^{3x}$ ... $(i)$ Differentiating with respect to $x$, we get: $y^{\prime} = A e^x + 2B e^{2x} + 3C e^{3x}$ ... (ii) Differentiating again with respect to $x$, we get: $y^{\prime \prime} = A e^x + 4B e^{2x} + 9C e^{3x}$ ... (iii) Differentiating again with respect to $x$, we get: $y^{\prime \prime \prime} = A e^x + 8B e^{2x} + 27C e^{3x}$ ... (iv) The given equation is a linear combination of $e^x, e^{2x}, e^{3x}$. The characteristic equation is $(m-1)(m-2)(m-3) = 0$. Expanding this, we get: $(m^2 - 3m + 2)(m - 3) = 0$ $m^3 - 3m^2 - 3m^2 + 9m + 2m - 6 = 0$ $m^3 - 6m^2 + 11m - 6 = 0$ Replacing $m^k$ with $y^{(k)}$, we get the differential equation: $y^{\prime \prime \prime} - 6y^{\prime \prime} + 11y^{\prime} - 6y = 0$.

$y=A e^x+B e^{2 x}+C e^{3 x}$ satisfies the differential equation

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