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

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(A) Given,$y = A e^x + B e^{2x} + C e^{3x} \quad \dots(i)$Differentiating with respect to $x$:$y' = A e^x + 2B e^{2x} + 3C e^{3x} \quad \dots(ii)$Diff

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

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(A) Given,$y = A e^x + B e^{2x} + C e^{3x} \quad \dots(i)$Differentiating with respect to $x$:$y' = A e^x + 2B e^{2x} + 3C e^{3x} \quad \dots(ii)$Differentiating again with respect to $x$:$y'' = A e^x + 4B e^{2x} + 9C e^{3x} \quad \dots(iii)$Differentiating again with respect to $x$:$y''' = A e^x + 8B e^{2x} + 27C e^{3x} \quad \dots(iv)$Alternatively,since the roots of the auxiliary equation are $m = 1, 2, 3$,the characteristic equation is $(m-1)(m-2)(m-3) = 0$.$(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:$y''' - 6y'' + 11y' - 6y = 0$.

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

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