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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The differential equation of the family of lines having $x$-intercept as $a$ and $y$-intercept as $b$ is:

$y = ax + b$ is

The order of the differential equation corresponding to the family of parabolas whose axes are along the $X$-axis and whose foci are at the origin,is

The differential equation of all circles,passing through the origin and having their centres on the $X$-axis,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module