y=A e^x+B e^{-2 x} satisfies which of the fol | vedclass

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

Question

(D) Given the equation $y=A e^x+B e^{-2 x}$.First,differentiate with respect to $x$:$\frac{d y}{d x} = A e^x - 2B e^{-2x}$.Second,differentiate again

Vedclass · Accepted Answer

$\frac{d^2 y}{d x^2}+\frac{d y}{d x}-2 y=0$

Vedclass · Answer

(D) Given the equation $y=A e^x+B e^{-2 x}$.First,differentiate with respect to $x$:$\frac{d y}{d x} = A e^x - 2B e^{-2x}$.Second,differentiate again with respect to $x$:$\frac{d^2 y}{d x^2} = A e^x + 4B e^{-2x}$.Now,we eliminate the constants $A$ and $B$. From the first derivative,$A e^x = \frac{d y}{d x} + 2B e^{-2x}$.Substitute this into the second derivative equation:$\frac{d^2 y}{d x^2} = (\frac{d y}{d x} + 2B e^{-2x}) + 4B e^{-2x} = \frac{d y}{d x} + 6B e^{-2x}$.Alternatively,consider the characteristic equation for the roots $m_1 = 1$ and $m_2 = -2$. The differential equation is $(D-1)(D+2)y = 0$,where $D = \frac{d}{dx}$.Expanding this: $(D^2 + 2D - D - 2)y = 0$,which simplifies to $\frac{d^2 y}{d x^2} + \frac{d y}{d x} - 2y = 0$.

$y=A e^x+B e^{-2 x}$ satisfies which of the following differential equations?

Similar Questions

Let $c_1, c_2, c_3, c_4$ be arbitrary constants. The order of the differential equation,corresponding to $y=c_1 e^x+c_2 e^{\log _{e} x}+c_3 \sin ^2 x-c_4\left(\cos ^2 x-1\right)$ is

The differential equation whose solution is $y=e^{ax}$ is

Form the differential equation representing the family of parabolas having vertex at origin and axis along the positive direction of the $x$-axis.

Form the differential equation representing the family of curves $y=a \sin (x+b),$ where $a$ and $b$ are arbitrary constants.

$y = a{e^{mx}} + b{e^{ - mx}}$ satisfies which of the following differential equations?

Vedclass Test Series

Exam Paper Generator

Online Exam Module