If y=e^{4x}+2e^{-x} satisfies the equation fr | vedclass

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

Question

(B) Given the function $y = e^{4x} + 2e^{-x}$.First,find the first derivative: $\frac{dy}{dx} = 4e^{4x} - 2e^{-x}$.Next,find the second derivative: $\

Vedclass · Accepted Answer

$-3, -4$

Vedclass · Answer

(B) Given the function $y = e^{4x} + 2e^{-x}$.First,find the first derivative: $\frac{dy}{dx} = 4e^{4x} - 2e^{-x}$.Next,find the second derivative: $\frac{d^2y}{dx^2} = 16e^{4x} + 2e^{-x}$.Substitute these into the given equation $\frac{d^2y}{dx^2} + A\frac{dy}{dx} + By = 0$:$(16e^{4x} + 2e^{-x}) + A(4e^{4x} - 2e^{-x}) + B(e^{4x} + 2e^{-x}) = 0$.Group the terms by $e^{4x}$ and $e^{-x}$:$(16 + 4A + B)e^{4x} + (2 - 2A + 2B)e^{-x} = 0$.For this to hold for all $x$,the coefficients must be zero:$16 + 4A + B = 0$ (Equation $1$)$2 - 2A + 2B = 0 \Rightarrow 1 - A + B = 0 \Rightarrow B = A - 1$ (Equation $2$)Substitute $B = A - 1$ into Equation $1$:$16 + 4A + (A - 1) = 0 \Rightarrow 5A + 15 = 0 \Rightarrow A = -3$.Then $B = -3 - 1 = -4$.Thus,$A = -3$ and $B = -4$.

If $y=e^{4x}+2e^{-x}$ satisfies the equation $\frac{d^2y}{dx^2}+A\frac{dy}{dx}+By=0$,then the values of $A$ and $B$ are respectively:

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 of the simple harmonic motion given by $x=A \cos (n t+\alpha)$ is

The differential equation of all parabolas whose axis is the $y$-axis is:

The differential equation of the family of circles touching the $y$-axis at the origin is

The differential equation of the family of circles touching $y$-axis at the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module