Question
Form the differential equation by eliminating arbitrary constants from the relation
$y= Ae ^{ 5r }+ B e ^{- 5r \text {. }}$
$y= Ae ^{ 5r }+ B e ^{- 5r \text {. }}$
$y=A e^{5 x}+B e^{-5 x}$
Differentitating w.r.t. x
$\begin{aligned} & \frac{d y}{d x}=A \cdot e^{5 x} \cdot(5)+B e^{-5 x}(-5) \\ & \therefore \frac{d y}{d x}=5 A \cdot e^{5 x}-5 B e^{-5 x}\end{aligned}$
Again differentitating w.r.t. x
$\begin{aligned} & \frac{d^2 y}{d x^2}=5 A e^{5 x} \cdot(5)-5 B e^{-5 x} \cdot(-5) \\ & \frac{d^2 y}{d x^2}=25 A e^{5 x}+25 B e^{-5 x} \\ & \frac{d^2 y}{d x^2}=25\left(A e^{5 x}+B e^{-5 x}\right) \\ & \frac{d^2 y}{d x^2}=25 y \\ & \frac{d^2 y}{d x^2}-25 y=0 \text { is the required differential equation }\end{aligned}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\int_0^1 \log \left(\frac{1}{x}-1\right) \cdot d x$
$\int_0^{\pi / 4} \sin 4 x \sin 3 x \cdot d x$