The general solution of the differential equa | vedclass

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

Question

(C) The given differential equation is $100 \frac{d^2 y}{dx^2}-20 \frac{dy}{dx}+y=0$. To find the general solution, we write the auxiliary equation as

Vedclass · Accepted Answer

$y=(c_1+c_2 x) e^{\frac{x}{10}}$

Vedclass · Answer

(C) The given differential equation is $100 \frac{d^2 y}{dx^2}-20 \frac{dy}{dx}+y=0$. To find the general solution, we write the auxiliary equation as $100 m^2 - 20 m + 1 = 0$. This can be factored as $(10 m - 1)^2 = 0$. Solving for $m$, we get $m = \frac{1}{10}$ as a repeated root. For a second-order linear differential equation with a repeated root $m$, the general solution is given by $y = (c_1 + c_2 x) e^{mx}$. Substituting $m = \frac{1}{10}$, we get $y = (c_1 + c_2 x) e^{\frac{x}{10}}$.

The general solution of the differential equation $100 \frac{d^2 y}{dx^2}-20 \frac{dy}{dx}+y=0$ is

Similar Questions

The solution of the differential equation $x \frac{dy}{dx} + y = x^3 y^6$ is:

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

The solution of $\frac{dy}{dx} + y = e^x$ is

If $\int\limits_a^x {t\,y(t)dt} = x^2 + y(x)$,then $y$ as a function of $x$ is

The general solution of the differential equation $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module