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 integrating factor of the differential equation $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$ is

The solution of $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$, $y(0)=1, y(1)=2 e^{1 / 5}$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$ is

If the form of the solution of the differential equation $(y^3+x) \frac{dy}{dx} = y$ with the condition $y(4) = 2$ is $y^3 = ax + b$,then $4a + 12b^2 = $

The general solution of the differential equation $x^{2} dy - 2xy dx = x^{4} \cos x dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module