The solution of 25 frac{d^{2} y}{d x^{2}}-10 | vedclass

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

Question

(C) The given differential equation is $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$.The auxiliary equation is $25 m^{2}-10 m+1=0$.This can be w

Vedclass · Accepted Answer

$y=(1+x) e^{x / 5}$

Vedclass · Answer

(C) The given differential equation is $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$.The auxiliary equation is $25 m^{2}-10 m+1=0$.This can be written as $(5 m-1)^{2}=0$, which gives $m=\frac{1}{5}, \frac{1}{5}$.Since the roots are real and equal, the general solution is $y=(c_{1}+c_{2} x) e^{x / 5} \quad \dots(i)$.Given $y(0)=1$, substituting $x=0$ in $(i)$ gives $1=(c_{1}+0) e^{0} \Rightarrow c_{1}=1$.Given $y(1)=2 e^{1 / 5}$, substituting $x=1$ and $c_{1}=1$ in $(i)$ gives $2 e^{1 / 5}=(1+c_{2}) e^{1 / 5}$.Dividing by $e^{1 / 5}$, we get $2=1+c_{2} \Rightarrow c_{2}=1$.Substituting $c_{1}=1$ and $c_{2}=1$ in $(i)$, the particular solution is $y=(1+x) e^{x / 5}$.

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

Similar Questions

If the curve $y=y(x)$ is the solution of the differential equation $2(x^{2}+x^{5/4}) dy - y(x+x^{1/4}) dx = 2x^{9/4} dx, x > 0$ which passes through the point $(1, 1-\frac{4}{3} \log_{e} 2)$,then the value of $y(16)$ is equal to :

The integrating factor of $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$ is

If $\cos x \frac{dy}{dx} - y \sin x = 6x$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then find $y(\frac{\pi}{6})$.

Let the solution curve $y=y(x)$ of the differential equation $(1+e^{2x})(\frac{dy}{dx}+y)=1$ pass through the point $(0, \frac{\pi}{2})$. Then,$\lim_{x \rightarrow \infty} e^{x} y(x)$ is equal to.

The solution of $(y-3 x^2) d x+x d y=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module