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

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y = \frac{x+3}{x+1}$,where $x > -1$,which passes through the point $(0,1)$. Then $y(1)$ is equal to:

If $y+\frac{d}{d x}(x y)=x(\sin x+\log x)$,then find $y$.

The integrating factor of the first order differential equation $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$ is

The general solution of the differential equation $\frac{dy}{dx} + y g'(x) = g(x) g'(x)$ is

Let $y=y(x)$ be the solution of the differential equation $x dy = (y + x^3 \cos x) dx$ with $y(\pi) = 0$. Then $y(\frac{\pi}{2})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module