Question

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

• A. $y=e^{5x}+e^{-5x}$
• B. $y=\left(1+x\right)e^{5x}$
• C. $y=\left(1+x\right)e^{\frac{x}{5}}$
• D. $y=\left(1+x\right)e^{\frac{-x}{5}}$

Answer 1

Answer: (C)

Let $y=e^{mx}$ be the solution of given differential equation,
$\Rightarrow \frac{dy}{dx}=m{e}^{mx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=m^2{e}^{mx}$
$\therefore 25\frac{d^{2}y}{d{x}^2}-10\frac{dy}{dx}+y=0$
$\Rightarrow 25m^2{e}^{mx}-10m{e}^{mx}+e^{mx}=0$
$\Rightarrow e^{mx}\left(25m^2-10m+1\right)=0$
$\Rightarrow$ Auxiliary equation
$\Rightarrow 25m^2-10m+1=0$
$e^{mx}\ne 0$
$\Rightarrow (5m{)}^2-2(5m)\times 1+1=0$
$\Rightarrow (5m-1{)}^2=0$
$\Rightarrow m=\frac{1}{5},\frac{1}{5}$
Since, roots are real and equal.
$\therefore$ General solution is $y=\left(c_1+c_{2}x\right)e^{x/5}$ ... (i)
$y(0)=1\Rightarrow c_1=1$
$y(1)=2e^{1/5}\Rightarrow 2e^{1/5}=\left(c_1+c_2\right)e^{1/5}$
$\Rightarrow c_1+c_2=2$
$\Rightarrow c_1=1$
Putting the value of $c_1$ and $c_2$ in Eq. (i), we get particular solution
$y=(1+x)e^{x/5}$