Question

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

$y=\left(1+x\right)e^{\frac{x}{5}}$

Answer 2

The correct answer is C:$y=(1+x)^{e^{\frac{x}{5}}}$
Let $y=e^{m x}$ be the solution of given differential equation,
$\Rightarrow \frac{d y}{d x} =m e^{m x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2} e^{m x}$
$\therefore 25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$
$\Rightarrow 25 m^{2} e^{m x}-10 m e^{m x}+e^{m x}=0$
$\Rightarrow e^{m x}\left(25 m^{2}-10 m+1\right)=0$
$\Rightarrow$ Auxiliary equation
$\Rightarrow 25 m^{2}-10 m+1=0$
$e^{m x} \neq 0$
$\Rightarrow (5 m)^{2}-2(5 m) \times 1+1=0$
$\Rightarrow (5 m-1)^{2}=0$
$\Rightarrow m=\frac{1}{5}, \frac{1}{5}$
Since, roots are real and equal.
$\therefore$ General solution is $y=\left(c_{1}+c_{2} x\right) e^{x / 5}$ ... (i)
$y(0)=1 \Rightarrow c_{1}=1$
$y(1)= 2 e^{1 / 5} \Rightarrow 2 e^{1 / 5}=\left(c_{1}+c_{2}\right) e^{1 / 5}$
$\Rightarrow c_{1}+c_{2}=2$
$\Rightarrow c_{1}=1$
Putting the value of $c_{1}$ and $c_{2}$ in E (i), we get particular solution
$y=(1+x) e^{x / 5}$