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

The Correct Option is C

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