The solution of 25d²y/dx²-10dy/dx+y=0, y(0)=1, y(1)=2 e¹/5} is

Correct Answer is : y=(1+x)ex5y=(1+x) e^{x/5}y=(1+x)e5x

Correct Answer is : y=(1+x)ex5y=(1+x) e^{x/5}y=(1+x)e5x

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WBJEE 2012 Math Solved Paper

Question : 53 of 80

Marks: +1, -0

25\frac{d^2y}{dx^2}-10\frac{dy}{dx}+y=0,

y(0)=1, y(1)=2 e^{\frac{1}{5}}

Solution:

Let

y=e^{mx}

be the solution of given differential equation,

\Rightarrow\;\frac{dy}{dx}=m e^{m x} \Rightarrow\;\frac{d^{2}y}{dx^{2}}=m^{2} e^{m x}

\therefore\;25\frac{d^{2}y}{dx^{2}}-10\frac{dy}{dx}+y=0

\Rightarrow\;25 m^{2} e^{m x}-10 m e^{m x}+e^{m x}=0

\Rightarrow\;e^{mx}(25 m^{2}-10 m+1)=0

⇒ Auxiliary equation

\Rightarrow 25 m^{2}-10 m+1=0

e^{\pi 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

∴ General solution is

y=(c_{1}+c_{2} x) e^{\frac{x}{5}}

...(i)

y(0)=1 \Rightarrow c_{1}=1

y(1)=2 e^{\frac{1}{5}} \Rightarrow 2 e^{\frac{1}{5}}=(c_{1}+c_{2}) e^{\frac{1}{5}}

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^{\frac{x}{5}}

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