The solution of 25 frac{d^{2} y}{d x^{2}}-10 | vedclass

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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

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