Solution:
The given differential equation is,
[tex]y=C_{1}e^{4 x}+C_{2}e^{-x}[/tex]------(A)
Differentiating once,with respect to x,
[tex]y'=4C_{1}e^{4 x}-C_{2}e^{-x}[/tex]-------(1)
Differentiating again with respect to x,
[tex]y"=16C_{1}e^{4 x}+C_{2}e^{-x}[/tex]-------(2)
Equation (1) + Equation (2)
y' +y" [tex]=20 C_{1}e^{4 x}[/tex]
[tex]C_{1}=\frac{y'+y"}{20e^{4 x}}[/tex]
4 ×Equation (1) - Equation (2)
4 y'- y"[tex]=-5 C_{2}e^{-x}[/tex]
[tex]C_{2}=\frac{4y'-y"}{-5e^{-x}}[/tex]
Substituting the value of [tex]C_{1},C_{2}[/tex] in A,we get
[tex]y=\frac{y'+y"}{20}+\frac{4 y'-y"}{-5}\\\\ 20 y=y'+y"-16 y'+4 y"\\\\ 20 y=-15 y'+5y"\\\\ 4 y+3 y'-y"=0[/tex]
As, y(0)=1 , and y'(0)=2, gives
[tex]C_{1}+C_{2}=1\\\\ 4C_{1}-C_{2}=2[/tex]
gives , [tex]5C_{1}=3\\\\ C_{1}=\frac{3}{5}\\\\ 5 C_{2}=2\\\\ C_{2}=\frac{2}{5}[/tex]
So, member of the family that is a solution of the initial-value problem, [tex]y=C_{1}e^{4 x}+C_{2}e^{-x}[/tex] is
[tex]5 y=3 e^{4 x}+2 e^{-x}[/tex]
