Question

The given family of functions is the general solution of the differential equation on the indicated interval. find a member of the family that is a solution of the initial-value problem. y = c1e4x + c2e−x, (−∞, ∞); y'' − 3y' − 4y = 0, y(0) = 1, y'(0) = 2

Answer 1

Solution:

The given differential equation is,

$y=C_{1}e^{4 x}+C_{2}e^{-x}$------(A)

Differentiating once,with respect to x,

$y'=4C_{1}e^{4 x}-C_{2}e^{-x}$-------(1)

Differentiating again with respect to x,

$y"=16C_{1}e^{4 x}+C_{2}e^{-x}$-------(2)

Equation (1) + Equation (2)

y' +y" $=20 C_{1}e^{4 x}$

$C_{1}=\frac{y'+y"}{20e^{4 x}}$

4 ×Equation (1) - Equation (2)

4 y'- y"$=-5 C_{2}e^{-x}$

$C_{2}=\frac{4y'-y"}{-5e^{-x}}$

Substituting the value of $C_{1},C_{2}$ in A,we get

$y=\frac{y'+y"}{20}+\frac{4 y'-y"}{-5}\\\\ 20 y=y'+y"-16 y'+4 y"\\\\ 20 y=-15 y'+5y"\\\\ 4 y+3 y'-y"=0$

As, y(0)=1 , and y'(0)=2, gives

$C_{1}+C_{2}=1\\\\ 4C_{1}-C_{2}=2$

gives , $5C_{1}=3\\\\ C_{1}=\frac{3}{5}\\\\ 5 C_{2}=2\\\\ C_{2}=\frac{2}{5}$

So, member of the family that is a solution of the initial-value problem, $y=C_{1}e^{4 x}+C_{2}e^{-x}$ is

$5 y=3 e^{4 x}+2 e^{-x}$

Answer 2

Try this option (see the attachment), if it is possible check result in other sources.