Question

In this problem, y = c1ex + c2eâx is a two-parameter family of solutions of the second-order DE y'' â y = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. y(â1) = 3, y'(â1) = â3

Answer 1

Answer:

$y(t) = 2e^t -e^{-t}$

Step-by-step explanation:

Assuming this complete problem: "In this problem,

y = c1ex + c2e−x

is a two-parameter family of solutions of the second-order DE

y'' − y = 0.

Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

y(0) = 1,    y'(0)= 3"

Solution to the problem

For this case we have a homogenous, linear differential equation with order 2, and with the general form:

$ay'' +by' +cy=0$

Where $a =1, b=0, c=-1$

And we can rewrite the differential equation in terms $y = e^{rt}$ like this:

$[e^{rt}]'' -e^{rt}=0$

And applying the second derivate we got:

$r^2 e^{rt} -e^{rt}=0$

We can take common factor $e^{rt}$ and we got:

$e^{rt} (r^2-1) =0$

And for this case the two only possibel solutions are $r=1, r=-1$

And the general solution for this case is given by:

$y = c_1 e^{r_1 t} + c_2 e^{r_2 t}$

Replacing the roots that we found we got:

$y = c_1 e^{t} +c_2 e^{-t}$

Now we can find the derivates for this last espression

$y' = c_1 e^t -c_2 e^{-t}$

$y'' = c_1 e^t +c_2 e^{-t}$

From the initial conditions we have this:

$y(0)=1 =c_1 e^{0} +c_2 e^{-0}= c_1 +c_2$   (1)

$y'(0) =3= c_1 e^{0} -c_2e^{-0}= c_1 -c_2$   (2)

If we add equations (1) and (2) we got:

$4 = 2c_1 , c_1 = 2$

And solving for $c_2$ we got:

$c_2=3-c_1= 3-2 = 1$

So then our general solution is given by:

$y(t) = 2e^t -e^{-t}$