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fenix001 [56]
3 years ago
7

Determine if -10,-12,or -16is the solution of 2t-2=26

Mathematics
1 answer:
Dafna11 [192]3 years ago
3 0
2t-2=26
first you would add 2 to both sides
that equals 2t=28
then you would divide both sides by two
that would give you
t= 14
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Step-by-step explanation:

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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
SCORPION-xisa [38]

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}

8 0
3 years ago
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