Question

find a solution to r'(t)=2r(t) where r(t) is a vector valued function in three space

@ganeshie8 wheeeeeetttt?

Answer 1

try
\[r(t) = e^{2t}i + e^{2t}j+ e^{2t}k \]

Answer 2

interesting, you can get that by integrating componentwise
we have the vector valued function

r(t) = 2* r ' (t)

Let r(t) = < f(t) , g(t) , h(t) >
So r ' (t) = < f ' (t) , g ' (t) , h ' (t) >

So we have
< f ' (t) , g ' (t) , h ' (t) > = 2*< f(t) , g(t) , h(t) >

< f ' (t) , g ' (t) , h ' (t) > = < 2*f(t) , 2*g(t) , 2*h(t) >

Equating the first components we have

f ' (t) = 2*f(t)

f ' (t) / f(t) = 2

ln (f(t)) = 2t
f(t) =Ce^(2t)

similarly
g(t) = C*e^(2t)
h(t) = C*e^(2t)

then it asks for *a* solution, so set C = 1 .
but the general solution is

r (t) = < C*e^(2t) , C*e^(2t) , C*e^(2t) >

you can write this as a linear combination of vectors i,j,k

r(t) = C*e^(2t)*i + C*e^(2t)*j + C*e^(2t)* k

Answer 3

Thank you !

Answer 4

ok... wait..so am i just allowed to throw in the t next to the 2 and the C as well?

Answer 5

you have to plug in a number for C , a nonzero number

Answer 6

C = 1 is easiest

Answer 7

because it says find 'a' solution, singular

Answer 8

i was just showing you how to derive the general solution, fyi

Answer 9

oh i left out a step

Answer 10

we had
f ' (t) = 2*f(t)

f ' (t) / f(t) = 2

ln (f(t)) = 2t + c

raise both sides to e

f(t) =e^(2t + c)

by law of exponents

f(t) = e^(2t) * e^c

now e^constant is still a constant, so let e^c = C

f(t) = e^(2t) * C

f(t) = C*e^(2t)

Answer 11

you always have to justify wherever you place a constant (so i didn't just append a C from out of nowhere).