Question

Help Please!
My last Homework Problem!

Answer 1

@zepdrix @IrishBoy123

Answer 2

which one?

Answer 3

Number 7

Answer 4

@StudyGurl14 @AloneS @Directrix

Answer 5

So you bascially want to take the integral twice. Is that what you did?

Answer 6

\[25y''+40y'+16y=0\]
with the Initial Values of y(0)=7, y'(0)=0

Answer 7

Um, no. what do you mean by take the integral twice?

Answer 8

The initial problem, and they show y'', meaning they had found the second derivative. They want you to basically undo that, meaning you need to take the integral twice.

Answer 9

\[25r^2+40r+16=0 \rightarrow (5r+4)^2 \rightarrow r=-\frac{ 4 }{ 5 }\]
\[y=C_1e^{\frac{ -4t }{ 5 }}+C_2te^{\frac{ -4t}{ 5 }}\]

Thats what i did.

Answer 10

I plugged it into to the formula.

Answer 11

Wait no, sorry. I think I misread the question. Hold on asec

Answer 12

@StudyGurl14 Okay

Answer 13

Hmm...I think this is outside my expertise..it looks familar but I'm not 100% sure. Sorry
@mathmate @mathstudent55 @zepdrix @Directrix

Answer 14

Okie.

Answer 15

@jabez177 @Elsa213

Answer 16

When i plug in the initial values, C_2 always becomes 0, and i cant find it.

Answer 17

you've got repeated roots

So you have a solution in

\(\large y=e^{\frac{ -4t }{ 5 }}( A t + B)\)

Answer 18

this really a linear algebra issue

i know it best as this, using the differential (linear) operator:

Answer 19

Well, isn't that what i technically did?

Answer 20

@iTz_Sid
Just a reminder, when you finish and put your answer, remember the independent variable is x, NOT t.
Also you have got B correctly, but you still need to find A using the condition y'(0)=0.

Answer 21

oh!! i'm sorry :-) i didn't read it properly, again.

yes, you did that. and you're plugging stuff into solved DE's. fine.

i'd follow @mathmate's counsel.

Answer 22

@mathmate @IrishBoy123

After plugging in in y(0) = 7
\[7=C_1e^0+C_2(0)e^0 \rightarrow C_1 = 7\]

Answer 23

yes

Answer 24

And then after taking the Derivative...
\[y' = (7)(\frac{ -4 }{ 5 }) e^{-4/5}+C_2\frac{ -4t }{ 5 }e^{-4/5} \rightarrow y' =(\frac{ -28 }{ 5 }) e^{-4/5}+C_2\frac{ -4t }{ 5 }e^{-4/5}\]

Answer 25

Oops. there is suppose to be a variable in the exponent.
\[y' = (\frac{ -28 }{ 5 }) e^{-4t/5}+C_2\frac{ -4t }{ 5 }e^{-4t/5}\]

Answer 26

And after plugging in y'(0)=0

\[0 =(\frac{ -28 }{ 5 }) e^0+C_2\frac{ -4(0) }{ 5 }e^{0}\]
\[0=\frac{ -28 }{ 5 }\]

Which makes no sense. See?

Answer 27

@Nnesha @zepdrix @IrishBoy123

Answer 28

Wait... i figured it out. I forgot to do product rule when taking the derivative...

Answer 29

👌

Answer 30

Good, so what do you get for C1 and C2?
Once you have C1 and C2, then the best next step is to verify that:
1. f(0)=7
2. f'(0)=0
3. 25y''+40y'+16y=0