Question

Consider the differential equation give by

Answer 1

\[\left(\begin{matrix}dy \\ dx\end{matrix}\right)=\left(\begin{matrix}xy \\ 3\end{matrix}\right)\]

Answer 2

\[\frac{ dy }{ dx } = \frac{ xy }{ 3 }\]
this?

Answer 3

Yes, I'm sorry

Answer 4

Plug in the values into the equation and evaluate then..
You have (-1,1) for example.
\[\frac{ dy }{ dx } = \frac{ xy }{ 3 } = \frac{ (-1)(1) }{ 3 } = \frac{ -1 }{ 3 }\]

Answer 5

Make sense?

Answer 6

Yea, So I will do that for all of them? What about

Find the particular solution y=f(x) to the given differential equation with the initial condition f(0)=4

Answer 7

Yup just do that for all of those.

Answer 8

Is it using the same equation dy/dx = xy/3 ?

Answer 9

Yes it is

Answer 10

dy/dx = xy/3
dy 3 = xy dx
integrate both sides.

Answer 11

So that would be
y = ±√(x^2 + 1)

Answer 12

Actually wait my bad I set that up wrong..
\[\frac{ dy }{ dx } = \frac{ xy }{ 3 }\]
\[\frac{ 1 }{ y } dy = \frac{ x }{ 3 } dx\]
Then integrate both sides.

Answer 13

\[\int\limits_{}^{} \frac{ 1 }{ y } dy = \frac{ 1 }{ 3 } \int\limits_{}^{} x ~dx\]
Do the integration yourself and tell me what you get.

Answer 14

I'm not good with integration........ But
(1/2)1 = 0 + C
C = 1/2

Answer 15

No..

Answer 16

Then it would be
(1/2)y^2 = (1/2)x^2 + (1/2)
Right?

Answer 17

that's wrong.

Answer 18

Then how would you do it?

Answer 19

\[\int\limits_{}^{} \frac{ 1 }{ y } dy = \frac{ 1 }{ 3 } \int\limits_{}^{} x~dx\]
\[\ln(y) = \frac{ x^2 }{ 6 } + C\]
Find the general solution from here, I've done the integration, all you need to do is apply logarithm rules from here on...
f(0) = 4
Plug that in and you will see that C = 4, and then plug in C back into the general solution to find the particular solution.

Answer 20

Ok, so I will plug in the C but then could you also use the dy and dx equation for this part as well?

On the axes provided, sketch a slope field for the given differential equation at the 9 points on the table

Answer 21

I got 20/3 when you plug in 4 for c

Answer 22

All you do is graph the points from the first problem.

Answer 23

Those are the solutions to dy/dx ..

Answer 24

Alright, Thank you, about the graph portions.

Answer 25

I told you to use the logarithm rules to find y = f(x)
the ln(y) = x^2/6 + C
is NOT done..
you need to still solve for y.
THEN you plug f(0) in to solve for the constant C.

Answer 26

Ok that is when you have the C and E come into the equation right

Answer 27

yes, when you solve for y, "e" will be used in the equation.

Answer 28

y = x + C
f(0) = 0 + C

1 = C.

Answer 29

no....

Answer 30

No it is going to be
y = ℮^(x² / 6 + C)

Answer 31

yes

Answer 32

Leading the final part to be
f(x) = 4℮^(x² / 6)

Answer 33

yeah

Answer 34

Thank you for your help!!!!