Question

If dy/dx = 7x^2/y^3 and y(3)=2, find an equation for y in terms of x

Answer 1

\[\Large\bf\sf \frac{dy}{dx}=\frac{7x^2}{y^3},\qquad\qquad y(3)=2\]So this appears to be separable.
Do you understand how to get all of the y stuff on one side and all of the x stuff on the other? It will involve moving the differentials (namely dx) as well as the other stuff.

Answer 2

Is it y^3dy=7x^2dx? I don't know what to do after this

Answer 3

then you get the antiderivative?

Answer 4

Ok looks good so far.
Integrating each side,\[\Large\bf\sf \int\limits y^3\;dy\quad=\quad \int\limits 7x^2\; dx\]

Answer 5

Yes. Normal power rule on each side.

Answer 6

Integration Power Rule I mean*

Answer 7

y^4/4=7x^3/3?

Answer 8

Ok good, we have a +c somewhere also right?

Answer 9

yes, on the right side?

Answer 10

Yes, wherever you want it :)
Let's put it on the x side.

Answer 11

Let's also multiply both sides by 4, get the y^4 alone.\[\Large\bf\sf y^4\quad=\quad \frac{28}{3}x^3+c\]

Answer 12

can it also be written as y^4= 28x^3/3+C

Answer 13

Hehe yer too fast for me XD

Answer 14

Do you understand why it's c and not 4c?

Answer 15

And after that I am lost! C is a constant?

Answer 16

I mean because we multiplied both sides by 4.

Answer 17

Yes it's a constant, but it's the fact that it's an arbitrary value.
It represents a family of solutions.

It could be any number.
So 4 times any number is still any number.
We just absorb the 4 into the c.

If you already have a good understanding of that, then keep moving :D

Answer 18

We `could` solve for y at this point.
But I think we're in a good position to use our `initial data` to solve for c.

Answer 19

Oh yeah, I've learned that. I don't know what to do after that

Answer 20

What do we do with the y(3)=2?

Answer 21

\[\Large\bf\sf y^4\quad=\quad \frac{28}{3}x^3+\mathcal C\]
We plug it in.\[\Large\bf\sf y(3)=2,\qquad\qquad\to\qquad\qquad (2)^4\quad=\quad \frac{28}{3}(3)^3+\mathcal C\]Does that make sense how I plugged the x and y value in?

Answer 22

Make sure you're comfortable with function notation.
\(\Large\bf\sf y(3)=2 \quad\text{means}\qquad\qquad y=2\quad\text{when}\quad x=3\)

Answer 23

WOW. I can't believe I forgot about that

Answer 24

heh :o

Answer 25

We have to find C, right? C=-236

Answer 26

Mmm ya that looks right!

Answer 27

\[\Large\bf\sf y^4\quad=\quad \frac{28}{3}x^3-236\]

Answer 28

So how should we finish this up?
How do we write it terms of y=?

Answer 29

Do we have to write it in terms of y=? We cannot leave it like that?

Answer 30

`find an equation for y in terms of x`
Ya we want it written as y= if it's possible.

Answer 31

Agh. do you mind helping me out?

Answer 32

The inverse of the powers is going to be the root operation.
If we had y^2, we would apply the square root (2nd root) to isolate y.

The square and root would undo one another, leaving us with y.

Answer 33

But we have a y^4 right? :o
Hmm are there any other root operations we can use?

Answer 34

We can't use 4th root?

Answer 35

Yesss good good.

Answer 36

and can we just leave it as y=\[\sqrt[4]{(28x^3/3)-236}\]

Answer 37

Yes, good.
It may not look pretty, but it's correct! :)

Answer 38

That's all? Thank you soooooo much! This really helped a lot. The teacher taught this in like 10 minutes haha!

Answer 39

haha XD np