Question

If the derivative of a function is y (the function it self), then f(x) = e^x.
If the derivative of a function is y*y, what is f(x) ?

Answer 1

so you want to integral of e^(2x)?

Answer 2

\[\int{e^{2x}}dx=\frac{e^{2x}}{2}\]

Answer 3

no, he means dy/dx = y^2

Answer 4

so the function should be (x^3)/3 ?

Answer 5

No, you see that doesn't work as y' = x^2.

Answer 6

Hint: Note that for the equation dy/dx = y, we could solve using a method called separation of variables:
\[\frac{dy}{y} = dx\]
Integrating,

ln y = x + C

hence

y = A e^x, where A = e^C

Use the same method of separation of variables for dy/dx = y^2

@agd: don't write the answer; let him try and figure it out.

Answer 7

Hmm I don't really follow, but

dy/dx = y*y
dy/y² = dx

ln y² = x+ c
y² = A e^x
y = e^(c/2) + e^(x/2) ?

Answer 8

No, you didn't integrate correctly:
\[\int\limits \frac{dy}{y^2} = .... what?\]

Answer 9

I don't follow, could you explain some more

Answer 10

You said that integral is ln(y^2). That is incorrect.

Let me ask you this. What is
\[\frac{d \ }{dy} (1/y)\]
?

Answer 11

d/dy 1/y =
-1/y²

d/dy 1/y2
-2/y³ ?

Answer 12

Yes

Hence the integral of 1/y^2 is ... what?

Answer 13

\[\int\limits \frac{dy}{y^2}\]

Answer 14

1/-y

Answer 15

Yes. So plugging that back into the equation above with both x and y we have ...

-1/y = x + c

Hence y = ....

Answer 16

y = -1/x+c

Answer 17

Right. Let's set x = 0 for simplicity.

Verify now directly be differentiating that
\[\frac{dy}{dx} = y^2\]

Answer 18

...by differentiating ...

Answer 19

Sorry set C = 0

Answer 20

sigh ... typos, sorry.

Answer 21

So to clarify, the function we're now verifying does meet the equation dy/dx = y^2 is

y(x) = -1/x

Answer 22

You mean differentiating y = -1/x+c
which comes to

dy/dx -1 / x =
1/x²
... I'm doing something wrong here

Answer 23

Cmon. If f(x) = x^n, what is df/dx?

Answer 24

df/dx = n x^(n-1)

Answer 25

n x^x-1

Answer 26

n-1

Answer 27

Hence here, with y = -1/x = - x^(-1),
dy/dx = -(-1)x^(-1-1)
= x^(-2)
= 1/x^2

Now does that satisfy the equation dy/dx = y^2 ?

Answer 28

nope

Answer 29

YES, it does

Answer 30

Because y^2 = (-1/x)^2 = 1/x^2

Answer 31

Hence indeed

dy/dx = y^2

when y = -1/x

Answer 32

oh like that

Answer 33

I read the whole equation wrong

Answer 34

the equation says does the derivative of f(x) equate f(x)^2, right?

Answer 35

Yes

Answer 36

I.e., dy/dx = y^2

Answer 37

Unless you tell me otherwise.

Answer 38

Thanks man!

Answer 39

Do yourself a favor: Take a blank piece of paper and write out the solution to the problem. When you can do that without looking at this website or the last time you wrote it out, then you know that you actually know it.

Answer 40

yea thanks for the tip. Is that how you study ?

Answer 41

Yes.