How do integrate with dy/dx?

[Ghost3k]

Hello, please help me out with this problem out, my professor taught us how to integrate but the professor never mentioned how to do integration with dy/dx. Here is the problem:

Solve the differential equation

with the condition that

. The solution to the equation is



Thanks in advance

Steps would be nice.

Tkhunny you better help me out with this one!

Also, their is no work to show because I have no clue how to start it.

 

[SAMUELK]

Hello, please help me out with this problem out, my professor taught us how to integrate but the professor never mentioned how to do integration with dy/dx. Here is the problem:

Solve the differential equation

with the condition that

. The solution to the equation is



Thanks in advance

Steps would be nice.

You do SEPARATION OF VARIABLES:

From:

dy
-- = 3x^2y^2
dx

Get the x's on one side, the y's on the other. (My apologies for the non-mathematical language.)
dy
--- = 3x^2 dx
y^2

y^-2 dy = 3x^2 dx

Now integrate each side separately and stick a constant on one side. (My apologies for ....)

y^-1
---- = x^3 + C
-1

You should be able to finish up.

 

[Ghost3k]

You do SEPARATION OF VARIABLES:

From:

dy
-- = 3x^2y^2
dx

Get the x's on one side, the y's on the other. (My apologies for the non-mathematical language.)
dy
--- = 3x^2 dx
y^2

y^-2 dy = 3x^2 dx

Now integrate each side separately and stick a constant on one side. (My apologies for ....)

y^-1
---- = x^3 + C
-1

You should be able to finish up.

thanks mate, and dont apologies, your language is perfect for learning.

Edit: Forget it, I don't know what to do next.

Am I supposed to do this:

-1/y = x ^ 3 + C

1/y = -x^3 - C
y = -x^3y2 - Cy^2
y(0) = -(0)^3(1)^2 - C(1)
1 = -C
C = -1

??

 

[renegade05]

thanks mate, and dont apologies, your language is perfect for learning.

Edit: Forget it, I don't know what to do next.

Am I supposed to do this:

-1/y = x ^ 3 + C

1/y = -x^3 - C
y = -x^3y2 - Cy^2
y(0) = -(0)^3(1)^2 - C(1)
1 = -C
C = -1

??

Yes, you don't even need to solve for y explicitly. Just plug in y=1, x=0 and solve for C.

Now to get it into a nice familiar form:

\(\displaystyle Let~y=f(x)\)

\(\displaystyle f(x) = -\frac{1}{x^3-1}\)

...if you want.

 

[Ghost3k]

Yes, you don't even need to solve for y explicitly. Just plug in y=1, x=0 and solve for C.

Now to get it into a nice familiar form:

\(\displaystyle Let~y=f(x)\)

\(\displaystyle f(x) = -\frac{1}{x^3-1}\)

...if you want.

thanks for the response, but how did u put it into that nice familiar form? I think you just took the negative reciprocal correct?

was my C = -1 correct?

 

[renegade05]

thanks for the response, but how did u put it into that nice familiar form? I think you just took the negative reciprocal correct?

was my C = -1 correct?

Yes c= -1.

Oh, i just solved for y and replaced it with f(x).

This is not really necessary, i just like to clean things up and put it in function notation. I didn't mean to confuse you.

This is the final answer:

\(\displaystyle y = \frac{1}{1-x^3}\)