Integral with a Y

[cmhex]

I'm having issues due to 'knowledge gaps' that I either missed or didn't cover in Calc I. This is Calc II, with what I presume must be the introductory level Differential Equations.

EDIT: Thanks pka!

EDIT2: I've answered my questions. If you want to see the answers scroll down to my reply.

First, a question I have no problem with.

\(\displaystyle dy/dx = 2x - 5x^2 = \int (2x-5x^2) dx\)

\(\displaystyle y = x^2 - 5/3x^3+C\)

But when you throw a "y" to the right of the = sign, I get lost.

\(\displaystyle dy/dx = y + 8\) and \(\displaystyle dy/dx = (3+y)^2\)

I have the solution to the second one. Here are the 'real questions'.

\(\displaystyle \int\) (3+y)^-2 dy = x + C

Why did we change that to -2? I'm not sure how that is equal to x + C? I'm presuming that the - exponent is from the fact that its dy and not dx?

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

Okay obviously that's an integration, add 1 to the exponent then divide by the value of the new exponent, which is -1, thus making it (3+y)/-1 or -(3+y). I have this part.

Start isolation of the y

3+y = 1/(x+C) then y = -3-(1/(x+C)

I have no issue with any of the steps except that one bit where the exponent and how its equal to x + C. Just tell me what I need to review if nothing else.

I do not have the solution for the dy/dx = y + 8 problem but I think I know how to do it based on the other one if anyone wants to work it out.

 

[pka]

, is there a way to put an integral symbol here by the way?

[ tex]\int {f(x)dx} [/tex] gives \(\displaystyle \int {f(x)dx} \) , use [ tex] without the space.

[ tex]\int_a^b {f(x)dx} [/tex] gives \(\displaystyle \int_a^b {f(x)dx} \)

 

[cmhex]

EDIT: Got it now

\(\displaystyle dy/dx = (3+y)^2\)

Multiply both sides by dx

\(\displaystyle dy = (3+y)^2 dx\)

Divide both sides by \(\displaystyle (3+y)^2\)

\(\displaystyle dy / (3+y)^2 = dx\)

Invert the left portion
EDIT: Thanks again pka!
\(\displaystyle (3+y)^{-2} dy = dx\)

Integrate

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

Start simplification

3+y = -1/(x+C)
y = -3 + -1/(x+C)
y = -3 - 1/(x+C)

 

[pka]

Here is more useful code.
[ TEX]\frac{2}{3} [/TEX] gives \(\displaystyle \frac{2}{3} \) where as [ TEX]\dfrac{2}{3} [/TEX] gives \(\displaystyle \dfrac{2}{3} \)

[ TEX]x^{23} [/TEX] gives \(\displaystyle x^{23} \)

​[ tex]{\log _{2}}(3)[/tex] gives \(\displaystyle {\log _{2}}(3)\)

​[ tex]{\ln}(3)[/tex] gives \(\displaystyle {\ln}(3)\)

 

[lookagain]

\(\displaystyle dy/dx = 2x - 5x^2 =\int {(2x-5x^2) dx} = x^2 - 5/3x^3+C\)

cmhex, all of those quantities are not equal to each other, so you can't
have a train of equal signs all the way across. Try something like this instead:


\(\displaystyle dy/dx \ = \ 2x - 5x^2 \)


\(\displaystyle y \ = \ \int {(2x - 5x^2) dx} \)


\(\displaystyle y \ = \ x^2 - (5/3)x^3 \ + \ C\)


or


\(\displaystyle y \ = \ - (5/3)x^3 \ + \ x^2 \ + \ C\)

 

[cmhex]

Ah I left out the y =, sorry bout that. Fixed it.

 

[HallsofIvy]

I'm having issues due to 'knowledge gaps' that I either missed or didn't cover in Calc I. This is Calc II, with what I presume must be the introductory level Differential Equations.

EDIT: Thanks pka!

EDIT2: I've answered my questions. If you want to see the answers scroll down to my reply.

First, a question I have no problem with.

\(\displaystyle dy/dx = 2x - 5x^2 =\int (2x-5x^2) dx\)

But you certainly have a problem with the notation! \(\displaystyle 2x- 5x^2\) is NOT equal to \(\displaystyle \int (2x- 5x^2)dx\)!

\(\displaystyle y = x^2 - 5/3x^3+C\)

But when you throw a "y" to the right of the = sign, I get lost.

\(\displaystyle dy/dx = y + 8\) and \(\displaystyle dy/dx = (3+y)^2\)

I have the solution to the second one. Here are the 'real questions'.

\(\displaystyle \int\) (3+y)^-2 dy = x + C

Why did we change that to -2? I'm not sure how that is equal to x + C? I'm presuming that the - exponent is from the fact that its dy and not dx?

You divide both sides by \(\displaystyle (3+ y)^2\) in order to have only y on one side and only x on the other.
\(\displaystyle \frac{1}{(3+y)^2}= (3+ y)^{-2}\). If you do not know that you need to review algebra!
Of course \(\displaystyle \int dx= x+ C\).

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

Okay obviously that's an integration, add 1 to the exponent then divide by the value of the new exponent, which is -1, thus making it (3+y)/-1 or -(3+y). I have this part.

No, it isn't. You forgot the -1 power.

Start isolation of the y

3+y = 1/(x+C)

No, 3+ y= -1/(x+ C)

then y = -3-(1/(x+C)

I have no issue with any of the steps except that one bit where the exponent and how its equal to x + C. Just tell me what I need to review if nothing else.

I do not have the solution for the dy/dx = y + 8 problem but I think I know how to do it based on the other one if anyone wants to work it out.

\(\displaystyle \int \frac{dy}{y+ 8}= \int dx\).

Integrate.

And don't be so careless with notation.