Question

- Separable Equation

Answer 1

\[\frac{ dy }{ dx } = \frac{ 2y }{ x(y-5) }\]

Solve for general/particular solution.

Answer 2

\[
\int \frac{y-5}{y}\frac{dy}{dx}dx=\int\frac{2}{x}dx
\]

Answer 3

Remember \((dy/dx)dx = dy\)

Answer 4

All I did was:
1) Multiply both sides by \(y-5\)
2) Divide both sides by \(y\)
3) Integrate both sides with respect to \(x\)

No magic, no abuse of notation.

Answer 5

hey

Answer 6

The hardest ones I find easiest, and these ones I'm thinking too much on, may I ask you one other thing?

Answer 7

Go ahead.

Answer 8

\[(1+x^3).\frac{ dy }{ dx } = x^2y\]

Answer 9

How would you isolate dy and dx?

Answer 10

\[
\int \frac{1}{y}\frac{dy}{dx}dx=\int \frac{x^2}{1+x^3}dx
\]

Answer 11

Whenever you want to multiply by \(dx\), DON'T!
Instead integrate by \(x\) and it will go away nicely.

Answer 12

I can integrate etc, use indice rules, ln all the ''difficult'' operations. I just always find it tough to transpose in the beginning for some reason..

Answer 13

What you'll have is: \[
f(x)g(y)\frac{dy}{dx}=h(x)k(y)
\]And what you end up with is: \[
\int \frac{g(y)}{k(y)}\frac{dy}{dx}dx=\int \frac{h(x)}{f(x)}dx
\]

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @Asylum15
\[\frac{ dy }{ dx } = \frac{ 2y }{ x(y-5) }\]

Solve for general/particular solution.
\(\color{blue}{\text{End of Quote}}\)
\[(1)(1) \frac{dy}{dx}=\left(\frac{2}{x}\right)\left(\frac{y}{y-5}\right) \\
.\\
f(x)=1\\
g(y)=1\\
h(x)=\frac{2}{x}\\
k(y)=\frac{y}{y-5}
\]

Answer 15

So the integral of \[\frac{ x^2 }{ (1+x^3) }\] = \[\frac{ x^2 }{ 1 } + \frac{ x^2 }{ x^3}\]?

Answer 16

No, that isn't how fractions work.

Answer 17

My apologies :|

Answer 18

\[ \frac{ \ln(x^3+1) }{ 3 }\]

Answer 19

Yeah, whoops let me rewrite that.

Answer 20

Thanks for your help btw :)

Answer 21

\[
\int \frac{x^2}{1+x^3}dx = \int \frac{1}{1+x^3} \frac{d(1+x^3)}{3} = \int \frac{1}{3u}du
\]

Answer 22

ah!