# ODE seperable method [Solved!]

**Ahmed** 13 Oct 2016, 22:18

### My question

(y/x)*(dy/dx)=sqrt(1+x^2+y^2+x^2*y^2)

### Relevant page

Differential Equations - Separable Equations

### What I've done so far

replaced 1+x^2+y^2+x^2*y^2 with t

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**Ahmed** 13 Oct 2016, 22:18

(y/x)*(dy/dx)=sqrt(1+x^2+y^2+x^2*y^2)

Differential Equations - Separable Equations

replaced 1+x^2+y^2+x^2*y^2 with t

X

(y/x)*(dy/dx)=sqrt(1+x^2+y^2+x^2*y^2)

Relevant page <a href="http://tutorial.math.lamar.edu/Classes/DE/Separable.aspx">Differential Equations - Separable Equations</a> What I've done so far replaced 1+x^2+y^2+x^2*y^2 with t

**Murray** 14 Oct 2016, 20:33

Hello Ahmed

You are encouraged to use the Math Input system so your math is easier to read.

Then your question looks like this:

`(y/x)(dy/dx)=sqrt(1+x^2+y^2+x^2*y^2)`

The first hint is to factor the expression under the square root. Do that first, then we'll go from there.

X

Hello Ahmed You are encouraged to use the <a href="/forum/entering-math-graphs-images-41/how-to-enter-math:91">Math Input system</a> so your math is easier to read. Then your question looks like this: `(y/x)(dy/dx)=sqrt(1+x^2+y^2+x^2*y^2)` The first hint is to factor the expression under the square root. Do that first, then we'll go from there.

**Ahmed** 14 Oct 2016, 22:19

Ok I've factorised it now it looks like

(y/x)*(dy/dx)=sqrt((x^2+1)(y^2+1))

now what.

X

Ok I've factorised it now it looks like (y/x)*(dy/dx)=sqrt((x^2+1)(y^2+1)) now what.

**Murray** 16 Oct 2016, 20:10

Putting back ticks around your (correct) answer makes it appear as:

`(y/x)*(dy/dx)=sqrt((x^2+1)(y^2+1))`

OK, now you need to separate the variables. That is, put all the `y` and `dy` parts together on the left, and the `x` and `dx` parts together on the right.

X

Putting back ticks around your (correct) answer makes it appear as: `(y/x)*(dy/dx)=sqrt((x^2+1)(y^2+1))` OK, now you need to separate the variables. That is, put all the `y` and `dy` parts together on the left, and the `x` and `dx` parts together on the right.

**Murray** 26 Oct 2016, 23:38

It seems Ahmed has disappeared. Anyone else like to have a go at finishing it?

X

It seems Ahmed has disappeared. Anyone else like to have a go at finishing it?

**stephenB** 27 Jan 2018, 03:23

I'll have a go.

`(y/x)*(dy/dx)=sqrt(x^2+1) xx sqrt(y^2+1)`

Separating gives:

`(ydy)/sqrt(y^2+1)=xsqrt((x^2+1))dx`

Integrating one side at a time:

`int(ydy)/sqrt(y^2+1)`

Put `u=y^2+1`

then `du=2ydy`

So we can write it as

`int((d(u))/2)/sqrt(u) = (1/2)int u^(-1/2)du`

`=(1/2)(2)u^(1/2)`

`=u^(1/2)`

`= sqrt(y^2+1)`

Now for the right hand side:

`int xsqrt((x^2+1))dx`

Put `u=x^2+1`

so `du = 2xdx`

So the integral becomes

`int (1/2)sqrt(u)du = 1/2 int (u^(1/2))du`

`=1/2 2/3 u^(3/2) + K` (we need a constant on one side only)

`=1/3 (x^2+1)^(3/2)+K`

So putting both sides together, we have:

`sqrt(y^2+1) = 1/3 (x^2+1)^(3/2)+K`

We can get `y` in terms of `x` fairly easily:

Square both sides:

`(y^2+1) = 1/9 (x^2+1)^3+K_1` (This is different to `K` from earlier)

`y^2 = 1/9 (x^2+1)^3+K_2` (This is a different to `K_1`, but still a constant)

`y = +- 1/3 (x^2+1)^(3/2)+K_3`

X

I'll have a go. `(y/x)*(dy/dx)=sqrt(x^2+1) xx sqrt(y^2+1)` Separating gives: `(ydy)/sqrt(y^2+1)=xsqrt((x^2+1))dx` Integrating one side at a time: `int(ydy)/sqrt(y^2+1)` Put `u=y^2+1` then `du=2ydy` So we can write it as `int((d(u))/2)/sqrt(u) = (1/2)int u^(-1/2)du` `=(1/2)(2)u^(1/2)` `=u^(1/2)` `= sqrt(y^2+1)` Now for the right hand side: `int xsqrt((x^2+1))dx` Put `u=x^2+1` so `du = 2xdx` So the integral becomes `int (1/2)sqrt(u)du = 1/2 int (u^(1/2))du` `=1/2 2/3 u^(3/2) + K` (we need a constant on one side only) `=1/3 (x^2+1)^(3/2)+K` So putting both sides together, we have: `sqrt(y^2+1) = 1/3 (x^2+1)^(3/2)+K` We can get `y` in terms of `x` fairly easily: Square both sides: `(y^2+1) = 1/9 (x^2+1)^3+K_1` (This is different to `K` from earlier) `y^2 = 1/9 (x^2+1)^3+K_2` (This is a different to `K_1`, but still a constant) `y = +- 1/3 (x^2+1)^(3/2)+K_3`

X

Good answer, stephenB

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