IntMath Home » Forum home » Differential Equations » ODE seperable method

# ODE seperable method [Solved!]

### 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

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

## Re: ODE seperable method

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.

## Re: ODE seperable method

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.

## Re: ODE seperable method

(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.

## Re: ODE seperable method

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?

## Re: ODE seperable method

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

## Re: ODE seperable method

Good answer, stephenB