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Year 12s: what will be the first way you research your future options? >>

Maths question

(dy(x)/dx)+6xy(x)=f(x)

how do I find the homogenous equation?

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help

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help still needed

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Original post by ThiagoBrigido

Have you manage to do anything?

no, help would be appreciated.

OP

??

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OP

help

Hokogeneous would be when f(x) = 0

(dy(x)/dx)+6xy(x)=f(x)

how do I solve for the homogenous equation?

Original post by DavidB89

(dy(x)/dx)+6xy(x)=f(x)

how do I solve for the homogenous equation?

Himomgeneous is when f(x) = 0.
You then have a reasonably standard 1st oder, linear ODE. What have you learnt about solving them?

(edited 3 years ago)

f(x)=0.

(dy/dx)+6xy=f(x)

y'+6x=0

y'=-6x

y=integrate|-6x

y=-3x^2+C1

Im not sure if this is the way since I do integrate or is it where I separate the variables first. X on one side and y on the other, then integrate them.

(edited 3 years ago)

Original post by DavidB89

f(x)=0.

(dy/dx)+6xy=f(x)

The aim is to offer hints, not solutions.
Pls delete.

Study Forum Helper

Original post by DavidB89

f(x)=0.

(dy/dx)+6xy=f(x)

y'+6x=0

y'=-6x

y=integrate|-6x

y=-3x^2+C1

Im not sure if this is the way since I do integrate or is it where I separate the variables first. X on one side and y on the other, then integrate them.

Where did the y go in 6xy?

Study Forum Helper

Original post by mqb2766

The aim is to offer hints, not solutions.
Pls delete.

It's okay, they are the same person.

Original post by mqb2766

The aim is to offer hints, not solutions.
Pls delete.

This isn't a hint, this is what ive worked up to so far, need to know whether its correct or not.

Original post by DavidB89

This isn't a hint, this is what ive worked up to so far, need to know whether its correct or not.

No, the y term disappears.
Your username has changed.
Have a look at
http://www.maths.surrey.ac.uk/explore/vithyaspages/firstorder.html
Amongst others.

(edited 3 years ago)

Can I do:

(dy/dx)+6xy=0

Then move the 6xy to the other side:

(dy/dx)=-6xy

Then put y with dy and x with dx:

integrate(1/y)dy = integrate(-6x)dx

Original post by DavidB89

Can I do:

(dy/dx)+6xy=0

Then move the 6xy to the other side:

(dy/dx)=-6xy

Then put y with dy and x with dx:

integrate(1/y)dy = integrate(-6x)dx

Yes, there are a few ways to proceed.

Original post by DavidB89

Can I do:

(dy/dx)+6xy=0

Then move the 6xy to the other side:

(dy/dx)=-6xy

Then put y with dy and x with dx:

integrate(1/y)dy = integrate(-6x)dx

Once I have integrated both sides I end with:

ln(y)=-3x^2+C

I then make y the subject so:

y=Ae^(-3x^2+C)

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