Y=ax^3 + bx^2

[Warlock]

Hi there,

I've been trying to do a sum but I'm not succeeding in solving the sum no matter how much I try.

We have to form a differential equation by eliminating arbituary values from the given equation.

Given equation y=ax^3 + bx^2

The solution (it's given after the exercise) is :

$\displaystyle 6.\, x^2\,\dfrac{d^2y}{dx^2}\, -\, 4x\, \dfrac{dy}{dx}\, +\, 6y\, =\, 0$

Thank you for your concern.

 

[Deleted member 4993]

Hi there,

I've been trying to do a sum but I'm not succeeding in solving the sum no matter how much I try.

We have to form a differential equation by eliminating arbituary values from the given equation.

Given equation y=ax^3 + bx^2

The solution (it's given after the exercise) is :

$\displaystyle 6.\, x^2\,\dfrac{d^2y}{dx^2}\, -\, 4x\, \dfrac{dy}{dx}\, +\, 6y\, =\, 0$

Thank you for your concern.

So you need to eliminate 'a' and 'b' and replace the equation with a second order linear differential equation - correct?

y=ax^3 + bx^2

y' = 3ax^2 + 2bx

y" = 6ax + 2b → y" * x/2 = 3ax^2 + bx^2

Now continue.....

 

[Ishuda]

Hi there,

I've been trying to do a sum but I'm not succeeding in solving the sum no matter how much I try.

We have to form a differential equation by eliminating arbituary values from the given equation.

Given equation y=ax^3 + bx^2

The solution (it's given after the exercise) is :

$\displaystyle 6.\, x^2\,\dfrac{d^2y}{dx^2}\, -\, 4x\, \dfrac{dy}{dx}\, +\, 6y\, =\, 0$

Thank you for your concern.

I would just start 'playing'
y=ax^3 + bx^2
So
y' = 3 a x² + 2 b x
and
b = y/x² - a x
or
y' = 3 a x² + 2 x (y/x² - a x) = a x² + 2 y/x.

So
y'' = ...
and
a = ...
or ...

 

[Warlock]

So you need to eliminate 'a' and 'b' and replace the equation with a second order linear differential equation - correct?

y=ax^3 + bx^2

y' = 3ax^2 + 2bx

y" = 6ax + 2b → y" * x/2 = 3ax^2 + bx^2

Now continue.....

That's the exact spot where I'm stuck.

 

[Deleted member 4993]

That's the exact spot where I'm stuck.

y=ax^3 + bx^2

y' = 3ax^2 + 2bx ................................................................(1)

y" = 6ax + 2b → y" * x/2 = 3ax^2 + bx ............edited...................(2)

(1) - (2)

y' - y" * x/2 = b* (x) → b = ??? → a = ????

 

[Warlock]

y=ax^3 + bx^2

y" = 6ax + 2b → y" * x/2 = 3ax^2 + bx^2 ...............................(2)

If you're multiplying by x/2 then shouldn't it be y" * x/2 = 3ax^2 + bx. I'm I missing something? I don't get it, how did you got bx^2.

 

[stapel]

That's the exact spot where I'm stuck.

This is why it was requested (here) that you provide a clear listing of all of your efforts so far: precisely so we don't provide you a "hint" that just so happens to be exactly as far as you'd already gotten.

 

[Deleted member 4993]

If you're multiplying by x/2 then shouldn't it be y" * x/2 = 3ax^2 + bx. I'm I missing something? I don't get it, how did you got bx^2.

You are correct - fixed it in the original post.