Can V and x be Separated in this Differential Equation?

[hotjohn]

Homework Statement

i am asked to form a differential equation using dy/dx = 1 + y + (x^2 ) + y(x^2) , but i gt stucked here , homework to proceed? as we can see , the V and x are not separable

Homework Equations

The Attempt at a Solution

 

[Buzz Bloom]

Hi hotjohn:

You may have a typo. You say

V and x are not separable​

but the is no "V" in the equation.

Regards,
Buzz

 

[hotjohn]

Hi hotjohn:

You may have a typo. You say

V and x are not separable​

but the is no "V" in the equation.

Regards,
Buzz

sorry , i mean y and x . How to continue ?

 

[Buzz Bloom]

Hi hotjohn:

If you factor the 2nd equation in your attachment, and make a substitution for y in terms of a new variable, say z, you can get a separable equation involving z and x.

Hope this helps.

Regards,
Buzz

 

[hotjohn]

Hi hotjohn:

If you factor the 2nd equation in your attachment, and make a substitution for y in terms of a new variable, say z, you can get a separable equation involving z and x.

Hope this helps.

Regards,
Buzz

sorry , i didnt get you , can you explain further ?

 

[Buzz Bloom]

Hi hotjohn:

dy/dx = (1+y) × (1+x²)
y = z-1

Regards,
Buzz

 

[hotjohn]

Hi hotjohn:

dy/dx = (1+y) × (1+x²)
y = z-1

Regards,
Buzz

can you expalin why there is a need to sub y = z-1 ?? and how do u knw why should sub y = z-1 ? why can't be y = z-2 ? or others ?

 

[Crush1986]

can you expalin why there is a need to sub y = z-1 ?? and how do u knw why should sub y = z-1 ? why can't be y = z-2 ? or others ?

You don't have to sub if you don't want to. Once it's separated just solve it like you would any seperable equation.

 

[hotjohn]

You don't have to sub if you don't want to. Once it's separated just solve it like you would any seperable equation.

how to determine the value of number or new constant to be substituted into the original equation ?

 

[Crush1986]

Remember, unless you are given initial conditions you will have an infinite amount of answers to most differential equations.

 

[hotjohn]

Remember, unless you are given initial conditions you will have an infinite amount of answers to most differential equations.

can it be y = z-2 , y = z-3 and etc ??

 

[Buzz Bloom]

can you expalin why there is a need to sub y = z-1 ?? and how do u knw why should sub y = z-1 ? why can't be y = z-2 ? or others ?

Hi hotjohn:

It is not a need, but a convenience.

y=z-1 → z=y+1 → dz/dx =z × (1+x²) →dz/z = (1+x²) dx​

This in now the standard form for a separable equation.

http://mathwiki.ucdavis.edu/Analysi...al_Equations/Separable_Differential_Equations​

Regards,
Buzz

 

[hotjohn]

Hi hotjohn:

It is not a need, but a convenience.

y=z-1 → z=y+1 → dz/dx =z × (1+x²) →dz/z = (1+x²) dx​

This in now the standard form for a separable equation.

http://mathwiki.ucdavis.edu/Analysis/Ordinary_Differential_Equations/1:_First_Order_Differential_Equations/Separable_Differential_Equations​

Regards,
Buzz

??
i found that if substituition of y=z-1 is not used , the whole equation become inseparable ...

how do we know that y must be replaced with y=z-1 ?

 

[Buzz Bloom]

how do we know that y must be replaced with y=z-1 ?

Hi hotjohn:

As I said previously, using y = z-1 is not a necessity, and not something that must be done. I thought that making that substitution might help you see the separability more easily.

Can you complete the solution of the problem from

dy/dx = (1+y) × (1+x²) ?​

If so, you are done.

Regards,
Buzz