dy/dx -5y =(-5/2) * x * y^3. Bernoulli. y^ -3 * dy/dx-5y^ -2=(-5/2) * x

Integrate

Integrate

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1.) dy/dx -5y =(-5/2) * x

2.) y^ -3 * dy/dx-5y^ -2=(-5/2) * x

3.) v=y^-2

4.) dv/dx=-2y^ -3 * dy/dx


Shouldn't y^-2 go to zero since there is no x in this expression?



Something I remember from implicit differentiation is that when we say "respect to" it means we are trying to find the relation of rate of change between variables dependent on each other.


Is that how I should see it here?
 
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Steven G said:
What exactly do you mean by Shouldn't y^-2 go to zero since there is no x in this expression??
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If we take the derivative of something with respect to x and there is no x doesn’t it go to zero?

or am I thinking of partial derivatives?
 
Integrate said:
Shouldn't y^-2 go to zero since there is no x in this expression?
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When I read that, I think of a limit. I think you mean that its derivative should be zero? You should say that.

Integrate said:
And if it is we derive it. Why?
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Do you mean "take the derivative"? In my experience, the verb is not "derive" but "differentiate", as odd as that is. To derive means something very different.

Integrate said:
If we take the derivative of something with respect to x and there is no x doesn’t it go to zero?

or am I thinking of partial derivatives?
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Yes, I believe you are thinking of partial derivatives, in which you think of other variables as being held constant.

That is not what is being done here. Since y is a function of x (that's what a differential equation is all about), you have to use the chain rule, as they did.
 
Dr.Peterson said:
When I read that, I think of a limit. I think you mean that its derivative should be zero? You should say that.


Do you mean "take the derivative"? In my experience, the verb is not "derive" but "differentiate", as odd as that is. To derive means something very different.


Yes, I believe you are thinking of partial derivatives, in which you think of other variables as being held constant.

That is not what is being done here. Since y is a function of x (that's what a differential equation is all about), you have to use the chain rule, as they did.
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Okay after watching a few videos on implicit differentiation I’ve got my bearings on it and how differentiation in this context is just the chain rule. Differentiate the function and then differentiate its argument which we don’t know so we leave it as dy/dx or whatever the variables may be.

Math is like swimming underwater. You’re fine one second then all of sudden you lose all sense of direction until you find up again.
 
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