- #1
PFStudent
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Hey,
1. Homework Statement .
I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph.
<br /> {y} = {f(x)}<br />
Where,
<br /> {{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} = {{{f}^{\prime}}{(x)}}<br />
so that it is understood that {{{f}^{\prime}}{(x)}} like {f(x)} is a function of only {x}. However, in the above paragraph it is insisted that {{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} must be a function of {x} and {y}. In other words {{y}^{\prime}} = {{F}{(x, y)}}, why is that?
2. Homework Equations .
Knowledge Differential Equations and Slope Fields.
3. The Attempt at a Solution .
If we begin from the conventional notation that,
<br /> {z} = {f(x, y)}<br />
I run into the problem that I don't know how to explicitly find {{z}^{\prime}}. Specifically, I don't know how to differentiate {f(x, y)} with respect to {x} and {y} simultaneously. So that the derivative is actually a function of {x} and {y}, like {f(x, y)}. How would I differentiate {z} with respect to {x} and {y} simultaneously?
I note however, that in this particular case we're talking about {{y}^{\prime}} as opposed to {{z}^{\prime}}.
Noting this I recall that we can rewrite,
<br /> {\frac {dy}{dx}} = {{{f}^{\prime}}{(x)}}<br />
as
<br /> {dy} = {{{{f}^{\prime}}{(x)}}{dx}}<br />
Where {y} is found by integrating both sides of the above equation.
However, when I try to do this with the equation given,
<br /> {{y}^{\prime}} = {F(x, y)}<br />
Which can be rewritten as,
<br /> {dy} = {F(x, y)dx}<br />
and when integrated is,
<br /> {y} = {{\int_{}^{}}{F(x, y)dx}}<br />
I find that I do not know how to evaluate the RHS. How would I evaluate it?
Thanks,
-PFStudent
1. Homework Statement .
I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph.
The above paragraph seemed a little confusing since, conventionally,From Textbook said:
<br /> {y} = {f(x)}<br />
Where,
<br /> {{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} = {{{f}^{\prime}}{(x)}}<br />
so that it is understood that {{{f}^{\prime}}{(x)}} like {f(x)} is a function of only {x}. However, in the above paragraph it is insisted that {{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} must be a function of {x} and {y}. In other words {{y}^{\prime}} = {{F}{(x, y)}}, why is that?
2. Homework Equations .
Knowledge Differential Equations and Slope Fields.
3. The Attempt at a Solution .
If we begin from the conventional notation that,
<br /> {z} = {f(x, y)}<br />
I run into the problem that I don't know how to explicitly find {{z}^{\prime}}. Specifically, I don't know how to differentiate {f(x, y)} with respect to {x} and {y} simultaneously. So that the derivative is actually a function of {x} and {y}, like {f(x, y)}. How would I differentiate {z} with respect to {x} and {y} simultaneously?
I note however, that in this particular case we're talking about {{y}^{\prime}} as opposed to {{z}^{\prime}}.
Noting this I recall that we can rewrite,
<br /> {\frac {dy}{dx}} = {{{f}^{\prime}}{(x)}}<br />
as
<br /> {dy} = {{{{f}^{\prime}}{(x)}}{dx}}<br />
Where {y} is found by integrating both sides of the above equation.
However, when I try to do this with the equation given,
<br /> {{y}^{\prime}} = {F(x, y)}<br />
Which can be rewritten as,
<br /> {dy} = {F(x, y)dx}<br />
and when integrated is,
<br /> {y} = {{\int_{}^{}}{F(x, y)dx}}<br />
I find that I do not know how to evaluate the RHS. How would I evaluate it?
Thanks,
-PFStudent