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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.
[tex]
{y} = {f(x)}
[/tex]
Where,
[tex]
{{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} = {{{f}^{\prime}}{(x)}}
[/tex]
so that it is understood that [itex]{{{f}^{\prime}}{(x)}}[/itex] like [itex]{f(x)}[/itex] is a function of only [itex]{x}[/itex]. However, in the above paragraph it is insisted that [itex]{{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}}[/itex] must be a function of [itex]{x}[/itex] and [itex]{y}[/itex]. In other words [itex]{{y}^{\prime}} = {{F}{(x, y)}}[/itex], 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,
[tex]
{z} = {f(x, y)}
[/tex]
I run in to the problem that I don't know how to explicitly find [itex]{{z}^{\prime}}[/itex]. Specifically, I don't know how to differentiate [itex]{f(x, y)}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously. So that the derivative is actually a function of [itex]{x}[/itex] and [itex]{y}[/itex], like [itex]{f(x, y)}[/itex]. How would I differentiate [itex]{z}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously?
I note however, that in this particular case we're talking about [itex]{{y}^{\prime}}[/itex] as opposed to [itex]{{z}^{\prime}}[/itex].
Noting this I recall that we can rewrite,
[tex]
{\frac {dy}{dx}} = {{{f}^{\prime}}{(x)}}
[/tex]
as
[tex]
{dy} = {{{{f}^{\prime}}{(x)}}{dx}}
[/tex]
Where [itex]{y}[/itex] is found by integrating both sides of the above equation.
However, when I try to do this with the equation given,
[tex]
{{y}^{\prime}} = {F(x, y)}
[/tex]
Which can be rewritten as,
[tex]
{dy} = {F(x, y)dx}
[/tex]
and when integrated is,
[tex]
{y} = {{\int_{}^{}}{F(x, y)dx}}
[/tex]
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:Slope Fields
Solving a differential equation analytically can be difficult or even impossible.
However, there is a graphical approach you can use to learn a lot about the solution of a differential equation. Consider a differential equation of the form
[tex]
{{y}^{\prime}} = {{F}{(x, y)}}
[/tex]
At each point [itex]{(x, y)}[/itex] in the [itex]{xy}[/itex]-plane where [itex]{F}[/itex] is defined, the differential equation determines the slope [itex]{{y}^{\prime}} = {{F}{(x, y)}}[/itex] of the solution at that point. If you draw a short line segment with slope [itex]{{F}{(x, y)}}[/itex] at selected points [itex]{(x, y)}[/itex] in the domain of [itex]{F}[/itex], then these line segments form a slope field or a direction field for the differential equation
[itex]{{y}^{\prime}} = {{F}{(x, y)}}[/itex]
Each line segment has the same slope as the solution curve through that point. A slope field shows the general shape of all the solutions.
[tex]
{y} = {f(x)}
[/tex]
Where,
[tex]
{{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} = {{{f}^{\prime}}{(x)}}
[/tex]
so that it is understood that [itex]{{{f}^{\prime}}{(x)}}[/itex] like [itex]{f(x)}[/itex] is a function of only [itex]{x}[/itex]. However, in the above paragraph it is insisted that [itex]{{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}}[/itex] must be a function of [itex]{x}[/itex] and [itex]{y}[/itex]. In other words [itex]{{y}^{\prime}} = {{F}{(x, y)}}[/itex], 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,
[tex]
{z} = {f(x, y)}
[/tex]
I run in to the problem that I don't know how to explicitly find [itex]{{z}^{\prime}}[/itex]. Specifically, I don't know how to differentiate [itex]{f(x, y)}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously. So that the derivative is actually a function of [itex]{x}[/itex] and [itex]{y}[/itex], like [itex]{f(x, y)}[/itex]. How would I differentiate [itex]{z}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously?
I note however, that in this particular case we're talking about [itex]{{y}^{\prime}}[/itex] as opposed to [itex]{{z}^{\prime}}[/itex].
Noting this I recall that we can rewrite,
[tex]
{\frac {dy}{dx}} = {{{f}^{\prime}}{(x)}}
[/tex]
as
[tex]
{dy} = {{{{f}^{\prime}}{(x)}}{dx}}
[/tex]
Where [itex]{y}[/itex] is found by integrating both sides of the above equation.
However, when I try to do this with the equation given,
[tex]
{{y}^{\prime}} = {F(x, y)}
[/tex]
Which can be rewritten as,
[tex]
{dy} = {F(x, y)dx}
[/tex]
and when integrated is,
[tex]
{y} = {{\int_{}^{}}{F(x, y)dx}}
[/tex]
I find that I do not know how to evaluate the RHS. How would I evaluate it?
Thanks,
-PFStudent