[RESOLVED]Diff - differential

[jrozsas]

Hi,
How can I solve this differential calculation in HP Prime?
z= 3x^2 - xy +y^2
Considering that (x, y) goes from (2,1) to (1.96,0.95).

I've made several attempts, but I can not enter the data correctly.

In this other POST (http://www.hpmuseum.org/forum/thread-7346.html), something similar happened, but Tim was able to enter the data correctly in LIMIT

[Han]

Your equation z=3x^2-xy+y^2 is not a differential equation. Are there differentials in there that you left out, perhaps?

The diff() command computes the derivative (or partial derivative) of an expression. Usually it takes two parameters: an expression, and a variable of differentiation.

For example: diff(x^2*y^3,x) returns 2x*y^3 whereas diff(x^2*y^3,y) returns 3x^2*y^2.

If you do not specify a variable, then the default variable 'x' is used. For example, diff(x^2*y^3) returns 2x*y^3. So in your second attempt, it is not clear what you meant by the diff(diff(..)) operation since you did not specify a variable. Did you mean to compute \( \frac{\partial^2}{\partial x^2} (3x^2-xy+y^2) \) or did you intend to have something else?

In particular, if you have questions on how to use a command, consider using the built-in help:

1) type the command
2) with the cursor on the typed command name, press the help button

Usually, the help also provides examples of how to use the commands. For desolve, the help says:

Code:

Syntax:
desolve(Eq,[TimeVar],Var)

Returns the solution to a differential equation.


Examples:
desolve(y''+y=0,y) → G_0*cos(x)+G_1*sin(x)
desolve((y''+y=sin(x)) and (y(0)=1) and (y'(0)=2),y)

[jrozsas]

I can not understand. The equation z= 3x^2 - xy +y^2 (where(x, y) from (2,1) to (1.96,0.95)) can not be solved with desolve?

[Han]

Usually a "derivative" is involved when solving a differential equation. For example, \( y = G_0 \cos(x) + G_1 \sin(x) \) is a general solution to the equation \( \frac{d^2}{dx^2} y(x) + y(x) = 0 \). The equation \( \frac{d^2}{dx^2} y(x) + y(x) = 0 \) can also be written as \( y''(x) + y(x) = 0 \). So in this example, the differential equation \( y''(x) + y(x) = 0 \) involves the second derivative: \( y''(x) \). To solve the equation \( y''(x) + y(x) = 0 \) we use:

desolve(y''(x)+y(x) = 0, y)

However, the equation you provided: \( z = 3x^2-xy+y^2\) does not involve any (partial) derivative or differential.

Google translate:
Por lo general, un "derivado" está involucrado cuando se resuelve una ecuación diferencial. Por ejemplo, \(y = G_0 \cos (x) + G_1 \sin (x) \) es una solución general a la ecuación \(\frac {d^2} {dx^2} y (x) + y(x) = 0 \). La ecuación \(\frac {d^2} {dx^2} y (x) + y(x) = 0 \) también se puede escribir como \(y '' (x) + y (x) = 0\). Así, en este ejemplo, la ecuación diferencial \(y '' (x) + y (x) = 0 \) implica la segunda derivada: \(y '' (x) \). Para resolver la ecuación \(y '' (x) + y (x) = 0 \) utilizamos:

Desolve (y '' (x) + y (x) = 0, y)

Sin embargo, la ecuación que proporcionó: \(z = 3x ^ 2-xy + y ^ 2 \) no implica ninguna derivada (parcial) o diferencial.

[roadrunner]

Leo,

Is this what you are trying to do?

   

-road

[Han]

Are you trying to compute:
\[ \int_{1.96}^2 \int_{0.95}^1 \left( 3x^2-xy+y^2\right) \ dx\ dy \]
Is that what you mean?

[jrozsas]

(12-06-2016 12:27 PM)roadrunner Wrote:  Leo,

Is this what you are trying to do?



-road

Roadrunner,
YESSSSSSS!!!!! Great!
Thank you all for your help!

[jrozsas]

(12-06-2016 12:36 PM)Han Wrote:  Are you trying to compute:
\[ \int_{1.96}^2 \int_{0.95}^1 \left( 3x^2-xy+y^2\right) \ dx\ dy \]
Is that what you mean?

No, but roadrunner sent the screen that solved the doubt. Thank you!