Linear Differential Equation: when x=x(y)

In summary, the conversation discusses rewriting the equation (x+y^2)dy=ydx to make it exact. This involves using the partial derivative method and dividing throughout by y^2. The concept of d(xy) = xdy + ydx is also mentioned.
  • #1
rygza
38
0
(x+y^2)dy=ydx
rewrote as: dx/dy - x/y = y
Realized I had P(y)x and Q(y) rather than the P(x) and Q(x) from equations where y is a function of x. My problem now is after I multiply by the Integrating factor (-1/y):

-1 - x/(y^2) + 1/y(dx/dy)

I tried to make exact but i don't know the proper variables to use. I used

(partial derivative. M/partial deriv. x) = - 1/(y^2) = (partial N/partial y)

Is this proper? Usually for y=y(x) functions it's (partial M/partial y) but if i use that for this problem it doesn't make the equation exact. OR Am I supposed to rewrite so i have dy/dx (and then use the y=y(x) method)?
 
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  • #2
How about if I write it as:

[tex]xdy+y^2dy=ydx[/tex]

[tex]xdy-ydx=-y^2dy[/tex]

Ok, we know:

[tex]d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}[/tex]

So that left side could be written as:

[tex]x^2d\left(\frac{y}{x}\right)=-y^2dy[/tex]

Now what happens if I divide throughout by y^2?
 
  • #3
jackmell said:
How about if I write it as:

[tex]xdy+y^2dy=ydx[/tex]

[tex]xdy-ydx=-y^2dy[/tex]

Ok, we know:

[tex]d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}[/tex]

So that left side could be written as:

[tex]x^2d\left(\frac{y}{x}\right)=-y^2dy[/tex]

Now what happens if I divide throughout by y^2?

lost me on the d(y/x) = ... part
 
  • #4
rygza said:
lost me on the d(y/x) = ... part

hi rygza! :smile:

i suppose you're happy with d(xy) = xdy + ydx ?

that's the equivalent of (xy)' = xy' + yx'.

ok, now start with (y/x)' = (xy' - yx')/x2,

and you get d(y/x) = (xdy - ydx)/x2.
 
  • #5


I would suggest that you consult with a mathematics expert or refer to a textbook on differential equations for a more accurate and comprehensive response to your question. However, based on my understanding of differential equations, your approach seems correct. Since the equation is in terms of y, it would be appropriate to use the partial derivative with respect to y. However, if this does not lead to an exact equation, you may need to rewrite the equation in terms of x and use the y=y(x) method. It is also possible that there may be other methods or techniques that could be applied to solve this particular differential equation. It would be best to consult with a mathematics expert for a more in-depth analysis and solution.
 

Related to Linear Differential Equation: when x=x(y)

1. What is a linear differential equation?

A linear differential equation is a mathematical equation that involves a function and its derivatives. It is called linear because the function and its derivatives appear in a linear manner, meaning there are no products or powers of the function or its derivatives.

2. How is a linear differential equation different from a non-linear differential equation?

A linear differential equation involves a function and its derivatives in a linear manner, while a non-linear differential equation involves products or powers of the function and its derivatives. This makes linear differential equations easier to solve compared to non-linear ones.

3. What is the general form of a linear differential equation?

The general form of a linear differential equation is: y' + p(x)y = q(x), where y' represents the derivative of the function y, p(x) is a function of x, and q(x) is a function of x.

4. How do you solve a linear differential equation?

To solve a linear differential equation, you can use techniques such as separation of variables, integrating factors, or the method of undetermined coefficients. The specific method used depends on the form of the equation and the given initial conditions.

5. What is the significance of when x=x(y) in a linear differential equation?

In a linear differential equation, when x=x(y), it means that x and y are inversely related. This can be useful in solving the equation, as it allows for the use of substitution and simplification to find a solution.

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