First order differential equations

[jonnburton]

Hi,

I've just started this topic and don't quite understand it. I was wondering whether anyone could explain a couple of things.

The worked example my book gives is as follows:

By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to \(\displaystyle x^2 + y^2 = Ay\)



Solution:

Differentiating both sides of \(\displaystyle x^2 + y^2 = Ay\) with respect to x gives:

\(\displaystyle 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}\)

Multiplying throughout by y (I presume this is done in order to get the term 'Ay' back)

\(\displaystyle 2xy+2y^2 \frac{dy}{dx}=Ay\frac{dy}{dx}\)

And using the given equation: \(\displaystyle 2xy + 2y^2\frac{dy}{dx} = (x^2+y^2)\frac{dy}{dx}\)

so the required first order differential equation is:

\(\displaystyle (x^2 - y^2) \frac{dy}{dx} = 2xy\)


In this example it's really the last step that I don't understand. How does \(\displaystyle (x^2+y^2)\) become \(\displaystyle (x^2-y^2)\)? and how does the \(\displaystyle 2y^2\frac{dy}{dx}\) disappear?


I began to try the first question that follows this worked example: By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to \(\displaystyle y = Ax\)

Trying to follow a similar procedure, first I took the derivative with respect to x:

\(\displaystyle y \frac{dy}{dx} = (1)\frac{dy}{dx}A\)

Then I multiplied by x, in order to get Ax back:
\(\displaystyle xy\frac{dy}{dx} = \frac{dy}{dx}Ax\) but then this led to:

\(\displaystyle xy\frac{dy}{dx}=y\frac{dy}{dx}\)

I know that the answer is \(\displaystyle \frac{dy}{dx}=\frac{x}{y}\) because \(\displaystyle \frac{dx}{dt} = kx\) But I can't see how to get there following this approach.


Any help would be much appreciated!

 

[stapel]

\(\displaystyle 2xy+2y^2 \frac{dy}{dx}=Ay\frac{dy}{dx}\)

And using the given equation: \(\displaystyle 2xy + 2y^2\frac{dy}{dx} = (x^2+y^2)\frac{dy}{dx}\)

so the required first order differential equation is:

\(\displaystyle (x^2 - y^2) \frac{dy}{dx} = 2xy\)

In this example it's really the last step that I don't understand. How does \(\displaystyle (x^2+y^2)\) become \(\displaystyle (x^2-y^2)\)? and how does the \(\displaystyle 2y^2\frac{dy}{dx}\) disappear?

Do for yourself the steps that they skipped:

. . . . .\(\displaystyle 2xy\, +\, 2y^2\dfrac{dy}{dx}\, =\, \left(x^2\, +\, y^2\right)\dfrac{dy}{dx}\)

. . . . .\(\displaystyle 2xy\, +\, 2y^2\dfrac{dy}{dx}\, =\, x^2\dfrac{dy}{dx}\, +\, y^2 \dfrac{dy}{dx}\)

. . . . .\(\displaystyle 2xy\, =\, x^2\dfrac{dy}{dx}\, +\, y^2\dfrac{dy}{dx}\, -\, 2y^2\dfrac{dy}{dx}\)

Simplify the right-hand side. What do you get?

 

[jonnburton]

Do for yourself the steps that they skipped:

. . . . .\(\displaystyle 2xy\, +\, 2y^2\dfrac{dy}{dx}\, =\, \left(x^2\, +\, y^2\right)\dfrac{dy}{dx}\)

. . . . .\(\displaystyle 2xy\, +\, 2y^2\dfrac{dy}{dx}\, =\, x^2\dfrac{dy}{dx}\, +\, y^2 \dfrac{dy}{dx}\)

. . . . .\(\displaystyle 2xy\, =\, x^2\dfrac{dy}{dx}\, +\, y^2\dfrac{dy}{dx}\, -\, 2y^2\dfrac{dy}{dx}\)

Simplify the right-hand side. What do you get?

Ah, OK, thanks Stapel..!

 

[MarkFL]

Another method would be to divide through by \(\displaystyle y\) to obtain:

\(\displaystyle \displaystyle \frac{x^2}{y}+y=A\)

Implicitly differentiate with respect to \(\displaystyle x\):

\(\displaystyle \displaystyle \frac{2xy-x^2y'}{y^2}+y'=0\)

Multiply through by \(\displaystyle y^2\):

\(\displaystyle 2xy-x^2y'+y'y^2=0\)

Arrange as:

\(\displaystyle \left(x^2-y^2 \right)y'=2xy\)

 

[jonnburton]

Thanks MarkFL, it's always good to see different ways of doing things...

Another method would be to divide through by \(\displaystyle y\) to obtain:

\(\displaystyle \displaystyle \frac{x^2}{y}+y=A\)

Implicitly differentiate with respect to \(\displaystyle x\):

\(\displaystyle \displaystyle \frac{2xy-x^2y'}{y^2}+y'=0\)

Multiply through by \(\displaystyle y^2\):

\(\displaystyle 2xy-x^2y'+y'y^2=0\)

Arrange as:

\(\displaystyle \left(x^2-y^2 \right)y'=2xy\)

 

[JeffM]

Hi,

I've just started this topic and don't quite understand it. I was wondering whether anyone could explain a couple of things.

The worked example my book gives is as follows:

By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to \(\displaystyle x^2 + y^2 = Ay\)



Solution:

Differentiating both sides of \(\displaystyle x^2 + y^2 = Ay\) with respect to x gives:

\(\displaystyle 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}\)

Multiplying throughout by y (I presume this is done in order to get the term 'Ay' back)

\(\displaystyle 2xy+2y^2 \frac{dy}{dx}=Ay\frac{dy}{dx}\)

And using the given equation: \(\displaystyle 2xy + 2y^2\frac{dy}{dx} = (x^2+y^2)\frac{dy}{dx}\)

so the required first order differential equation is:

\(\displaystyle (x^2 - y^2) \frac{dy}{dx} = 2xy\)


In this example it's really the last step that I don't understand. How does \(\displaystyle (x^2+y^2)\) become \(\displaystyle (x^2-y^2)\)? and how does the \(\displaystyle 2y^2\frac{dy}{dx}\) disappear?


I began to try the first question that follows this worked example: By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to \(\displaystyle y = Ax\)

Trying to follow a similar procedure, first I took the derivative with respect to x:

\(\displaystyle y \frac{dy}{dx} = (1)\frac{dy}{dx}A\) Your mistake is here

\(\displaystyle y = Ax \implies \dfrac{d(y)}{dy} * \dfrac{dy}{dx} = \dfrac{d(Ax)}{dx} \implies 1 * \dfrac{dy}{dx} = A \implies \dfrac{dy}{dx} = A.\)

Then I multiplied by x, in order to get Ax back:
\(\displaystyle xy\frac{dy}{dx} = \frac{dy}{dx}Ax\) but then this led to:

\(\displaystyle xy\frac{dy}{dx}=y\frac{dy}{dx}\)

I know that the answer is \(\displaystyle \frac{dy}{dx}=\frac{x}{y}\) because \(\displaystyle \frac{dx}{dt} = kx\) But I can't see how to get there following this approach.


Any help would be much appreciated!

.

 

[jonnburton]

.

Thanks Jeff for pointing that out. I think I'm slowly getting there now...