Question

Hello I got this question from Binmore and Davies' text on Calculus, Chapter 12 Differential Equations.

If the following differential equation is not separable but is homogeneous of degree 3:

(x^3 + y^3) = 3xy^2 (dy/dx)

[Qn1. How do I know I should substitute in "y=vx"?]
So dy/dx = v + x (dv/dx)

Then
(x^3 + v^3 x^3) = 3x * v^2 x^2 (v + x dv/dx)
(1+v^3) - 3v^3 - 3v^2 *x (dv/dx) = 0
(1-2v^3) = 3v^2 *x (dv/dx)

Separating the variables
1/x *dx = (3v^2)/(1-2v^3) *dv

On Integrating
ln |x| = -1/2 ln |1-2v^3| + ln c
x * (1-2v^3)^1/2 = c

Substitute back "y=vx", ie "v=y/x"

Answer 1

We get
x (1- 2(y/x)^3)^1/2 = c
OR if we square BOTH sides
x^2 * (1- 2(y/x)^3) = c

x^2 - 2y^3 /x = c
x^3 - 2y^3 = cx

If boundary condition given is (1,0),
1 - 0 = c
c=1
So the solution should be x^3 - 2y^3 = x

Answer 2

Q2. I was wondering whether you guys know how to do questions about "Change of variables".

I do not understand why are we changing the roles of x and y (for example 10), and how do I go about changing them.

Answer 3

And my third and last question for now. For example 13, I understand that we have to differentiate both sides with respect to x. But I do not get how the outcome of the differentiation is
\[1/x * deltaf/deltax - 1/x^2 (f-y) =0\]

And also, how do I know when to differentiate the whole equation partially with respect to x, and when do I do this wrt y?

Answer 4

Please have you got the initial condition of the correspondent Cauchy problem?

Answer 5

For the first 2 posts, the initial condition is y(1)=0.

Answer 6

Ok! in order to solve your differential equation, we can use the theorem of implicit functions. From your differential equation (DE for short), we can write:
\[dx(x ^{3}+y ^{3})-3xy ^{2}dy\]
whic is not an exact differential form, our idea is getting from the above differential form an exact differential form, then using the theorem of implicit function, we get our solution.
If we multiply the above differential form, by the function \[\frac{ 1 }{ x ^{2} }\]
we get an exact differential form, namely:
the new differential form which is exact, is
\[\frac{ x ^{3} +y ^{3}}{ x ^{2} }dx-3\frac{ y ^{2} }{ x }dy=dF\]
now all primitives of the above exact differential form, are:
\[F(x,y)=\frac{ x ^{2} }{ 2 }-\frac{ y ^{3} }{ x }+k, k \in \mathbb{R} \]
but we want the form one, which is equal to zero, at the point (x,y(x))=(1,0),
so our differential form is:
\[F(x,y)=\frac{ x ^{3} }{ 2 }-\frac{ y ^{3} }{ x }-\frac{ 1 }{ 2 }\]
equating to zero, the above exact differential form, we get the subsequent expression for y:
\[y(x)=\sqrt[3]{\frac{ x ^{3} -x}{ 2 }}\]
that's your solution

Answer 7

@subbie

Answer 8

sorry:
\[F(x)=\frac{ x ^{2} }{ 2 }-\frac{ y ^{3} }{ x }-\frac{ 1 }{ 2 }\]

Answer 9

@Michele_Laino while you did find the solution to the DE, @subbie was more concerned with solving the equation using the homogeneous substitution, \(y=vx\), rather than approaching it as an exact equation.

"Qn1. How do I know I should substitute in 'y=vx'?"
For me, it's usually a matter of trial and error, and sometimes just a matter of pattern recognition.

In the given DE, \(x^3+y^3=3xy^2\dfrac{dy}{dx}\), the substitution works nicely because it eliminates factors of \(x^3\), while making an equation separable in \(v\) and \(x\):
\[\begin{align*}
x^3+y^3&=3xy^2\frac{dy}{dx}\\\\
1+\left(\frac{y}{x}\right)^3&=3\left(\frac{y}{x}\right)^2\frac{dy}{dx}&(*)
\end{align*}\]
I'm not a fan of the way your text describes homogeneous equations. I think a more useful way to identify them is if you can write the DE as a function of powers of the ratio between the dependent and independent variables, i.e.
\[\frac{dy}{dx}=f\left(\frac{y}{x}\right)\]
and this is readily seen in the form of equation \((*)\). From here, I find the transition of \(y,x\) to \(v,x\) more intuitive, letting \(y=vx\), or \(v=\dfrac{y}{x}\).

Answer 10

"Q2. I was wondering whether you guys know how to do questions about Change of variables'."

Referring to the equation given in the text:
\[y\ln y+(x-\ln y)\frac{dy}{dx}=0\]
As the text mentions, the equation is nonlinear in \(y\), and nonlinear ODEs are generally much harder to solve than linear ones. Fortunately, we can convert from one variable to another. To do this (without involving another variable through substitution), we solve for \(\dfrac{dx}{dy}\) instead of \(\dfrac{dy}{dx}\):
\[\begin{align*}y\ln y+(x-\ln y)\frac{dy}{dx}&=0\\\\
(x-\ln y)\frac{dy}{dx}&=-y\ln y\\\\
(\ln y-x)\frac{dy}{dx}&=y\ln y\\\\
\frac{1}{\ln y-x}\frac{dx}{dy}&=\frac{1}{y\ln y}\\\\
\frac{dx}{dy}&=\frac{\ln y-x}{y\ln y}\\\\
\frac{dx}{dy}&=\frac{1}{y}-\frac{x}{y\ln y}\\\\
\frac{dx}{dy}+\frac{x}{y\ln y}&=\frac{1}{y}
\end{align*}\]
I don't see much usefulness for this approach outside of this nonlinear-to-linear conversion, but I might be wrong.

Answer 11

As for your third question, PDEs are not my forte, sorry!

Answer 12

For your 3rd question on the PDE. Here is a more explicit differentiation:\[f(x,y) = g(y)x+y\]
\[\frac{f-y}{x}=g(y)\]
\[\frac{∂}{∂x}[\frac{f(x,y)-y}{x}]=\frac{∂}{∂x}[g(y)]=0\]
\[\frac{∂}{∂x}[\frac{f(x,y)}{x}]-\frac{∂}{∂x}[\frac{y}{x}]\]
\[-\frac{f(x,y)}{x^{2}}+\frac{1}{x}\frac{∂f(x,y)}{∂x}+\frac{y}{x^{2}}=0\]
\[\frac{1}{x^{2}}[y-f(x,y)]+\frac{1}{x}\frac{∂f(x,y)}{∂x}=0\]
\[-\frac{1}{x^{2}}[-(y-f)]+\frac{1}{x}\frac{∂f}{∂x}=0\]
\[\frac{1}{x}\frac{∂f}{∂x}-\frac{1}{x^{2}}(f-y)=0\]
Hence (by multiplying through by x^2 and moving (f-y) to right side):\[x\frac{∂f}{∂x}=f-y\]