ODE

[jasoncurious]

Convert the
following differential equation into separable form by using suitable substitution:

yy’=x³+(y²/x)

When I try
to prove it’s being homogeneous or not,
I found that

yy’=x³+

y’=

(lx)⁴+(ly)²/ (lx)(ly)

Factorise it:

(l^2x²)+(y^2) / xy

Is this
considered homogeneous?

the l stands for lambda

[Nehushtan]

The differential equation is clearly not homogeneous and I don’t see what homogeneity has got to do with the solution.

Hint: Write the equation as [latex]2yy'-\frac{2}xy^2=2x^3[/latex] and find an integrating factor for the left-hand side. This will tell you exactly what substitution you need to use.

Edited March 14, 2013 by Nehushtan

[jasoncurious]

Thank you, I think I was caught in the thought that "only homogeneous differential equations can be solved".

[Nehushtan]

You’re welcome.

Solving differential equations often requires thinking outside the box. If a particular equation does not fit into any classifed category, it doesn’t mean it can’t be solved – you may need to try some other method.

Edited March 15, 2013 by Nehushtan

[jasoncurious]

So, even if the differential equation is not homogeneous, we can use y=u*v as well?

 

This is howthe solution starts:

yy'=x^3+(y^2)/x

Let y=vx

dy/dx=v+x dv/dx

Then the rest is substitution.

[Nehushtan]

yy'=x^3+(y^2)/x

Let y=vx

dy/dx=v+x dv/dx

 

Then the rest is substitution.

No, [latex]y=vx[/latex] doesn’t work here. Did you find the integrating factor I told you to find earlier in the thread?

 

In other words, multiply the expression

 

[latex]2yy'-\frac2xy^2[/latex]

 

by a function [latex]u(x)[/latex] of [latex]x[/latex] only such that

 

[latex]u(x)\cdot2yy'-u(x)\frac2xy^2\,=\,\frac{\mathrm d}{\mathrm dx}\!\left(u(x)y^2\right)[/latex]

 

[latex]u(x)[/latex] is called an integrating factor. (Hint for finding it: Clearly you want [latex]u'=-\frac2xu[/latex].) When you’ve found [latex]u(x)[/latex], the substitution you require for the original problem should be obvious.

Edited March 15, 2013 by Nehushtan

[jasoncurious]

y=vx
dy/dx=v+x dv/dx
vx(v+x dv/dx)=x^3+(v^2 x^2)/x
v^2*x+vx^2 dv/dx=x^3+v^2*x
Eliminate the v^2*x:
vx^2 dv/dx=x^3
Divide both sides with x^2:
v dv/dx=x
vdv=xdx
Continue the integration:
y^2=x^2(x^2+c), where c is a constant

[jasoncurious]

I thought the d.e has to be in the form of dy/dx+P(x)y=Q(x) for integrating factor to be implemented?

[Nehushtan]

Well, it looks like your substitution works perfectly after all.

I must have made a careless mistake and didn’t get the [latex]v^2x[/latex] to cancel, ending up by going through a slightly more complicated path to solve the ODE. Just as well you didn’t take my advice.

Edited March 17, 2013 by Nehushtan

[jasoncurious]

I saw my friend solving it using Bernoulli equation, guess everything is possible in mathematics