Can we use y=vx for non-homogeneous differential equation?

jasoncurious
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Homework Statement


Can we use y=vx for non-homogeneous differential equation?

Example:
yy'=x^3+(y^2/x)→not homogeneous


Homework Equations


y=vx
dy/dx=v+x(dv/dx)

The Attempt at a Solution


By substituting the equation above:
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
 
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The substitution y=vx makes it easier to find the solution, because a homogenous differential equation takes the form:

##\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}##

which then equals ##\frac{f(y/x)}{g(y/x)}## or ##\frac{f(x/y)}{g(x/y)}## by taking ##x^n## or ##y^n## common, if f1 and f2 are homogenous in degree n.
 
jasoncurious said:

Homework Statement


Can we use y=vx for non-homogeneous differential equation?

Example:
yy'=x^3+(y^2/x)→not homogeneous


Homework Equations


y=vx
dy/dx=v+x(dv/dx)

The Attempt at a Solution


By substituting the equation above:
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

Easier: ##v \, dv/dx = d(v^2/2)/dx.## And, there are two solutions.
 
May I know the hint to the other solution? Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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