What is the general solution of the differential equation # (x^2+y^2)dx+(x^2-xy)dy = 0 #?

Question

Paisley Johnson · Accepted Answer

# y/x-2ln(y/x+1) = lnx + C #

We have: # (x^2+y^2)dx+(x^2-xy)dy = 0 # We can rearrange this Differential Equation as follows: # dy/dx = - (x^2+y^2)/(x^2-xy) # # " " = - ((1/x^2)(x^2+y^2))/((1/x^2)(x^2-xy)) # # " " = - (1+(y/x)^2)/(1-y/x) # So Let us try a substitution, Let: # v = y/x => y=vx# Then: # dy/dx = v + x(dv)/dx # And substituting into the above DE, to eliminate #y#: # v + x(dv)/dx = - (1+v^2)/(1-v) # # " " = (1+v^2)/(v-1) # # :. x(dv)/dx = (1+v^2)/(v-1) - v# # :. " " = {(1+v^2) - v(v-1)}/(v-1)# # :. " " = {(1+v^2 - v^2+v)}/(v-1)# # :. " " = (v+1)/(v-1)# # :. (v-1)/(v+1) \ (dv)/dx = 1/x # This is now a separable DIfferential Equation, and so "separating the variables" gives us: # int \ (v-1)/(v+1) \ dv = \ int \ 1/x \ dx # This is now a trivial integration problem, thus: # int \ (v+1-2)/(v+1) \ dv = \ int \ 1/x \ dx # # int \ 1-2/(v+1) \ dv = \ int \ 1/x \ dx # # v-2ln(v+1) = lnx + C # And restoring the substitution we get: # y/x-2ln(y/x+1) = lnx + C #

Samantha Adkison · Answer

See below.

Making the substitution #y = lambda x# we have #dy = lambda dx + x d lambda# then #(x^2+y^2)dx+(x^2-xy)dy equiv x^2(1+lambda^2)dx+x^2(1-lambda)(lambda dx + x dlambda)# or assuming #x ne 0# #(1+lambda^2+lambda(1-lambda))dx+x(1-lambda) dlambda = 0# or #(1+lambda)dx+x(1-lambda)dlambda=0# This is a separable differential equation so #((1-lambda)/(1+lambda))dlambda = -dx/x # and integrating both sides #2log(lambda+1)-lambda = log x + C# or #(lambda+1)^2/lambda = C_1 x# and solving for #lambda# #lambda = y/x = 1/2(C_1 x pm sqrt(C_1 x) sqrt(C_1x-4))# then #y = x /2(C_1 x pm sqrt(C_1 x) sqrt(C_1x-4))#

Isaiah Allen · Answer

undefined To solve the differential equation $(x^2+y^2)dx + (x^2 - xy)dy = 0$, we can start by rearranging it into a more manageable form. The goal is to find a solution $y(x)$ or to express the relationship between $x$ and $y$ implicitly.

Given:
$$ (x^2 + y^2)dx + (x^2 - xy)dy = 0 $$

Rearrange it to separate variables or find an integrating factor that simplifies the equation. This equation is not immediately separable, nor is it exact as is. However, we can attempt to make it exact by identifying an integrating factor or through manipulation.

In this case, it's helpful to notice that both terms involve $x^2$, suggesting we might want to look for symmetry or a way to combine terms efficiently. However, directly solving this as presented without a clear method to separate variables or an obvious integrating factor suggests we look at alternative methods such as substitution or transformation.

Given the symmetry in the equation and the mix of $x^2$ and $y^2$ terms, one might be tempted to use polar coordinates ($x = r\cos(\theta)$, $y = r\sin(\theta)$), but the mixed $xy$ term complicates this approach directly.

Another approach could be to try and identify a function or form that this equation represents directly, but without clear separability, exactness, or a linear form, we fall back on intuition or advanced methods that could involve looking at the equation from the perspective of differential forms or leveraging more complex transformations.

To solve this equation directly here without stepping into methods that require a deeper dive into advanced mathematical theories or transformations that aren't immediately apparent from the equation as given, we would need more context or to apply a specific technique tailored to the equation's structure. In standard undergraduate courses, an equation of this form might lead to a teaching moment about recognizing when and how to apply certain techniques like finding an integrating factor for exact equations, using substitutions effectively, or recognizing patterns that suggest a move to polar coordinates or another coordinate system might simplify the problem.

Without applying a specific, advanced method or a clever substitution that isn't immediately apparent from the form of the equation given, we acknowledge this problem's complexity and the potential need for a deeper exploration of mathematical techniques beyond the basic separation of variables, integration factors, or simple transformations.