Show that the equation (x + 2y) dx + y dy = 0 has an…
Show that the equation \[ (x + 2y)\,dx + y\,dy = 0 \] has an integrating factor of the form \(\mu = \mu(x+y)\) and find its general integral.
Explanation of Method
With \(M=x+2y\), \(N=y\), \(u=x+y\), the exactness condition for \(\mu(u)M\,dx+\mu(u)N\,dy=0\) reduces to the separable ODE \(\mu'(u)/\mu(u)=-2/u\). Every algebraic step of this derivation — the exactness failure of the original equation, the coefficient PDE for \(\mu\), its solution, the exactness of the multiplied equation, the potential function, and finally an implicit-differentiation check of the general integral against the original ODE — was independently reproduced and confirmed with Sympy. No error was found anywhere in the source's derivation; it is presented here in the same structure with the verification made explicit.
1. The equation is not exact
\(M=x+2y\), \(N=y\): \(M_y=2\), \(N_x=0\). Since \(M_y\neq N_x\), \((x+2y)\,dx+y\,dy=0\) is not exact.
2. Derive the ODE for \(\mu(u)\), \(u=x+y\)
Requiring \(\partial_y[\mu(u)M]=\partial_x[\mu(u)N]\) with \(u_x=u_y=1\) gives \[ \mu'(u)(x+2y)+2\mu(u) = \mu'(u)y \;\Longrightarrow\; \mu'(u)(x+y) = -2\mu(u) \;\Longrightarrow\; \frac{\mu'(u)}{\mu(u)} = -\frac{2}{u}. \] (Sympy, differentiating \(\mu(u)M\) and \(\mu(u)N\) directly with \(\mu\) an unspecified \(\texttt{Function}\) and \(u=x+y\), reproduces this identity exactly.)
3. Solve for \(\mu\)
\(d\mu/\mu=-2\,du/u \Rightarrow \ln|\mu|=-2\ln|u| \Rightarrow \mu(u)=1/u^2\) (taking the constant of integration to be \(0\), the standard normalization). So \[ \boxed{\mu(x+y)=\frac{1}{(x+y)^2}}. \]
4. Confirm exactness of the multiplied equation
With \(\tilde M=\dfrac{x+2y}{(x+y)^2}\), \(\tilde N=\dfrac{y}{(x+y)^2}\), Sympy confirms \(\partial\tilde M/\partial y-\partial\tilde N/\partial x=0\) identically.
5. Find the potential function
Integrating \(\tilde N\) in \(y\) and matching \(\partial F/\partial x=\tilde M\) (both checked directly with Sympy, giving residual \(0\)) gives \[ F(x,y) = \ln|x+y| + \frac{x}{x+y}. \]
6. Verify against the original ODE by implicit differentiation
Differentiating \(F(x,y)=C\) implicitly (Sympy: solve \(F_x+F_yy'=0\) for \(y'\)) gives \(y'=-x/y-2\), which is exactly \(-(x+2y)/y\) — i.e. the original equation \((x+2y)+yy'=0\) rearranged. The residual is \(0\), confirming the general integral solves the original ODE, not just the multiplied one.
Result
Final answer:
Integrating factor: \[ \boxed{\ \mu(x+y)=\frac{1}{(x+y)^2}\ } \] General integral (confirmed by implicit differentiation to reproduce the original ODE): \[ \boxed{\ \ln|x+y| + \frac{x}{x+y} = C\ } \]