A particular solution of y' + y^2 - 3ytan x + tan^2 x - 1 =…

A particular solution of \[ y' + y^2 - 3y\tan x + \tan^2 x - 1 = 0 \] has the form \(y=k\tan x\) for a constant \(k\). Find the general solution.

1. Determine \(k\)

Substituting \(y=k\tan x\), \(y'=k\sec^2x=k(1+\tan^2x)\) into the equation and collecting powers of \(\tan^2x\) gives \[ (k-1) + (k-1)^2\tan^2x = 0 \quad\Longrightarrow\quad (k-1)\big[1+(k-1)\tan^2x\big]=0, \] which must hold for all \(x\), forcing \(k=1\). (Confirmed with Sympy's \(\texttt{Poly}\) on the substituted expression: the coefficient of \(\tan^2x\) is \((k-1)^2\) and the constant term is \(k-1\), both vanishing only at \(k=1\).) So \(y_1=\tan x\) is the particular solution, and direct back-substitution gives \(0=0\).

2. Riccati substitution \(y=\tan x + 1/v\)

Substituting \(y=\tan x+1/v\) into the original equation and simplifying (using \(\sec^2x=1+\tan^2x\)) — verified symbolically with Sympy by expanding \(y'+y^2-3y\tan x+\tan^2x-1\) with \(y=\tan x+1/v(x)\) and simplifying — the result collapses exactly to \[ \frac{1 - v\tan x - v'}{v^2} = 0 \quad\Longleftrightarrow\quad \frac{dv}{dx} + v\tan x = 1, \] confirming the source's reduction step is correct.

3. Solve the linear equation for \(v\)

With integrating factor \(\mu=e^{\int\tan x\,dx}=\sec x\), \[ \frac{d}{dx}(v\sec x) = \sec x \quad\Longrightarrow\quad v\sec x = \ln|\sec x+\tan x| + C, \] so \(v = \cos x\,\big(\ln|\sec x+\tan x|+C\big)\). (Sympy confirms \(\tfrac{d}{dx}\ln(\sec x+\tan x)=\sec x\) and that this \(v\) satisfies \(v'+v\tan x=1\) identically.)

4. Back-substitute and verify against the ORIGINAL equation

\(y=\tan x+1/v\) gives \[ y = \tan x + \frac{\sec x}{\ln|\sec x+\tan x|+C}. \] Because step 2 showed the substitution \(y=\tan x+1/v\) turns the original nonlinear equation into \(v'+v\tan x-1=0\) exactly (with the same nonzero factor \(1/v^2\) throughout, introducing no extraneous branches), and step 3 showed this specific \(v\) satisfies \(v'+v\tan x=1\) exactly, chaining these two verified facts together proves the final \(y\) solves the original Riccati equation — this closes the gap the source left (it never checked the final answer against the original nonlinear ODE).