Solve the differential equation x(2x-1)y' + y^2 - (4x+1)y +…

Solve the differential equation $x(2x-1)y' + y^2 - (4x+1)y + 4x = 0$, given that one particular solution is $y=1$.

Explanation of Method

This is a Riccati equation. Given the particular solution $y_1=1$, the substitution $y = 1 + 1/v$ reduces it to a first-order linear equation for $v(x)$, which is solved using an integrating factor found by partial fractions.

Verify the particular solution

For $y=1$, $y'=0$, and the left side becomes $0 + 1 - (4x+1) + 4x = 0$. So $y=1$ indeed solves the equation.

Substitute $y = 1+1/v$

With $y = 1+\frac1v$, $y' = -\frac{1}{v^2}v'$. Substituting into the ODE and simplifying (the constant terms cancel because $y_1=1$ is a solution) leads, after multiplying through by $v^2$, to

\[x(2x-1)v' = v(1-4x) + 1,\]

a first-order linear equation for $v$.

Find the integrating factor

Writing the equation as $v' - \dfrac{1-4x}{x(2x-1)}v = \dfrac{1}{x(2x-1)}$, partial fractions give $\dfrac{1-4x}{x(2x-1)} = -\dfrac1x - \dfrac{2}{2x-1}$, so $\displaystyle\int\frac{1-4x}{x(2x-1)}dx = -\ln|x(2x-1)|$, and the integrating factor is $\mu(x) = x(2x-1)$.

Integrate the linear equation

Multiplying by $\mu$ makes the left side exact: $\dfrac{d}{dx}\bigl[vx(2x-1)\bigr] = 1$, so $vx(2x-1) = x+C$, giving

\[v = \frac{x+C}{x(2x-1)}.\]

Back-substitute for $y$

Since $y = 1+1/v = 1 + \dfrac{x(2x-1)}{x+C} = \dfrac{x+C+2x^2-x}{x+C} = \dfrac{2x^2+C}{x+C}$.

Verify with sympy

Substituting $y=\dfrac{2x^2+C}{x+C}$ and its derivative directly into $x(2x-1)y'+y^2-(4x+1)y+4x$ and simplifying symbolically yields exactly $0$ for all $x$ and $C$, confirming the result. (The special case $C=0$ gives $y=2x$, which also checks directly.)

Result

Final answer:

\[\boxed{y = \frac{2x^2 + C}{x+C}}\]

where $C$ is an arbitrary constant. This was confirmed by direct symbolic substitution into the original Riccati equation, which reduces identically to $0$. The source's derivation and final answer are correct as given.