For the Riccati equation…

For the Riccati equation $y'-\dfrac{1}{1-x^3}y^2+\dfrac{x^2}{1-x^3}y+\dfrac{2x}{1-x^3}=0$: (1) determine $a$ so that $y=ax^2$ is a particular integral and find the general solution; (2) determine $a,b$ so that $y=ax+b$ is a particular integral and find the general solution.

Explanation of Method

Substitute each trial form, match coefficients to fix the constants, then apply the Riccati reduction $y=y_1+1/u$ to obtain and solve a linear equation for $u(x)$ in each case.

Part 1, Step 1: Find $a$

Substituting $y=ax^2$ and collecting powers of $x$ forces $a+1=0$, so $a=-1$, i.e. $y_1=-x^2$.

Part 1, Step 2: Reduce and solve for $u$

With $R=\dfrac{1}{1-x^3}$, $Q=-\dfrac{x^2}{1-x^3}$: $Q+2Ry_1=-\dfrac{3x^2}{1-x^3}$, giving $u'-\dfrac{3x^2}{1-x^3}u=-\dfrac{1}{1-x^3}$. With $\mu(x)=1-x^3$: $\dfrac{d}{dx}[(1-x^3)u]=-1$, so $(1-x^3)u=-x+C$.

Part 1, Result

\[y=-x^2+\frac{1-x^3}{C-x}=\boxed{\frac{1-Cx^2}{C-x}}.\]

Part 2, Step 1: Find $a,b$

Substituting $y=ax+b$ and matching coefficients of $x^0,x^1,x^2$ gives $a=b^2$, $b=a^2$, $ab=1$, whose only consistent solution is $a=b=1$, i.e. $y_1=x+1$.

Part 2, Step 2: Reduce and solve for $u$ (note: this differs from Part 1)

Here $Q+2Ry_1=\dfrac{-x^2+2x+2}{1-x^3}$, giving $u'+\dfrac{-x^2+2x+2}{1-x^3}u=-\dfrac{1}{1-x^3}$. Solving this linear equation (its integrating factor is not the simple power $1-x^3$ used in Part 1, since $y_1$ is different) yields $u=\dfrac{C(x-1)-1}{x^2+x+1}$.

Result

Final answer:

Part 1 ($y_1=-x^2$): \[\boxed{\,y=\frac{1-Cx^2}{C-x}}\]

Part 2 ($y_1=x+1$):

\[\boxed{\,y=x+1+\frac{x^2+x+1}{C(x-1)-1},\qquad C\in\mathbb{R}}\]

Note: a commonly circulated version of Part 2's answer simply reuses Part 1's formula ($y=x+1+\frac{1-x^3}{C-x}$); this does not satisfy the equation — the correct linear equation for $u$ in Part 2 has different coefficients (since $y_1$ differs), and the boxed formula above has been verified symbolically to satisfy the original Riccati equation exactly.