Q.(a) Solve the equation $6x^4-5x^3-38x^2-5x+6=0$ if it is known that $\dfrac13$ is a solution. OR

(a) Uses the reciprocal (palindromic) symmetry of the coefficients to get a second root from the given one, divides it out, and factorises the remaining quadratic; (b) reduces the homogeneous ODE with $y=vx$ to a separable equation and integrates. Both alternatives answered below.

(a) Solve $6x^4-5x^3-38x^2-5x+6=0$, given $x=\dfrac13$ is a root

1. Verify $x=\dfrac13$. $6\left(\dfrac1{81}\right)-5\left(\dfrac1{27}\right)-38\left(\dfrac19\right)-5\left(\dfrac13\right)+6=\dfrac{6-15-342-135+486}{81}=\dfrac0{81}=0$ ✓.

2. Reciprocal equation. The coefficients $6,-5,-38,-5,6$ read the same forwards and backwards, so this is a reciprocal (palindromic) equation of even degree: if $r$ is a root, so is $\dfrac1r$. Since $\dfrac13$ is a root, $x=3$ must also be a root.

3. Verify $x=3$. $6(81)-5(27)-38(9)-5(3)+6=486-135-342-15+6=0$ ✓.

4. Form the quadratic factor from these two roots: $(3x-1)(x-3)=3x^2-10x+3$.

5. Divide the quartic by $3x^2-10x+3$.

$6x^4-5x^3-38x^2-5x+6=(3x^2-10x+3)(2x^2+5x+2)$

(by polynomial long division: quotient terms $2x^2,5x,2$, each subtraction leaving zero remainder.)

6. Factor the remaining quadratic. $2x^2+5x+2=(2x+1)(x+2)$, giving roots $x=-\dfrac12,\,x=-2$.

7. All four roots. $x=\dfrac13,\ 3,\ -\dfrac12,\ -2$.

(b) Solve $(x^2-3y^2)\,dx+2xy\,dy=0$

1. Rewrite as $\dfrac{dy}{dx}$. $2xy\,dy=-(x^2-3y^2)dx\Rightarrow\dfrac{dy}{dx}=\dfrac{3y^2-x^2}{2xy}$ — homogeneous (degree $0$ in $y/x$).

2. Substitute $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$:

$v+x\dfrac{dv}{dx}=\dfrac{3v^2x^2-x^2}{2x(vx)}=\dfrac{3v^2-1}{2v}$

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