Form the differential equation from the given cartesian equation:

Question

Form the differential equation from the given cartesian equation:
$$y=ax^3+bx^2$$

My Attempt:
Given,
$$y=ax^3+bx^2$$
Differentiating both sides with respect to $x$
$$\dfrac {dy}{dx}=3ax^2+2bx$$
Again differentiating both sides with respect to $x$
$$\dfrac {d^2 y}{dx^2}=6ax+2b$$

Answer 1

$$y=ax^3+bx^2\tag1$$ $$y'=3ax^2+2bx\tag2$$ $$y''=6ax+2b\tag3$$ From $(3)$, $~xy''=6ax^2+2bx=2y'-2bx\tag4$

From $(2)$, $$xy'=3ax^3+2bx^2=3y-bx^2$$ $$\implies xy'=3y+\dfrac{1}{2}(x^2y''-2xy')\qquad \text{by $~(4)$}$$ $$\implies x^2y''-4xy'+6y=0$$ This is the required differential equation.