Calculus problem No.1

Let $y'(x) =z$

Then the given equation reduces to :

$z'(x) -\dfrac zx=x$

Integrating factor of this equation is

$e^{-\displaystyle \int \frac{dx}{x}}=\dfrac 1x$

Solution to the equation is

$z=y'(x) =x^2+C_1x$

$\implies y=\dfrac {x^3}{3}+\dfrac {C_1}{2}x^2+\zeta_2$

$=\boxed {\dfrac 13 x^3+\zeta_1x^2+\zeta_2}$ .

$\begin{aligned} y''(x) - \frac {y'(x)}x & = x \\ \frac {d^2 y}{dx^2} - \frac 1x \cdot \frac {dy}{dx} & = x & \small \blue{\text{Let }\frac {dy}{dx} = v(x) \implies \frac {d^2 y}{dx^2} = \frac {dv(x)}{dx}} \\ \frac {dv}{dx} - \frac vx & = x & \small \blue{\text{Multiply both sides by }\mu(x) = e^{\int -1/x \ dx} = \frac 1x} \\ \frac 1x \cdot \frac {dv}{dx} - \frac v{x^2} & = 1 \\ \frac d{dx} \left(\frac {v(x)}x \right) & = 1 \\ \int \frac d{dx} \left(\frac {v(x)}x \right) dx & = \int 1 \ dx \\ \frac {v(x)}x & = x + c_1 \\ v(x) & = x^2 + c_1 x \\ \frac {dy}{dx} & = x^2 + c_1x \\ \implies y(x) & = \int \left(x^2 + c_1x \right) dx \\ & = \boxed{\frac {x^3}3 + \zeta_1 x^2 + \zeta_2} \end{aligned}$