Find the general solution of y''' + x y'' = 2y'

Find the general solution of $y''' + x\,y'' = 2y'$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$ and $p'' = y'''$. The equation becomes: \[p'' + x\,p' = 2p. \tag{1}\]

Step 2: Try a polynomial solution

Look for a polynomial particular solution. Try $p = ax^2 + bx + c$: $p' = 2ax + b,\qquad p'' = 2a.$

Substitute into (1): \[2a + x(2ax + b) = 2(ax^2 + bx + c),\qquad 2a + 2ax^2 + bx = 2ax^2 + 2bx + 2c.\]

Equate coefficients: $x^2$: $2a=2a$ $\checkmark$;\ \ $x^1$: $b = 2b \Rightarrow b=0$;\ \ $x^0$: $2a = 2c \Rightarrow c = a$.

So, up to a constant factor: $p_1 = x^2 + 1.$

Step 3: Reduce the order

Since $p_1 = x^2+1$ is one solution of (1), use reduction of order. Set $p = (x^2+1)\,v(x)$: $p' = 2x\,v + (x^2+1)\,v',\qquad p'' = 2v + 4x\,v' + (x^2+1)\,v''.$

Substitute into (1) and simplify — the $v$ terms cancel: \[(x^2+1)\,v'' + x(x^2+5)\,v' = 0. \tag{2}\]

Step 4: Solve the reduced equation

Let $w = v'$, so $w' = v''$. Equation (2) becomes separable: $\frac{dw}{w} = -\frac{x(x^2+5)}{x^2+1}\,dx.$

Decompose the right-hand side: $\frac{x(x^2+5)}{x^2+1} = x + \frac{4x}{x^2+1}.$

Integrate: \[\ln|w| = -\frac{x^2}{2} - 2\ln(x^2+1) + \mathrm{const},\qquad w = \frac{A\,e^{-x^2/2}}{(x^2+1)^2},\] where $A$ is a constant. So $v' = \dfrac{A\,e^{-x^2/2}}{(x^2+1)^2}$.

Step 5: Recover $p$ and $y$

\[ p = y' = C_1(x^2+1) + C_2(x^2+1)\int_0^x \frac{e^{-t^2/2}}{(t^2+1)^2}\,dt. \]

Since $\displaystyle\int \frac{e^{-x^2/2}}{(x^2+1)^2}\,dx$ has no closed elementary form, the general solution is expressed with a definite integral.

Step 6: General solution

The general solution is recorded in the result below, where $C_1$, $C_2$, and $C_3$ are arbitrary constants.

Step 7: Verification of the particular solution

For $y = \dfrac{x^3}{3} + x$: $y' = x^2+1$, $y'' = 2x$, $y''' = 2$.

\[ y''' + x\,y'' - 2y' = 2 + x\cdot 2x - 2(x^2+1) = 2 + 2x^2 - 2x^2 - 2 = 0.\ \checkmark \]