Solve the differential equation y''' = 2x y''

Solve the differential equation $y''' = 2x\,y''$.

Step 1: Substitution

Let $p = y''$, so $p' = y'''$. The equation becomes $p' = 2x\,p$, a first-order separable ODE in $p(x)$.

Step 2: Solve for $p$

Separating variables, $\dfrac{dp}{p} = 2x\,dx$. Integrating gives $\ln|p| = x^2 + \text{const}$, so $p = C_1 e^{x^2}$, i.e.

\[y'' = C_1 e^{x^2}.\]

Step 3: First integration

\[y' = C_1\int e^{x^2}\,dx + C_2.\]

Since $\int e^{x^2}\,dx$ has no elementary closed form, write it as a definite integral: $y' = C_1\int_0^x e^{t^2}\,dt + C_2.$

Step 4: Second integration

\[y = C_1\int_0^x\left(\int_0^s e^{t^2}\,dt\right)ds + C_2 x + C_3.\]

Integration by parts with $u = \int_0^s e^{t^2}\,dt$ and $dv = ds$ yields

\[\int_0^x\left(\int_0^s e^{t^2}\,dt\right)ds = x\int_0^x e^{t^2}\,dt - \int_0^x t\,e^{t^2}\,dt,\]

and $\displaystyle\int_0^x t\,e^{t^2}\,dt = \dfrac12(e^{x^2}-1)$.

Step 5: General solution

\[y = C_1\left(x\int_0^x e^{t^2}\,dt - \frac{e^{x^2}-1}{2}\right) + C_2 x + C_3,\] where $C_1$, $C_2$, $C_3$ are arbitrary constants.

Step 6: Verification

Differentiating twice: $y' = C_1\int_0^x e^{t^2}\,dt + C_1 x\,e^{x^2} - C_1 x\,e^{x^2} + C_2 = C_1\int_0^x e^{t^2}\,dt + C_2$.

\[y'' = C_1 e^{x^2},\qquad y''' = 2C_1 x\,e^{x^2} = 2x\,y''.\]