Solve the differential equation x y^((4)) + y''' = e^x

Solve the differential equation $x y^{(4)} + y''' = e^x$.

Step 1: Recognize a perfect derivative

\[\frac{d}{dx}(x y''') = y''' + x y^{(4)},\] so the equation is $\dfrac{d}{dx}(x y''') = e^x$.

Step 2: First integration

\[x y''' = e^x + C_1,\]

hence $y''' = \frac{e^x}{x} + \frac{C_1}{x}.$

Step 3: Second integration

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

The integral $\int\dfrac{e^x}{x}\,dx$ is the exponential integral $\mathrm{Ei}(x) = \displaystyle\int_{-\infty}^{x}\dfrac{e^t}{t}\,dt$ for $x\neq 0$, so

\[y'' = \mathrm{Ei}(x) + C_1\ln|x| + C_2.\]

Step 4: Third integration

Using $\int\mathrm{Ei}(x)\,dx = x\,\mathrm{Ei}(x) - e^x$, $y' = x\,\mathrm{Ei}(x) - e^x + C_1(x\ln|x| - x) + C_2 x + C_3.$

Step 5: Fourth integration

Using $\int x\,\mathrm{Ei}(x)\,dx = \dfrac{x^2}{2}\mathrm{Ei}(x) - \dfrac12(x-1)e^x$, $\int\bigl(x\,\mathrm{Ei}(x) - e^x\bigr)dx = \frac{x^2}{2}\mathrm{Ei}(x) - \frac{x+1}{2}e^x.$

Also $\int(x\ln|x|-x)\,dx = \dfrac{x^2}{2}\ln|x| - \dfrac{3x^2}{4}$.

Step 6: General solution

\[y = \frac{x^2}{2}\,\mathrm{Ei}(x) - \frac{x+1}{2}e^x + C_1\left(\frac{x^2}{2}\ln|x| - \frac{3x^2}{4}\right) + C_2 x^2 + C_3 x + C_4,\] valid for $x\neq 0$.

Step 7: Verification

\[y''' = \frac{e^x}{x} + \frac{C_1}{x},\qquad y^{(4)} = \frac{e^x(x-1)}{x^2} - \frac{C_1}{x^2},\]

\[x y^{(4)} + y''' = \frac{e^x(x-1)}{x} - \frac{C_1}{x} + \frac{e^x}{x} + \frac{C_1}{x} = e^x.\]