How to solve $y-y''=x^a$

Question

I need to solve this differential equation $$y-y''=x^a$$ I am not well-versed in differential equations. I found the complementary solution first, and that was simple $$y-y''=0$$ $$y_c=C_1\cosh x+C_2\sinh x$$ I could not find the particular solution. I tried to induce a pattern based on specific cases of $a$ $$y_p=\frac{7!}{7!}x^7+\frac{7!}{5!}x^5+\frac{7!}{3!}x^3+\frac{7!}{1!}x, \space a=7$$ $$y_p=\frac{6!}{6!}x^6+\frac{6!}{4!}x^4+\frac{6!}{2!}x^2+\frac{6!}{0!},\space a=6$$ $$y_p=\sum_{n=0}^{⌊a/2⌋}\frac{a!}{(a-2n)!}x^{a-2n}$$ I think this solution works only when $a$ is a natural number. When I plug in $a=1/2$, I get 0, but WA says the solution requires the incomplete gamma function, so the solution is clearly not a polynomial. How can I solve this differential equation?

a = 0.5

general case

Answer 1

Look at the equation in $2D$,

Let $a>0$ and $\mathbf Y = \begin{bmatrix} y'\\ y \end{bmatrix}$, $\mathbf A = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$ and $\mathbf b(x) = \begin{bmatrix} x^a\\ 0 \end{bmatrix}$ then $$\mathbf Y' = \mathbf A \mathbf Y + \mathbf b(x)$$

The solution of the this DE will be:

$$\mathbf Y(x) = \int_{0}^{x} e^{-(x-t)\mathbf A} \mathbf b(t)\mathrm dt + e^{-x\mathbf A}\mathbf Y_0$$

Since $$\mathbf A = \mathbf U \mathbf D \mathbf U^\intercal$$ with $\mathbf U = \frac1{\sqrt 2}\begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix}$ and $\mathbf D = \begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix}$ then,

$$e^{t\mathbf A} = \mathbf U e^{t\mathbf D}\mathbf U^{\intercal} = \begin{bmatrix}\cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t)\end{bmatrix}$$

So,

\begin{align} \mathbf Y(x) &= \int_0^x \begin{bmatrix}\cosh(x-t) & -\sinh(x-t) \\ -\sinh(x-t) & \cosh(x-t)\end{bmatrix}\begin{bmatrix} t^a \\ 0 \end{bmatrix}\mathrm d t + \begin{bmatrix}\cosh(x) & -\sinh(x) \\ -\sinh(x) & \cosh(x)\end{bmatrix}\mathbf Y_0\\ &= \int_0^x \begin{bmatrix} t^a\cosh(x-t) \\ -t^a\sinh(x-t) \end{bmatrix}\mathrm dt + \begin{bmatrix}\cosh(x) & -\sinh(x) \\ -\sinh(x) & \cosh(x)\end{bmatrix}\mathbf Y_0 \end{align}

Finally,

$$y(x) = -\int_0^x t^a \sinh(x-t)\mathrm d t + y_0\cosh(x) - y'_0\sinh(x)$$

Since,

\begin{align} \int_0^x t^{a} \sinh(x-t)\mathrm d t &= \frac12 e^{x}\int_0^x t^a e^{-t}\mathrm dt -\frac12e^{-x}\int_0^x t^a e^{t}\mathrm dt \end{align}

Your expression can be expressed your solution using incomplete gamma function.

Answer 2

\begin{gather*} \boxed{y-y^{\prime \prime}-x^{\alpha}=0} \end{gather*} Let the solution be $$ y = y_h + y_p $$ Where $y_h$ is the solution to the homogeneous ODE and $y_p$ is a particular solution to the non-homogeneous ODE. The homogeneous solution $y_h$ can be easily found to be $$ y_h = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x} $$ The particular solution $y_p$ can be found using the method of variation of parameters. Let \begin{align*} y_p(x) &= u_1 y_1 + u_2 y_2\tag{1} \end{align*} Where $u_1,u_2$ to be determined, and $y_1,y_2$ are the two basis
solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{x} \end{align*} In the Variation of parameters $u_1,u_2$ are found using \begin{align*} u_1 &= -\int \frac{y_2 f(x)}{a W(x)}\tag{2} \\ u_2 &= \int \frac{y_1 f(x)}{a W(x)} \end{align*} Where $W(x)$ is the Wronskian and $a$ is the coefficient in front of $y''$ in the given ODE. The Wronskian is given by $W= \begin{vmatrix} y_1 & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{vmatrix} $. Hence $$ W = \begin{vmatrix} {\mathrm e}^{-x} & {\mathrm e}^{x} \\ \frac{d}{dx}\left({\mathrm e}^{-x}\right) & \frac{d}{dx}\left({\mathrm e}^{x}\right) \end{vmatrix} $$ Which simplifies to \begin{align*} W = 2 \end{align*} Therefore Eq. (2) becomes \begin{align*} u_1 &= - \int -\frac{{\mathrm e}^{x} x^{\alpha}}{2}d x \end{align*} Hence \begin{align*} u_1 &= -\frac{\left(-1\right)^{-\alpha} \left(x^{\alpha} \left(-1\right)^{\alpha} \alpha \Gamma \left(\alpha \right) \left(-x \right)^{-\alpha}-x^{\alpha} \left(-1\right)^{\alpha} {\mathrm e}^{x}-x^{\alpha} \left(-1\right)^{\alpha} \alpha \left(-x \right)^{-\alpha} \Gamma \left(\alpha , -x \right)\right)}{2}\\ &= \frac{x^{\alpha} \left(\left(\Gamma \left(\alpha , -x \right) \alpha -\Gamma \left(\alpha +1\right)\right) \left(-x \right)^{-\alpha}+{\mathrm e}^{x}\right)}{2}\\ u_2 &= \int \frac{{\mathrm e}^{-x} x^{\alpha}}{-2}\,dx\\ &= -\frac{x^{\frac{\alpha}{2}} {\mathrm e}^{-\frac{x}{2}} \operatorname{WhittakerM}\left(\frac{\alpha}{2}, \frac{\alpha}{2}+\frac{1}{2}, x\right)}{2 \left(\alpha +1\right)}\\ u_2 &= -\frac{x^{\frac{\alpha}{2}} {\mathrm e}^{-\frac{x}{2}} \operatorname{WhittakerM}\left(\frac{\alpha}{2}, \frac{\alpha}{2}+\frac{1}{2}, x\right)}{2 \alpha +2} \end{align*} Therefore the particular solution, from equation (1) is \begin{align*} y_p(x) = \frac{{\mathrm e}^{-x} \left(\left(\alpha +1\right) \left(\left(\Gamma \left(\alpha , -x \right) \alpha -\Gamma \left(\alpha +1\right)\right) \left(-x \right)^{-\alpha}+{\mathrm e}^{x}\right) x^{\alpha}-x^{\frac{\alpha}{2}} \operatorname{WhittakerM}\left(\frac{\alpha}{2}, \frac{\alpha}{2}+\frac{1}{2}, x\right) {\mathrm e}^{\frac{3 x}{2}}\right)}{2 \alpha +2} \end{align*} The general solution is \begin{align*} y &= y_h + y_p\\ &= \left(c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}\right) + \left(\frac{{\mathrm e}^{-x} \left(\left(\alpha +1\right) \left(\left(\Gamma \left(\alpha , -x \right) \alpha -\Gamma \left(\alpha +1\right)\right) \left(-x \right)^{-\alpha}+{\mathrm e}^{x}\right) x^{\alpha}-x^{\frac{\alpha}{2}} \operatorname{WhittakerM}\left(\frac{\alpha}{2}, \frac{\alpha}{2}+\frac{1}{2}, x\right) {\mathrm e}^{\frac{3 x}{2}}\right)}{2 \alpha +2}\right) \end{align*}

Answer 3

Proposing for the particular $y_p(x) = c(x)\cosh x$ after substitution in the complete ODE we obtain

$$ \sinh(x)c''(x)+2\cosh(x)c'(x)+x^a=0 $$

now calling $z(x) = c'(x)$ we follow with

$$ \sinh(x)z'(x)+2\cosh(x)z(x)+x^a=0 $$

The homogeneous have solution

$$ z_h(x) = c_1 \text{csch}^2(x) $$

then choosing as particular $z_p(x) = c_1(x)\text{csch}^2(x)$ and substituting into the $z(x)$ complete ODE we have

$$ \text{csch}(x) c_1'(x)+x^a=0 $$

so we have

$$ c_1(x) = \frac{1}{2} x^{a+1} (E_{-a}(-x)-E_{-a}(x)) $$

then going backwards

$$ z_h = \left(\frac{1}{2} x^{a+1} (E_{-a}(-x)-E_{-a}(x))\right)\text{csch}^2(x) $$

and

$$ c(x) = \int z_h(\eta)d\eta = \frac{1}{2} \left(x^a (-x)^{-a} (\coth (x)-1) \Gamma (a+1,-x)+(\coth (x)+1) \Gamma (a+1,x)\right) $$

and thus we have the particular

$$ y_p = \left(\frac{1}{2} \left(x^a (-x)^{-a} (\coth (x)-1) \Gamma (a+1,-x)+(\coth (x)+1) \Gamma (a+1,x)\right)\right)\cosh x $$

NOTE

$E_a(x)$ is the exponential integral function.