Difficult to understand incomplete gamma function in the solution of 1st order Non-Homogeneous linear ODE.

Question

I have solved manually this 1st order non-homogeneous linear ODE.$\frac{dy}{dx}+x^2*y(x)=5$

The solution is $\large e^{\frac{-x^3}{3}}*\int5*\large e^{\frac{x^3}{3}}dx$. HP 50g graphing calculator also gives this solution.

But my differential equation solver gives the answer $y(x)=\large e^{-\frac{x^3}{3}}*\large(C_1+\frac{5\sqrt[3]{3}\gamma incomplete(\frac13,-\frac{x^3}{3})}{3}\large)\ldots (1)$

Numerical answer:

(x,y):(-10.0,0.75)

(-7.77777,0.08230539)

(-5.5555,0.1601631122)

(-3.3333,0.428426947)

(-1.1111,2.28138067)

(1.11111,8.053819154)

(3.3333,0.478712609038)

(5.5555,0.16394741)

(7.777777,0.08300819583)

(10.0,0.050100504).

Now suppose i plug in 1.111111 in the solution (1) as x, i must get the y=8.053819154.

But i am not getting this y value.

Note:- My answer to the lower incomplete gamma function is non-real result which is (1.3069748148+2.263746789599i).

When i numerically integrate$\displaystyle\int_{-\frac{1000}{2187}}^0 x^{-\frac23}*e^{-x}dx$, my integral calculator gives the answer 2.61394963.

I want to know where is i am wrong?

Any member may reply to this question, stating the correct answer to this question pointing out where is i am wrong.

Answer 1

Computing an integrating factor:$$\mu(x)=e^{\in x^2dx}=e^{x^3/3}$$ so we get $$e^{x^3/3}y'(x)+e^{x^3/3}x^2y(x)=5e^{x^3/3}$$ Note that $$\frac{d}{dx}\left(e^{x^3/3}y(x)\right)=5e^{x^3/3}$$ so we have $$\int\frac{d}{dx}\left(e^{x^3/3}y(x)\right)dx=\int5e^{x^3/3}dx$$ The first integral gives $$-\frac{5x\Gamma(1/3;-x^3/3)}{3^{2/3}\sqrt[3]{-x^3}}$$ so our solutions is $$y(x)=e^{-x^3/3}\left(\frac{-5x\Gamma(1/3;-x^3/3)}{3^{2/3}\sqrt[3]{-x^3}}+C\right)$$ P.s.: Maple gives the following solution: $$y \left( x \right) ={{\rm e}^{-1/3\,{x}^{3}}}{\it \_C1}+5/9\,{\frac { \sqrt [3]{3}x \left( 2\,\sqrt {3}\pi-3\,\Gamma \left( 1/3,-1/3\,{x}^{3 } \right) \Gamma \left( 2/3 \right) \right) {{\rm e}^{-1/3\,{x}^{3}}} }{\Gamma \left( 2/3 \right) \sqrt [3]{-{x}^{3}}}}$$

Answer 2

The problem is that $$\Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right)$$ is real only for $x <0$. A few values $$\left( \begin{array}{cc} x & \Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right) \\ -1.0 & 0.75689 \\ -0.9 & 0.91312 \\ -0.8 & 1.08254 \\ -0.7 & 1.26317 \\ -0.6 & 1.45290 \\ -0.5 & 1.64960 \\ -0.4 & 1.85132 \\ -0.3 & 2.05631 \\ -0.2 & 2.26320 \\ -0.1 & 2.47095 \\ 0.0 & 2.67894 \\ 0.1 & 2.57493 -0.18016\, i \\ 0.2 & 2.47079 -0.36052\, i \\ 0.3 & 2.36622 -0.54164\, i \\ 0.4 & 2.26069 -0.72443\, i \\ 0.5 & 2.15344 -0.91020\, i \\ 0.6 & 2.04345 -1.10071\, i \\ 0.7 & 1.92940 -1.29824\, i \\ 0.8 & 1.80960 -1.50573\, i \\ 0.9 & 1.68185 -1.72700\, i \\ 1.0 & 1.54329 -1.96701\, i \end{array} \right)$$

Concerning

$$\displaystyle\int_{-\frac{1000}{2187}}^0 x^{-\frac23}*e^{-x}\,dx=\Gamma \left(\frac{1}{3},-\frac{1000}{2187}\right)-\Gamma \left(\frac{1}{3}\right)\approx -1.30697 - 2.26375 \,i$$ which is confirmed using https://www.integral-calculator.com/#