A very lengthy ODE. Can you give another way around?

[Knight2023]

The equation is the following:

$$ (3x+2)^2 y'' + 5(3x+2)y' -3y= x^2+x+1$$

(it came in UPSC CSE 2022 Maths Optional Paper-I, Q.8)

I went with the usual procedure, that is first solving the associated homogeneous equation, and then using the variation of parameters. I must confess, it was very difficult to come up with a particular solution. The overall work got so lengthy that I have to post the link:

https://1drv.ms/f/c/eff33ac169bd871a/Ep6lKSWIuztGkJbuRd1ZmLYB0W_IhId4YtRLOrUC5wlNIQ?e=ZyLm4k (Microsoft OneDrive link)

https://drive.google.com/drive/folders/1TUQN13nNLKkSUVv4kucKVzL5L4MILYg1?usp=sharing (Google Drive link)

I have not shown the actual evaluation of integrals, nor the multiplication process of $y_1 \int u dx$.

My questions:
1. I mean, this question came in exam, could it have been possible to solve it there?
2. Can you please offer some other method for solving it without making it as lengthy as above?
3. Is there a way to check if my general solution is correct? (possibly, someone can give me Mathematica code for checking it?)

Thank you.

[drzewo]

first change the variable $3x+2=e^t$ and write down the result here

-- 29.07.2024, 16:02 --

I am not reading you references and I am only looking at the text here

-- 29.07.2024, 16:24 --

Knight2023 в сообщении #1647724 писал(а):

Is there a way to check if my general solution is correct?

sure! just substitute your solution into the equation

-- 29.07.2024, 16:36 --

try to find a particular solution as $ax^2+bx+c$

[Knight2023]

drzewo в сообщении #1647726 писал(а):

first change the variable $3x+2=e^t$ and write down the result here

$
3x+2=e^t$

$3= e^t dt/dx$
$dx = 1/3 e^t dt$
$dy/dx= 3 e^{-t} \dot{y}$
$d^2 y/dx^2= 9 e^{-2t} \ddot{y}$
Re-writing the equation:
$e^{2t} 9 e^{-2t} \ddot{y} + 5e^{t} \cdot 3e^{-t} \dot{y} -3y=0$
$9\ddot{y} +15\dot{y}-3y=0$
$(3D^2 +5D-1)y=0$
But that doesn't give out any neat solution for y, as the quadratic operator doesn't have integral roots.

-- 29.07.2024, 19:49 --

Цитата:

try to find a particular solution as $ax^2+bx+c$

But manually plugging it in will involve so much work, can some computer algebra system do it? I'm not well-versed in Mathematica.

[mihiv]

The equation can be rewritten as follows:

$$\left [(3x+2)^2y'-(3x+2)y\right ]'=x^2+x+1$$

[drzewo]

The topicstarter  continues to use the word "integral" instead of "integer" despite people having already corrected him. 

[Markiyan Hirnyk]

The command of Maple 2024

dsolve((3*x + 2)^2*diff(y(x), x, x) + 5*(3*x + 2)*diff(y(x), x) - 3*y(x) = x^2 + x + 1)

produces

$$y\! \left(x\right)=\frac{c_{1}}{3 x+2}+\left(3 x+2\right)^{\frac{1}{3}} c_{2}+\frac{4 x^{3}+3 x^{2}-48 x-24}{60 \left(3 x+2\right)}$$

The command of Maple 2024

odetest(y(x) = (c[1] - x/4 - x^2/8 - x^2/12)/(3*x + 2) - x^2/4 - x/4 + (3*x + 2)*(x/48 + 1/8) - 1/144*(3*x + 2)^2 + c[2]*(3*x + 2)^(1/3), (3*x + 2)^2*diff(y(x), x, x) + 5*(3*x + 2)*diff(y(x), x) - 3*y(x) = x^2 + x + 1)

results in $-\frac{49}{4} x^{2}-\frac{71}{6} x-\frac{41}{12}$ , not confirming your result by hand.

[Knight2023]

drzewo в сообщении #1647868 писал(а):

The topicstarter  continues to use the word "integral" instead of "integer" despite people having already corrected him. 

https://www.ck12.org/flexi/cbse-math/pe ... counting./