MHB How can we solve the differential equation $y^\prime - 2y = x^2 e^{-2x}$?

[karush]

ok going to try the next de!

$$\displaystyle y^\prime - 2y = x^2 e^{-2x}$$

I would assume that first

$$u(x) = e^{-2x}$$

book answer is:

$$y=\color{red}{\displaystyle ce^{2x}+\frac{x^3}{3}e^{2x}}$$

 

[MarkFL]

You've correctly computed the integrating factor, so what does the ODE become after you multiply though by this integrating factor?

 

[MarkFL]

I think we just multiply every term

$$\displaystyle (e^{-2x})y^\prime - 2(e^{-2x})y = x^2 e^{-2x}(e^{-2x})$$

if so the next would be to simplify

Yes, rewrite the LHS as the derivative of a product, and simplify the RHS. :)

 

[MarkFL]

$$\frac{d}{dx}\left(e^{-2x}2y\right)=e^x(xe^{2x}+1)$$

really?

No, you would have:

$$\frac{d}{dx}\left(e^{-2x}y\right)=x^2e^{-4x}$$

 

[karush]

No, you would have:

$$\frac{d}{dx}\left(e^{-2x}y\right)=x^2e^{-4x}$$

ok it was a product
what happened to the 2?
presumed next step...

$$\displaystyle y=e^{2x}\int x^2e^{-4x} \, dx + ce^{2x}$$

IBP??

 

[MarkFL]

ok it was a product
what happened to the 2?

I'm not sure which 2 you're talking about...

presumed next step...

$$\displaystyle e^{-2x}y=\int x^2e^{-4x} \, dx$$

IBP??

Yes, IBP for the RHS. :)

 

[MarkFL]

Personally, I would integrate first and then solve for $y(x)$. Either way though.

 

[karush]

Personally, I would integrate first and then solve for $y(x)$. Either way though.

W|A returned this I don't see this heading towards the answer

$$\displaystyle\int x^2 e^{-4x} \, dx = -\frac{1}{32}e^{-4x} (8x^2+4x+1)+c$$

then

$$\displaystyle e^{-2x}y= -\frac{1}{32}e^{-4x} (8x^2+4x+1)+c$$

and

$$\displaystyle y= -\frac{1}{32}e^{-2x}(8x^2+4x+1)+ce^{2x}$$

But ? W|A says

$$y(x)=c_1 e^{2e} -\frac{1}{2} e^{-2e}x^2-\frac{1}{2}e^{-2e}x-\frac{ e^{-2e}}{4}$$

 

[MarkFL]

W|A returned this I don't see this heading towards the answer

$\displaystyle\int x^2 e^{-4x} \, dx = -\frac{1}{32}e^{-4x} (8x^2+4x+1)+c$

The answer you cited as being given by the book is incorrect. Are you sure the ODE is given correctly?

 

[karush]

answer #2
https://www.physicsforums.com/attachments/8206
problem #2
View attachment 8205

W|A returned this:

$$\displaystyle y(x)=c_1 e^{2e} -\frac{1}{2} e^{-2e} x^2-\frac{1}{2} e^{-2e}x-\frac{ e^{-2e}}{4}$$