Find the general solution for the differential equation

[GoodEngineering]

So in my previous math class I spotted on my book an exercise that I couldn't solve. We had to find the general solution for the differential equation. This was the exercise: 4y'' - 4y' + y = e^x/2√(1-x²)
Can anyone tell me how to solve this step by step?

 

[fresh_42]

You first solve the homogeneous version.

 

[HallsofIvy]

The "associated homogeneous equation" is -4y'+ y= 0 so y'= -y/4. We can write that as dy/y= -dx/4. Integrating, ln|y|= -x/4+ C. Taking the exponential of both sides $$y= C' e^{-x/4}$$ where $$|C'|= e^C$$ and we can remove the absolute value by allowing C' to be either positive or negative.
To solve the entire equation use "variation of parameters": Look for a solution of the form $$y(x)= u(x)e^{-x/4}$$. Then $$y'= u'e^{-x/4}- (u/4)e^{-x/4}= (u'- u/4)e^{-x/4}$$ so the equation becomes $$(4u'- u)e^{-x/4}+ ue^{-x/4}= 4u'e^{-x/4}= e^{x/2}\sqrt{1- x^2}$$ so that $$u'= \frac{1}{4}e^{3x/4}(1- x^2)^{1/2}$$. Can you integrate that?

 

[Mark44]

We had to find the general solution for the differential equation. This was the exercise: 4y'' - 4y' + y = e^x/2√(1-x²)

The "associated homogeneous equation" is -4y'+ y= 0

No. You apparently missed the second-order term 4y''.

Can anyone tell me how to solve this step by step?

That's not what we do here. Per our rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/, under Homework Guidelines), you must first make an attempt. Also, homework-type questions should be posted in one of the sections under Homework & Coursework.

 

[HallsofIvy]

Yes, I did miss that! Thanks.