I Find the general solution for the differential equation

So in my previous math class I spotted on my book an exercise that I couldnt solve. We had to find the general solution for the differential equation. This was the exercise: 4y'' - 4y' + y = ex/2√(1-x2)
Can anyone tell me how to solve this step by step?
 

fresh_42

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You first solve the homogeneous version.
 

HallsofIvy

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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 [tex]y= C' e^{-x/4}[/tex] where [tex]|C'|= e^C[/tex] 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 [tex]y(x)= u(x)e^{-x/4}[/tex]. Then [tex]y'= u'e^{-x/4}- (u/4)e^{-x/4}= (u'- u/4)e^{-x/4}[/tex] so the equation becomes [tex](4u'- u)e^{-x/4}+ ue^{-x/4}= 4u'e^{-x/4}= e^{x/2}\sqrt{1- x^2}[/tex] so that [tex]u'= \frac{1}{4}e^{3x/4}(1- x^2)^{1/2}[/tex]. Can you integrate that?
 
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We had to find the general solution for the differential equation. This was the exercise: 4y'' - 4y' + y = ex/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

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Yes, I did miss that! Thanks.
 

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