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2 Answers

Problem:  Solve for y(x):   dy/dx + y = xe-x
 
Step 1:  multiply both sides of the equation by ex to get   ex(dy/dx) + yex = x
 
Step 2:   Substitute ex = d(ex)/dx  to get ex(dy/dx) + y(dex/dx) = x
 
Step 3:  Apply the product rule in reverse:  g(df/dx) + f(dg/dx) = d(fg)/dx to get
                 d(yex)/dx = x.
 
 
Step 5:  Integrate both sides with respect to x to get
                 yex = x2/2  + C
 
Step 5:  Divide both sides by ex,   giving y = e-x((x2/2) + C) 
 
Step 6:  Multiply by the distributive law to get y = (x2/2)e-x + Ce-x
 
 
 
 
There are several methods for solving first-order linear equations. Here's one of them:
Bring the equation into standard form, y'+p(x) y = q(x). Then find the function r(x)=e∫p(x)dx. This function is called an integrating factor. When you multiply the equation by this factor, the left-hand side will become a total derivative, which is then easy to integrate.
Your equation is already in standard form:
y'+y=xe-x, with p(x)=1, q(x)=xe-x.
Then r(x)=e∫1dx=ex is an integrating factor, so multiply the equation by ex:
 
exy'+exy=x
The left side is the derivative of exy:
(exy)'=x
so that when you integrate both sides of the equation,
exy=x2/2+c
y=(x2/2)e-x+ce-x.
 
 

Comments

Great general rule, Andre! 
Thanks, Kenneth! Wish I could claim ownership of this rule, but a guy named Euler beat me by some 250 years!

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