Find the general solution?

Question

Find the general solution for the equation dy/dx+y=xe^-x.

 

Answer: y=(x^2/2)e^(-x)+Ce^-x

 

Please show all your work.

Answer 1

Problem:  Solve for y(x):   dy/dx + y = xe^-x

 

Step 1:  multiply both sides of the equation by e^x to get   e^x(dy/dx) + ye^x = x

 

Step 2:   Substitute e^x = d(e^x)/dx  to get e^x(dy/dx) + y(de^x/dx) = x

 

Step 3:  Apply the product rule in reverse:  g(df/dx) + f(dg/dx) = d(fg)/dx to get

                 d(ye^x)/dx = x.

 

 

Step 5:  Integrate both sides with respect to x to get

                 ye^x = x²/2  + C

 

Step 5:  Divide both sides by e^x,   giving y = e^-x((x²/2) + C) 

 

Step 6:  Multiply by the distributive law to get y = (x²/2)e^-x + Ce^-x

 

 

 

 

Answer 2

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=e^x is an integrating factor, so multiply the equation by e^x:

 

e^xy'+e^xy=x

The left side is the derivative of e^xy:

(e^xy)'=x

so that when you integrate both sides of the equation,

e^xy=x²/2+c

y=(x²/2)e^-x+ce^-x.