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.

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.

Tutors, please sign in to answer this question.

Tarzana, CA

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

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

New Wilmington, PA

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^{x}y'+e^{x}y=x

The left side is the derivative of e^{x}y:

(e^{x}y)'=x

so that when you integrate both sides of the equation,

e^{x}y=x^{2}/2+c

y=(x^{2}/2)e^{-x}+ce^{-x}.

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