- #1

basty

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## Homework Statement

Solve:

##x \frac{dy}{dx} - 4y = x^6 e^x.##

Solution: Dividing by ##x##, we get the standard form

##\frac{dy}{dx} - \frac{4}{x}y = x^5 e^x.##

Hence the integrating factor is

##e^{\int -\frac{4}{x}dx}= e^{-4 \int \frac{1}{x}dx} = e^{-4 \ln x} = e^{\ln x^{-4}} = x^{-4}##

Now we multiply ##\frac{dy}{dx} - \frac{4}{x}y = x^5 e^x## by ##x^{-4}## and rewrite

##x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x## as ##\frac{d}{dx}[x^{-4}y] = x e^x##

It follows from integration by parts that the general solution defined on the interval ##(0, ∞)## is

##x^{-4}y = x e^x - e^x + c##

or

##y = x^5e^x - x^4e^x + cx^4.##

My question is, why integrate of ##x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x## is ##y = x^5e^x - x^4e^x + cx^4##?

Compare it with mine below.

Which one is correct?

## Homework Equations

## The Attempt at a Solution

##x^{-4} \frac{dy}{dx} - 4x^{-5}y = x e^x##

Multiplying by ##dx## yield

##x^{-4} dy - 4x^{-5}y \ dx = x e^x dx##

Integrating both sides

##\int x^{-4} dy = x^{-4}y + c_1##

##\int - 4x^{-5}y \ dx = -4 \int x^{-5}y \ dx = -4 \frac{1}{-4}x^{-4}y + c_2 = x^{-4}y + c_2##

##\int x e^x dx = xe^x - e^x + c_3##

So

##\int x^{-4} dy - \int 4x^{-5}y \ dx = \int x e^x dx##

##x^{-4}y + c_1 + x^{-4}y - c_2 = xe^x - e^x + c_3##

##x^{-4}y + x^{-4}y = xe^x - e^x - c_1 + c_2 + c_3##

##x^{-4}y + x^{-4}y = xe^x - e^x + c_4##

##2x^{-4}y = xe^x - e^x + c_4##

##2y = \frac{xe^x - e^x + c_4}{x^{-4}}##

##2y = x^5e^x - x^4e^x + c_4x^4##

##y = \frac{x^5e^x - x^4e^x + c_4x^4}{2}##