- #1
basty
- 95
- 0
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}##