Step-by-step explanation:
1. Rewrite this in the form y'(x)+p(x)y(x)=q(x)
x³f' - x = xf +1
x³f' -xf - x = 1
x³f' -xf = 1+x
f' -(x/x³)f=(1+x)/x³
f' -(1/x²)f=(1+x)/x³
So in our case, p(x)=-(1/x²) and q(x)=(1+x)/x³
2. Find the integrating factor
The integrating factor, μ, is equal to [tex]e^{\int\limits {P(x)} \, dx[/tex].
μ = [tex]e^{\int\limits {P(x)} \, dx[/tex]
μ = [tex]e^{\int\limits {-1/x^{2} } \, dx[/tex]
μ = [tex]e^{1/x}[/tex]
3. Multiply equation by the integrating factor
f' - [tex]\frac{1}{x^{2} }[/tex]f = [tex]\frac{1+x}{x^{3} }[/tex]
[tex]e^{1/x}[/tex]f' - [tex]e^{1/x}[/tex][tex]\frac{1}{x^{2} }[/tex]f = [tex]e^{1/x}[/tex] [tex]\frac{1+x}{x^{3} }[/tex]
4. Apply product rule
Consider the product rule: (a⋅b)'=a'⋅b+b'⋅a
In our case, we could say a=[tex]e^{1/x}[/tex] and b=f [Note that ([tex]e^{1/x}[/tex])'=[tex]e^{1/x}[/tex][tex]\frac{1}{x^{2} }[/tex]]
Therefore...
[tex]e^{1/x}[/tex]f' - [tex]e^{1/x}[/tex][tex]\frac{1}{x^{2} }[/tex]f = ([tex]e^{1/x}[/tex] f )'
5. Solve
([tex]e^{1/x}[/tex] f )' = [tex]e^{1/x}[/tex] [tex]\frac{1+x}{x^{3} }[/tex]
[tex]e^{1/x}[/tex] f = ∫ ([tex]e^{1/x}[/tex] [tex]\frac{1+x}{x^{3} }[/tex]) dx
[tex]e^{1/x}[/tex] f = ∫ ([tex]e^{1/x}[/tex]( [tex]\frac{1}{x^{3} }+\frac{1}{x^{2} }[/tex])) dx
[tex]e^{1/x}[/tex] f = ∫ [tex]\frac{e^{1/x} }{x^{3} }+\frac{e^{1/x}}{x^{2} }[/tex] dx
[note the sum rule: ∫(a+b)=∫a+∫b ]
[tex]e^{1/x}[/tex] f = ∫ [tex]\frac{e^{1/x} }{x^{3} }[/tex]dx+ ∫ [tex]\frac{e^{1/x}}{x^{2} }[/tex] dx
[tex]e^{1/x}[/tex] f = (-[tex]\frac{e^{1/x} }{x }[/tex]+[tex]e^{1/x}[/tex] )+(-[tex]e^{1/x}[/tex]) +C [step by step for integrating[tex]\frac{e^{1/x} }{x^{3} }[/tex]and[tex]\frac{e^{1/x}}{x^{2} }[/tex] below]
[tex]e^{1/x}[/tex] f = -[tex]\frac{e^{1/x} }{x }[/tex] +C
f = -[tex]\frac{1 }{x }[/tex] +[tex]\frac{C}{e^{1/x}}[/tex]
f = [tex]\frac{C}{e^{1/x}}[/tex] - [tex]\frac{1 }{x }[/tex]
C is just any constant, so for our purposes, let's let C equal 1.
f = [tex]\frac{1}{e^{1/x}}[/tex] - [tex]\frac{1 }{x }[/tex]
[note that [tex]\frac{1}{a^{b} }[/tex]=a⁻ᵇ]
f= [tex]e^{-1/x}[/tex] - [tex]\frac{1 }{x }[/tex]
QED
Below is a full explanation on integrating [tex]\frac{e^{1/x} }{x^{3} }[/tex]and[tex]\frac{e^{1/x}}{x^{2} }[/tex] , I didn't put it above as there was already a bunch of info and it's pretty long
Solving ∫[tex]\frac{e^{1/x} }{x^{3} }[/tex]dx (u-substitution)
Let u =1/x. Therefore [tex]\frac{du}{dx}[/tex] = 1/x² → du = 1/x² dx → dx = -x² du
Therefore, ∫[tex]\frac{e^{1/x} }{x^{3} }[/tex]dx = ∫[tex]\frac{e^{u} }{x^{3} }[/tex](-x²) du
∫[tex]\frac{e^{u} }{x^{3} }[/tex](-x²) du
∫- [tex]\frac{e^{u} }{x} }[/tex]du
Note that if u=1/x, x=1/u
∫- [tex]\frac{e^{u} }{x} }[/tex]du
∫- [tex]\frac{e^{u} }{1/u} }[/tex]du
- ∫ [tex]e^{u}[/tex]u du
Note that ∫ab'=ab-∫a'b. Here, a=u, b= [tex]e^{u}[/tex]. Therefore, a'=u'=1, b'= [tex]e^{u}[/tex].
So, - ∫ [tex]e^{u}[/tex]u du = - ([tex]e^{u}[/tex]u-- ∫ [tex]e^{u}[/tex] du) = -([tex]e^{u}[/tex]u-[tex]e^{u}[/tex]) = -[tex]e^{u}[/tex]u+[tex]e^{u}[/tex]
Now sub 1/x back in for u: -[tex]e^{u}[/tex]u+[tex]e^{u}[/tex] = -[tex]e^{1/x}[/tex](1/x) +[tex]e^{1/x}[/tex] = [tex]\frac{-e^{1/x} }{x}[/tex]+[tex]e^{1/x}[/tex]
So ∫[tex]\frac{e^{1/x} }{x^{3} }[/tex]dx = [tex]\frac{-e^{1/x} }{x}[/tex]+[tex]e^{1/x}[/tex]
Solving ∫ [tex]\frac{e^{1/x}}{x^{2} }[/tex] dx (u-substitution)
Let u =1/x again. Therefore, as we've seen above, [tex]\frac{du}{dx}[/tex] = 1/x² → du = 1/x² dx → dx = -x² du.
Therefore, ∫[tex]\frac{e^{1/x} }{x^{2} }[/tex]dx = ∫[tex]\frac{e^{u} }{x^{2} }[/tex](-x²) du = ∫-[tex]e^{u}[/tex] du = -∫[tex]e^{u}[/tex] du = - [tex]e^{u}[/tex].
Sub 1/x back in for u: - [tex]e^{u}[/tex]= - [tex]e^{1/x}[/tex]
So ∫ [tex]\frac{e^{1/x}}{x^{2} }[/tex] dx = [tex]e^{1/x}[/tex]