Integrating xe[itex]^{x}[/itex] without integration by parts

[cmajor47]

Homework Statement

Use the method of separation of variables or an integrating factor to find a particular solution of the differential equation that satisfies the given initial condition.

y'=x-y+2 ; y(0)=4


2. The attempt at a solution
I've used an integrating factor of e^{x} to obtain the following from y'=x-y+2:

\frac{d}{dx}e^{x}y=xe^{x}+2e^{x}

I know that I know have to integrate both sides of the equation. However, this is an issue since the equation contains the term xe^{x}. This book hasn't yet taught integration by parts which is commonly used to integrate xe^{x}. I don't think that separation of variables can be used to solve integrate this either.

Therefore, my question is: is there a way to integrate xe^{x} without using integration by parts?

 

[tiny-tim]

hi cmajor47!

… is there a way to integrate xe^{x} without using integration by parts?

hint: can you integrate (x + 1)e^x without using integration by parts?

 

[Mark44]

Homework Statement

Use the method of separation of variables or an integrating factor to find a particular solution of the differential equation that satisfies the given initial condition.

y'=x-y+2 ; y(0)=4


2. The attempt at a solution
I've used an integrating factor of e^{x} to obtain the following from y'=x-y+2:

\frac{d}{dx}e^{x}y=xe^{x}+2e^{x}

This doesn't look right to me. Can you show what you did to get it?

I know that I know have to integrate both sides of the equation. However, this is an issue since the equation contains the term xe^{x}. This book hasn't yet taught integration by parts which is commonly used to integrate xe^{x}. I don't think that separation of variables can be used to solve integrate this either.

Therefore, my question is: is there a way to integrate xe^{x} without using integration by parts?

 

[lurflurf]

^There problem is in your look.
we desire
(u y)'=u(y+y')
u' y+u y'=u y + u y'
u=u'
u=e^x
so
(e^x y)'=e^x (x+2)

to solve (e^x y)'=e^x(x+2)
we have a few choices all similar
1)Integration by parts
You want to avoid this one, but it is just a use of the product rule
(u v)'=u' v + u v'
u v'=(u v)'-u' v
suppose u=x+2 v'=e^x
then u'=1 v=e^x
and
u v'=(u v)'-u' v
becomes
(x+2)e^x=((x+2)e^x)'-e^x

2)Undetermined coefficients
assume
((a x+b)e^x)'=(x+2)e^x
determine a,b

3)repeated differentiation
(e^x y)'=e^x (x+2)
(e^-x (e^x y)')''=(x+2)''=0
this just reproduces the undetermined coefficients

4)mixed
As tiny-tim points out we can rewrite the equation in terms of
(x e^x)'