initial value problem
y' = xy +x^3
y(0) = -1

Question

initial value problem
 y' = xy +x^3 
 y(0) = -1

anonymous · Answer

y(0)=-1
y'=0(-1)+0^3
y'=0

anonymous · Answer

y' - xy=x^3
IF = e^(-xdx)
multiply by IF on both sides

anonymous · Answer

d(ye^(-x^2 / 2)) = x^3 * e^(-x^2 / 2)dx

anonymous · Answer

now someone remind me how to integrate the rhs..im growing old

amistre64 · Answer

gummage, that is not appropriate language for this forum; please keep it clean

anonymous · Answer

my bad

anonymous · Answer

take x^2 = u and then do integral by parts

anonymous · Answer

you mean u = x^3?

anonymous · Answer

was this integration

anonymous · Answer

u=x^2/2
du= xdx
RHS becomes
ue^(-u)du

anonymous · Answer

as it is x^3(e^(-x^2/2))

anonymous · Answer

it becomes ue^(-u)du/2 sorry

anonymous · Answer

now use integral by parts

mathteacher1729 · Answer

After integrating and using the initial condition, you obtain: $$y(x) = -x^2+e^{x^2/2}-2$$ Attached is a picture of the slope field with the initial condition drawn in for fun. :) http://www.math.rutgers.edu/~sontag/JODE/JOdeApplet.html

anonymous · Answer

he needs to learn to do himself....u guys are extremely dependent on websites...

anonymous · Answer

did u do it gummage?

anonymous · Answer

how does the x^3 change into u/2?

mathteacher1729 · Answer

NOTES: http://tutorial.math.lamar.edu/Classes/DE/IntroFirstOrder.aspx http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/readings/ VIDEOS: http://www.khanacademy.org/#differential-equations http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/ In particular "First order ODEs" seem to be right up your ally. :) If you have windows, download WINPLOT, it is awesome and free. http://math.exeter.edu/rparris/winplot.html A great way to visualize your work. HIM1628-- I agree, sometimes people just "Give answers". I try whenever possible to give helpful resources and guidance. It seemed that this discussion broke down the problem fairly well, so here are some final thoughts on ODEs. :)

anonymous · Answer

see gummage
havent u got$$\int\limits_{}^{}x^{3}e^{-x^{2}/2}dx$$

anonymous · Answer

yes

anonymous · Answer

now 
$$x^{2}/2=u$$

then
xdx=du

anonymous · Answer

yep got that

anonymous · Answer

now u can write the RHS as
$$\int\limits_{}^{}2ue^{-u}du$$

anonymous · Answer

xdx becomes du
x^2 becomes 2u
the divide by 2 ws my bad

anonymous · Answer

ive got x^4*e^-u du

anonymous · Answer

sorry got it now. thanks