How would i go about solving this equation using integrating factors x dy/dx + x^2 *y = x^2 ?

Question

would: p(x) = x
mu(x) = e^(x) ??
q(x) = x^2

i'm trying to use this equation: dy/dx + p(x) * y = q(x)

by mu i might the math symbol that looks similar to a "u"

i'm just really confused, i missed that class and just looking at notes to teach myself how to do it :(

i don't know if it helps to know the answer but its y(x) = 1 + C e^((-x^2)/2)

Answer 1

Divide the original equation by x, you find

dy/dx + p(x)y = q(x)
p(x) = q(x) = x.

I will derive the general solution, then substitute the known values.

An "integrating factor" has the form

u(x) = exp(integral of p(z)dz)

It is clear (chain rule) that

du/dx = u(x) * p(x)

Now multiply the original equation with u(x),

u(x) dy/dx + u(x)p(x)y = u(x)q(x)

The left hand side is the derivative of the product u(x)*y (product rule):

d[u(x) * y]/dx = u(x) q(x)

Take integrals:

u(x) * y = INT u(z) q(z) dz
y = [INT u(z) q(z) dx] / u(x)


In your case,

u(x) = exp(integral of x dx) = exp(x^2/2)

y = [INT exp(x^2/2) x dx] / exp(x^2 / 2)
... = [exp(x^2/2) + C] / exp(x^2/2)
... = 1 + C exp(-x^2/2)

Answer 2

the major element this equation is making an attempt to augment, i imagine, is how order of operations works so we are going to take it in steps. a million. change 8 for X on your equation: y = (8*2 / 4) - 2. Now sparkling up for y. 2. remember that you do the stuff interior parenthesis first. With the stuff in the parenthesis carry out multiplication and branch first, going from left to correctly. So do 8 * 2 first it really is 16. Then divide 16 with suggestions from 4 it really is 4. This leaves you with y = 4 - 2 and ... 3. y = 2. you could verify your answer with suggestions from substituting 2 for y on your unique equation: 2 = (8*2/4)-2. Do the mathematics and also you'll discover that 2 = 2 it really is actual and to that end proves your answer is actual. sturdy success, Amanda

Answer 3

x (dy/dx) + x^2 *y = x^2
divide all by x so that dy/dx would have a coefficient of 1.

so : (dy/dx) + xy = x
p(x) = x, and q(x) = x

f(x) = e^ (integration of p(x) )
= e^ (integration of x) = e^( [x^2] /2 )
now multiply all of the equation by f(x):

e^( [x^2] /2 ) (dy/dx) + e^( [x^2] /2 )*x y = e^( [x^2] /2 ) *x

integration of ( y *f(x) ) ' dx = integration of ( q(x) * f(x) ) dx

so y*f(x) = integration of ( q(x) * f(x) ) dx

y e^( [x^2] /2 ) = integration of [ x* e^( [x^2] /2 ) ]

}for the RHS:

what is the differentiation of e^[ (x^2)/2 ]?
I think that this would be x e^ [ (x^2)/2 ] , which is the derivative on the right hand side.

so: y e^( [x^2] /2 ) = e^[ (x^2)/2 ]

so y = 1
try it in ur equation: d1 /dx = 0

so x*0 + 1* x^2 = 0 + x^2 = x^2, so this is true.

Answer 4

wow...i dont know if anyone on yahoo answers can do that...wow...good luck...next time go to classes....

Answer 5

..........que?