Homework Statement
i have to prove that
∇x(FxG)=(G⋅∇)F-(F⋅∇)G+F(∇⋅G)-G(∇⋅F)
where F and G are vector fields with F=F1,F2,F3 and G=G1,G2,G3 ∇=d/dx,d/dy/d/dz
Homework Equations
The Attempt at a Solution
i have tried applying scalar multiplication and the cross product to...
Homework Statement
dy/dx=a^2/(x+y)^2
where a is a constant
need the answer in the form
x=f(y)
Homework Equations
The Attempt at a Solution
multiplying out (x+y)^2
gives dy/dx=a^2/(x^2+2xy+y^2)
setting u=y/x dy/dx can be rewritten as
dy/dx=a^2/((x^2)*(u+1)^2)...
i have attempted it with a different method since the last method did not seem to supply the right answer please could someone tell me where i am going wrong with this new method:
dy/dx=(a^2)/((x+y)^2)
where a^2 is a constant and the answer must be expressed in the form x=f(y)...
ok so i set f(u)=u^2
f'(u)=2u
g'(u)=1/(u^2+a^2)
g(u)=ArcTan[u/a]/a
applying integration by parts formula gives
(u^2)*ArcTan[u/a]/a-int(2uArcTan[u/a]/a du)
which gives a*arctan(u/a)-x=int a^2 dx
arctan(u/a)=ax+c
so (u/a)=tan(ax) +tan(c)
subbing back in for x+y gives...
cheers for the response
ive tried integrating:
u^2/((u^2)+(a^2))du
by parts but the answer i get is not the same as what i get when i use wolfram online integrator
is it correct to try integration by parts? where the equation is
int(f(u)g'(u))=f(u)g(u)-int(f'(u)g(u)du)...
Homework Statement
dy/dx=a^2/((x+y)^2)
where a is a constant
express answer in the form x=f(y)
Homework Equations
The Attempt at a Solution
let u=x+y
du/dx=1+dy/dx
du/dx=1+(a^2)/(u^2)
int(du/(((a^2)/(u^2))+1))=int(a^2 dx)
after integration substituting back in for u...
Homework Statement
dy/dx=-ylny/x initial conditions y(x=1)=1
express answer in the form f(x,y)=0
Homework Equations
The Attempt at a Solution
i let u=ln y
du/dy=1/y
y*du=dy
subbing into equation gives int(y/yu du)=-int dx
which is equal to int(1/u du)=-int(1/x) dx...
Homework Statement
dx/dt=x(1-2y) t(0)=0 x(t(0))=1
dy/dt=-y(1-2x), t(0)=0 y(t(0))=2
inerval of integration= [0,40]
Homework Equations
The Attempt at a Solution
i let x=p(t) so t=(p^-1)(x)
dy/dx=(∂y/∂(p^-1))∂(p^-1)/∂x
dy/dx=(-y(1-2x))/(x(1-2y))
since this equation is...
Homework Statement
integrate
dy/dx-y=cos(x)-2
Homework Equations
The Attempt at a Solution
dy/dx-y=cos(x)-2
is in the form
dy/dx+p(x)=q(x)
take the I.F as e^int(p(x))dx=e^-x
multiplying throughout by e^-x
d(e^-x)z/dx=-(cos(x)e^-x)-(2e^-x)...
hi thanks for the help
so i can take p(x)=2009 and q(x)=-2009cos(x)
using e^int(2009)dx=e^2009x
multiply throughout by e^2009x giving
d(e^2009x)z/dx= -2009cos(x)*e^2009x
so e^2009x*z= int -2009cos(x)*e^2009x dx
using the integrating factor method
but what i don't understand is why i need...
Homework Statement
solve
dy/dx=y+cos(x)y^2010
using variation of parameters/constant method
Homework Equations
The Attempt at a Solution
let z=1/(y^(n-1))=1/(y^(2009))
dz/dx=(dz/dy)(dy/dx)
dz/dy=-(n-1)*y^(-n)*dy/dx=-(2009*y^-2010)*(y+cos(x)y^2010)...
Homework Statement
find general solution of
dy/dx=xy+x
using integrating factor
Homework Equations
The Attempt at a Solution
rearange
dy/dx-xy=x
take integrating factor as
e^int(-x)=e^-(x^2)/2
multiply throughout
dy/dx(e^-(x^2)/2)y=(e^-(x^2)/2)x
integrate...
e^-(5x+c)=(1/y)-(1)
raise both sides to the power of minus 1 so 1/y=y^-1 becomes y and -1 becomes -1^-1 =1/-1 =-1 and e^-(5x+c)=1/e^(5x+c) becomes e^(5x+c)
e^(5x+c)=y-1