Recent content by dooogle

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...

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)...

cheers for the help i let u=-(x^2)/2 du/dx=-2x/2=-x du=-x dx so -int e^u du = int e^-(x^2)/2 dx =-e^u =-e^-(x^2)/2 +c dooogle

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...

ok i understand now thanks for your help i did what u said e^-(5x+c)=(1/y)-(1) 1/y=e^-(5x+c) +1 y=1/(e^-(5x+c) +1) thanks very much for help doogle

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

e^-(5x+c)=(1-y)/y e^-(5x+c)=(1/y)-(y/y) e^-(5x+c)=(1/y)-(1) e^(5x+c)=y-1 e^(5x+c)+1=y?