Recent content by BarackObama

Can someone help me out with variation of parameters? It's urgent!

Homework Statement 4y'' + y = cosx Solve using variation of parameters Homework Equations The Attempt at a Solution from a) -> yc(x) = c1cos(x/2) + c2sin(x/2) let y1 = cos(x/2) , y2 = sin(x/2) y1y2' - y2y1' = 1/2cosx/2 + 1/2sinx/2 = 1/2 u1' = ? How do I find this?

Homework Statement xy' - 4y = x4ex Homework Equations The Attempt at a Solution y' - 4x-1y = x3ex x-4y' - 4x-5y = x-1ex I'm not sure what to do next, I can't express the LS as a derivative

right, let u = sin^2x can you show me how to integrate the left side?

Ok, I can do that substitution and then integrate by parts. Still not sure how to integrate the left side though.

I'm still not sure how to proceed with this one... how do I integrate the right hand side?

Homework Statement dy/dx = e^ysin^2x/ysecx Stewart 6e 10.3 # 8 Homework Equations The Attempt at a Solution ydy/e^y = sin^2xdx/secx e^-ydy = sec^-1xsin^2xdx Integration by parts u = e^-y du = -e^-y dv = ydy v = y^2/2 ∫udv = e^-yy^2/2 + ∫y^2/2e^-y = y^2/2e^y +...

Homework Statement dy/dz = ycosx/(1+y^2), y(0) = 1 Stewart 6e, 10.3 # 12 Homework Equations The Attempt at a Solution ∫(1+y^2)dy/y = ∫cosxdy -------------- = sinx + C How do I find the integral of this product? Do I use integration by parts?

z = -ln(e^t + C) ... not exactly easy to verify

Thanks! y= 1/(cosx + C)

Good morning! ∫e^-zdz = ∫-e^tdt -e^-z = -e^t + C e^-z = e^t + C Is there any way to bring the variables down? Thanks!

Homework Statement dz/dt + e^(t+z) = 0 Homework Equations The Attempt at a Solution dz/dt = -e^te^z integral(dz/e^z) = integral(-e^tdt) let u = 1/e^z dv = dz du = -e^-zdz v= z integral(udv) = z/e^z + integral(ze^-zdz)

Homework Statement y' = y^2sinx Homework Equations The Attempt at a Solution dy/dx = y^2sinx dy/y^2 = dxsinx integral(dy/y^2) = integral(sinxdx) Aside: let u = y^2 du = 2ydy dy = du/2y ln(abs(y^2))/2y = -cosx+C integral(du/2y^3) = integral(sinxdx)