Topic: integrating factor found by inspection 1. t(st^2 + 2t +s)ds + s^2  (1-t^2)dt = 0

Find the general solution. You may use the -3/x or x^-3 for integrating factor. Note: prove that it is x^-3 y=1/2 x^2 +C is the answer in general solution.

Calculus: Sketch: 1. x^2 + 4z^2 - y = 0 2. y^2 = x^2 + z^2 3. x^2 + 2z^2 - 6x - y + 10 = 0

Solve the following equation. Note: the integrating factor must be x^-3 A. x^3 y=1/2 x^2+C B. 1/2x^2+C C. x^-3 y=1/2 x^2+C

Use Gauss's method to solve the cyclotomic equation x^6+x^5+x^4+x^3+x^2+x+1=0.

I need help solving the vertical component equation. where H=-16t^2+v0tsin 0+h0   t=4.3 h=4.8 angel is 65 degree's h0= 4.8. I need to find the v^0 and I'm having trouble figuring it out after I got =-16(4.3)^2+v0(4.3)sin(65)+4.8

The Area of a triangle with vertex (4, 3), (4, -3) and (7, -3) can be obtained through double integration.

Please show full work

I don't understand why T'+(a^2)*k=0 is equal to (c3)e^k(n^2)(pi^2)t+c4(e^-k(n^2)(pi^2)t).Can you please explain it to me. Thank you