Find the general solutions to the following differential equations:
a. (dy/dx)= cube-root (x/y)
b. (d^2x/dt^2)= -ω^2x
c. (d^2x/dt^2)= -g
These are problems from a practice exam, I'm very lost, please show your work! Thank you so much
by 
Anonymous
ANSWERS: 1
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Part a. (dy/dx)= cube-root (x/y) cube-root(x/y) = {x^(1/3)}/{y^(1/3)} (dy/dx) = {x^(1/3)}/{y^(1/3)} multiply both sides by {y^(1/3)}dx {y^(1/3)}dx(dy/dx) = {y^(1/3)}dx{x^(1/3)}/{y^(1/3)} on the left hand side dx/dx = 1 and on the right hand side {y^(1/3)}/{y^(1/3)} = 1 {y^(1/3)}dy = {x^(1/3)}dx integrate both sides: ∫{y^(1/3)}dy = ∫{x^(1/3)}dx (3/4){y^(4/3)} = (3/4){x^(4/3)} + c multiply both sides by 4/3 and let k = (4/3)c {y^(4/3)} = {x^(4/3)} + k raise both sides to the 3/4 power: y = [{x^(4/3)} + k]^(3/4) Part b. (d²x/dt²)= -ω²x multiply both sides by dt²/x: (d²x/dt²)(dt²/x) = -ω²x(dt²/x) on the left hand side dt²/dt² = 1 and on the right hand side x/x = 1: (d²x/x) = -ω²dt² integrate of both sides: ∫(d²x/x) = -∫ω²dt² (ln(x) + k)dx = (-ω²t + c)dt integrate both sides again: ∫(ln(x) + k)dx = ∫(-ω²t + c)dt x{ln(x) + k - 1} = (-ω²/2)t² + ct + m This one does not look good. There may be a mistake but I cannot find it. Sorry. Part c. (d²x/dt²) = -g integrate both sides: dx/dt = -gt + c integrate again: x = (-g/2)t² + ct + k
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