Differential Equations?

These are both homogeneous DE's.

1) Rewrite the DE as [x/y + √(1 - x/y)] dy – dx = 0.

Let x = uy, dx = u dy + y du.

Thus, [u + √(1 - u)] dy – (u dy + y du) = 0.

==> √(1 - u) dy – y du = 0.

Now, we solve by separating variables:

dy/y = du/√(1 - u)

==> ln|y| = 2√(1 - u) + C'

==> y = Ce^[2√(1 - u)], with C = e^(C')

==> y = Ce^[2√(1 - x/y)].

2) Rewrite as (1 + y/x) dx – (1 - y/x) dy = 0.

Let y = xu, dy = x du + u dx.

So, (1 + u) dx - (1 - u)(x du + u dx) = 0.

==> (u - 1) x du + (u^2 + 1) dx = 0.

Separating variables:

(u - 1) du/(u^2 + 1) + dx/x = 0.

Integrate both sides:

(1/2) ln(u^2 + 1) - arctan(u) + ln|x| = C.

Letting u = y/x:

(1/2) ln(y^2/x^2 + 1) - arctan(y/x) + ln|x| = C.

==> (1/2) ln(y^2/x^2 + 1) - arctan(y/x) + (1/2) * ln(x^2) = C.

==> (1/2) ln(y^2 + x^2) - arctan(y/x) = C.

(Multiplying both sides by -1 gives your answer above. Remember that C is arbitrary.)

I hope this helps!

dy/dx = (x^2)(8 + y) First, we separate the variables by multiplying by dx and dividing by (8 + y): dy/(8 + y) = (x^2)dx combine the two sides, remembering the left will use a organic log and the the final option might have a persevering with: ln(8 + y) = (a million/3)x^3 + C improve e to the capability of the two section: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final form: y = Ae^((a million/3)x^3) - 8 Now, we've the element (0, 3), so plug those in to discover a and the particular answer: 3 = Ae^((a million/3)(0)^3) - 8 clean up for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the particular answer is: y = 11e^((a million/3)x^3) - 8

Equation 1. is homogeneous. Substitute v = y/x.

Equation 2. is exact.

Consult the standard techniques for solving equations of these types.