Question

solve (x^2y^3+2xy)dy=dx

Answer 1

How to solve the differential equation (x+2y−4)dx−(2x+y−5)dy=0(x+2y−4)dx−(2x+y−5)dy=0. It's not separable, nor exact nor homogeneous. The solution is (x−y−1)3=C(x+y−3)(x−y−1)3=C(x+y−3). How can I achieve it?

The other equations similar to this are:

(1+x+y)dy−(1−3x−3y)dx=0(1+x+y)dy−(1−3x−3y)dx=0Answer: 3x+y+2ln(−x−y+1)=k3x+y+2ln(−x−y+1)=k

(3x−y+2)dx+(9x−3y+1)dy=0(3x−y+2)dx+(9x−3y+1)dy=0Answer: 2x+6y+C=ln(6x−2y+1)2x+6y+C=ln(6x−2y+1)

If someone point me out how to solve the first equation I will be likely to solve the others. Thank you very much.

Update: Given Orangutango and Chris help I moved to a solvable d.e. But didn't get the same answer my professor listed. Did I miss some step?

(X + 2Y)dX = (2X+Y)dY
dY/dX = (X + 2Y)/(2X + Y)

Making a substitution to solve the now homogenous: Y = VX, Y'= V + V'X

V+V'X = (X + 2VX)/(2X + VX)
V+V'X = (1 + 2V)/(2 + V)
V'X = ((1 + 2V)/(2 + V)) - V
V'X = (1 - V^2)/(2 + V)

(2 + V)dV/(1-V^2) = dX/X

Integrating the left side I got:

int (2 + V)dV/1-V^2 =
2 * int dV / (1-V^2) + int V dV / (1-V^2) =
log | (v + 1) / (v-1) | - 1/2 log | V^2 + 1 | + c1

Integrating the right side: log X + c2

Then replacing V=Y/X and then X=x-2 and Y=y-1 don't seems to yield the proposed answers. Where did my professor got this? Is the above solution correct? Thanks again.

Answer 2

Answer:

(Second form of Bernoulli's Equations)

dx/dy +Px=Q*x^n

Step-by-step explanation: