(x2+1)dy/dx+4xy=1/x2+1
Answers
Answered by
0
read and understand our Cookie Policy, Privacy Policy, and our Terms of Service.
Questions
Tags
Users
Badges
Ask
up vote4down votefavorite
How to solve (x+2y−4)dx−(2x+y−5)dy=0(x+2y−4)dx−(2x+y−5)dy=0
differential-equations
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=0 Answer: 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=0 Answer: 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.
Questions
Tags
Users
Badges
Ask
up vote4down votefavorite
How to solve (x+2y−4)dx−(2x+y−5)dy=0(x+2y−4)dx−(2x+y−5)dy=0
differential-equations
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=0 Answer: 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=0 Answer: 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.
Similar questions