Determine whether the following differential equations are exact or not

hvacwk

hvacwk

Answered question

2021-12-29

Determine whether the following differential equations are exact or not exact. if exact, solve for the general solution
(x+2y)dx(2x+y)dy=0

Answer & Explanation

turtletalk75

turtletalk75

Beginner2021-12-30Added 29 answers

Step 1
(x+2y)dx(2x+y)dy=0
The differential equation Mdx+Ndy=0 is a exact if
My=Nx
Step 2
From the given differential equation
M=x+2y
N=(2x+y)
N=(2x+y)
My=2
Nx=2
Since, MyNx
The given differential equation is not exact differential equation.
Buck Henry

Buck Henry

Beginner2021-12-31Added 33 answers

Simplifying
(x+2y)dx+1(2x+1y)dy=0
Reorder the terms for easier multiplication:
dx(x+2y)+1(2x+1y)dy=0
(xdx+2ydx)+1(2x+1y)dy=0
Reorder the terms:
(2dxy+dx2)+1(2x+1y)dy=0
(2dxy+dx2)+1(2x+1y)dy=0
Reorder the terms for easier multiplication:
2dxy+dx2+1dy(2x+1y)=0
2dxy+dx2+(2x1dy+1y1dy)=0
2dxy+dx2+(2dxy+1dy2)=0
Reorder the terms:
2dxy+2dxy+dx2+1dy2=0
Combine like terms: 2dxy+2dxy=0
0+dx2+1dy2=0
dx2+1dy2=0
Solving
dx2+1dy2=0
Solving for variable d.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), d.
d(x2+y2)=0
karton

karton

Expert2022-01-10Added 613 answers

It is a homogeneous equation
Put y=vxSo, v+xdvdx=x+2vx2xvxv+xdvdx=1+2v2vxdvdx=1+2v2vvxdvdx=1+2v2v+v22vxdvdx=1+v22vIntegrating both sides we get,2v1+v2dv=dxx21+v2dvv1+v2dv=dxx2tan1v12log|1+v2|=log|x|+logc2tan1v=log|xc|+log|1+v2|12etan1v={1+v2}12xcetan1yx={(y2+x2)x2}12xcetan1yx={(y2+x2)}12c

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