Determine whether the following differential equations are exact or not exact. if exact, solve for...

hvacwk

hvacwk

Answered

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

Expert

2021-12-30Added 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

Expert

2021-12-31Added 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

Expert

2022-01-10Added 439 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

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?