Solve both the differential equations. (y^{2} + 1)dx + (2xy+

Terrie Lang

Terrie Lang

Answered question

2021-12-30

Solve both the differential equations. (y2+1)dx+(2xy+1)dy=0

Answer & Explanation

Marcus Herman

Marcus Herman

Beginner2021-12-31Added 41 answers

dydx=x2+y22xy
2xydy=x2dx+y2dx
my=2y
mx=2y
(y2+1)dx+(2xy+1)dy=0
The differential equation is of the form
Mdx+Mdy=0
Where m=y2+1,N=2xy+1
Find my,Nx
my=2y,Nx=2y
my=Nx
The differential equation is Exact
Solution of Exact Differential Equation is
Mdx+Ndy=C
(y2+1)dx+1dy=c
(y2+1)x+y=c
y2x+x+y=c
is the required solution
Bubich13

Bubich13

Beginner2022-01-01Added 36 answers

Solution:
Differential equation Mdx+Ndy=0
is exact if Nn=My
(y2+1)dx+(2xy+1)dy=0
Compase to Mdx+Ndy=0
N=2xy+1 M=y2+1
Ny=2y My=2y
Nn=My
So given differential equation is exact
Solution of differential equation
Mdx+Ndy=c
(y2+1)dx+1dy=c
(y2+1)x+y=c
xy2+x+y=c
xy2+y+xc=0
y=1±(1)24x(xc)2×x
y=1±14x(xc)2x
So y=1+14x(xc)2x,114x(xc)2x
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

dydx=x2+y22xy2xydy=x2dx+y2dxmy=2ymx=2y(y2+1)dx+(2xy+1)dy=0Mdx+Mdy=0m=y2+1,N=2xy+1Find my,Nxmy=2y,Nx=2ymy=NxMdx+Ndy=C(y2+1)dx+1dy=c(y2+1)x+y=cy2x+x+y=c

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