Find the solution of the following Differential Equations by Exact (2y+xy)dx+2xdy=0,y(3)=2

Tiffany Russell

Tiffany Russell

Answered

2021-12-31

Find the solution of the following Differential Equations by Exact (2y+xy)dx+2xdy=0,y(3)=2

Answer & Explanation

esfloravaou

esfloravaou

Expert

2022-01-01Added 43 answers

Given:
(2x+xy)dx+2xdy=0,y(3)=2
this equation is form M(x,y)dx+N(x,y)dy=0
At this equation is exact if and only it
My=Nx
M=2y+xy My=y(2y+xy)=2+x
N=2x Nx=2
Hence MyNx
Now we find the integrating factor to made this equation exact.
MyNx=2+x2=x
MyNxN=x2x=12
Hence I.F.=e12dx=ex2
ex2(2y+xy)dx+2ex2xdy=0 (2)
=2e2/2y+xex/2y
N=2ex2x
My=2ex2+xex2
Nx=2ex2y+2xex212=2ex2+xex2
My=Mx
equation (2) is exact equation:
Now solution of equation
M dx+M dy=c
(2yex2+xyex2)dx+dy=c

eskalopit

eskalopit

Expert

2022-01-02Added 31 answers

(2y+xy)dx+2xdy=0
rearranging,
dydx=2y+xy2x=2+x2x.y=g(x).y
dydx+g(x).y=0
Integrating Factor
IF=exp(g(x)dx)
g(x)dx=2+x2xdx={1x+12}dx
g(x)dx=ln(x)+x2
Hence,
IF=exp(ln(x)+x2)=xex2
The DE now becomes,
d(IFy)=0
d(xex2.y)=0
Integrating,
xex2.y=const
Answer: xy.ex2=c
karton

karton

Expert

2022-01-09Added 439 answers

\(\begin{array}{}(2y+xy)dx+2ydy=0 \\Rightarrow 2y+xy+2y \frac{dy}{dx}=0 \\text{Substitute}\ \frac{dy}{dx}\ \text{with y}

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