Find the solution of the following Differential Equations (2x+e^{y})dx + xe^{y}

reproacht3

reproacht3

Answered question

2021-12-31

Find the solution of the following Differential Equations
(2x+ey)dx+xeydy=0

Answer & Explanation

nghodlokl

nghodlokl

Beginner2022-01-01Added 33 answers

Step 1
(2x+ey)dx+xeydy=0
2xdx+eydx+xeydy=0
2xdx+d(xey)=0
d(fg)=fd(g)+gd(f)
Step 2
2xdx+d(xey)=0
2x22+xey=c
x2+xey=c
Pademagk71

Pademagk71

Beginner2022-01-02Added 34 answers

Solution:
(2x+ey)dx+xeydy=0,y(1)=0
Compare above with Mdx+Ndy=0
M=2x+ey,N=xey
My=ey,Nx=ey
Since My=Ny=Nx, therefore differential equation is exact
fx=2x+ey (i)
fy=xey (ii)
integrate (i) wrt x, we get
f(x,y)=(2x+ey)dx
f(x,y)=x2+xey+h(y)
Differentiate above wrt y, we get
fy=xey+h(y)
compare above and (ii), we get
xey+h(y)=xey
h(y)=0
integrate both sides wrt y, we get h(y)=C
therefore, we get
f(x,y)=x2+xey+C
or we can write as x2+xey+C=0
karton

karton

Expert2022-01-09Added 613 answers

Consider the differential equation
(2x+ey)dx+xeydy=0
Compare it with M(x,y)dx+N(x,y)dy=0, get
M(x,y)=2x+ey and N(x,y)=xey
My=ey,Nx=ey
Observe that, My=Nx
So, the given differential equation is in exact form
(2x+ey)dx+xeydy=0
2xdx+eydx+xeydy=0
2xdx+d(xey)=0
on taking integration get
x2+xey=C
which is the required solution.

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