Solve for G.S. / P.S. for the following differential equations using the method of solution...

Painevg

Painevg

Answered

2021-12-31

Solve for G.S. / P.S. for the following differential equations using the method of solution for homogeneous equations.
(3x+25y)dx+(25x+3y)dy=0

Answer & Explanation

servidopolisxv

servidopolisxv

Expert

2022-01-01Added 27 answers

Step 1
Given differential equation is (3x+25y)dx+(25x+3y)dy=0
Given differential equation can be written as:
dydx=3x+25y25x+3y
This homogeneous differential equation.
Put y=vx
dydx=v+xdvdx
Step 2
Substituting the value, we get
v+xdvdx=3x+25vx25x+3vx
v+xdvdx=3+25v25+3v
xdvdx=3+25v25+3vv
xdvdx=3+25vv(25+3v)25+3v
xdvdx=33v225+3v
(25+3v33v2)dv=1xdx
Step 3
Integrating both sides, we get
(25+3v33v2)dv=1xdx
13[25(1v2)dv+3v(1v2)dv]=lnx+c
13[252ln(1+v1v)32ln(1v2)]=lnx+c
Jim Hunt

Jim Hunt

Expert

2022-01-02Added 45 answers

dydx=(3x+25y)(25x+3y)
Let y=vxdydx=v+xdvdx
v+xdvdx=(3x+25vx)(25x+3vx)=x(3+25v)x(25+3v)
=(3+25v)(25+3v)
xdvdx=(3+25v)(25+3v)v=3+25v25v3v2(25+3v)
xdvdx=33v225+3v
25+3v33v2dv=dxx
Take, integration both sides
PSK25+v33v2dv=dxx (1)
Take, 25+3v33v2dv=(vv21253(v21))dv
=vv21dv2531v21dv
Take, vv21dv=12dtt=12ln(|t|)+c1.
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

(3x+25y)dx+(25x+3y)dy=0dydx=3x+25y25x+3yy=vxdydx=v+xdvdxv+xdvdx=3x+25vx25x+3vxv+xdvdx=3+25v25+3vxdvdx=3+25v25+3vvxdvdx=3+25vv(25+3v)25+3vxdvdx=33v225+3v(25+3v33v2)dv=1xdx(25+3v33v2)dv=1xdx13[25(1v2)dv+3v(1v2)dv]=lnx+c13[252ln(1+v1v)32ln(1v2)]=lnx+c25ln(1+v1v)3ln(1v2)=6lnx+Cv=yx25ln(1+yx1yx)3ln(1y2x2)=6lnx+C25ln(x+yxy)3ln(x2y2x2)=6lnx+C

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