Solve for G.S. / P.S. for the following differential equations

Painevg

Painevg

Answered question

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

Beginner2022-01-01Added 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

Beginner2022-01-02Added 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

Expert2022-01-09Added 669 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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

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

Didn't find what you were looking for?