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

Name: 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
Uploaded: 2022-02-23T17:22:57+00:00
Description: 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

Question

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

servidopolisxv · Accepted Answer

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+3v−v  xdvdx=3+25v−v(25+3v)25+3v  xdvdx=3−3v225+3v  (25+3v3−3v2)dv=1xdx  Step 3  Integrating both sides, we get  ∫(25+3v3−3v2)dv=∫1xdx  13[∫25(1−v2)dv+∫3v(1−v2)dv]=ln⁡x+c  13[252ln⁡(1+v1−v)−32ln⁡(1−v2)]=ln⁡x+c

Jim Hunt · Accepted Answer

dydx=(3x+25y)(25x+3y)  Let y=vx⇒dydx=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+25v−25v−3v2(25+3v)  ⇒xdvdx=3−3v225+3v  ⇒25+3v3−3v2dv=dxx  Take, integration both sides  ⇒∫PSK25+v3−3v2dv=∫dxx (1)  Take, ∫25+3v3−3v2dv=∫(−vv2−1−253(v2−1))dv  =−∫vv2−1dv−253∫1v2−1dv  Take, ∫vv2−1dv=12∫dtt=12ln⁡(|t|)+c1.

Vasquez · Accepted Answer

−(3x+25y)dx+(25x+3y)dy=0dydx=3x+25y25x+3yy=vxdydx=v+xdvdxv+xdvdx=3x+25vx25x+3vxv+xdvdx=3+25v25+3vxdvdx=3+25v25+3v−vxdvdx=3+25v−v(25+3v)25+3vxdvdx=3−3v225+3v(25+3v3−3v2)dv=1xdx∫(25+3v3−3v2)dv=∫1xdx13[∫25(1−v2)dv+∫3v(1−v2)dv]=ln⁡x+c13[252ln⁡(1+v1−v)−32ln⁡(1−v2)]=ln⁡x+c25ln⁡(1+v1−v)−3ln⁡(1−v2)=6ln⁡x+Cv=yx25ln⁡(1+yx1−yx)−3ln⁡(1−y2x2)=6ln⁡x+C25ln⁡(x+yx−y)−3ln⁡(x2−y2x2)=6ln⁡x+C