Solve first order DE
x(x-y)dy=y(x+y)dx
Answers
Answer:
∴ The homogeneous equation is log(\frac{y}{x} )+\frac{y}{x} =-2logx-2clog(
x
y
)+
x
y
=−2logx−2c
Step-by-step explanation:
Given;
x(x-y)dy=y(x+y)dxx(x−y)dy=y(x+y)dx
⇒\frac{dy}{dx} =\frac{y(x+y)}{x(x-y)}
dx
dy
=
x(x−y)
y(x+y)
___equation-1
Let, y=vxy=vx
differentiate with respect to xx ;
⇒\frac{dy}{dx}=v+x\frac{dv}{dx}
dx
dy
=v+x
dx
dv
Plug \frac{dy}{dx}
dx
dy
value in equation-1;
∴v+x\frac{dv}{dx}=\frac{vx(x+vx)}{x(x-vx)}v+x
dx
dv
=
x(x−vx)
vx(x+vx)
⇒v+x\frac{dv}{dx}=\frac{v(1+v)}{(1-v)}v+x
dx
dv
=
(1−v)
v(1+v)
⇒x\frac{dv}{dx}=\frac{(v+v^{2} )}{(1-v)}-vx
dx
dv
=
(1−v)
(v+v
2
)
−v
⇒x\frac{dv}{dx}=\frac{(v+v^{2} -v+v^{2} )}{(1-v)}x
dx
dv
=
(1−v)
(v+v
2
−v+v
2
)
⇒x\frac{dv}{dx}=\frac{2v^{2}}{(1-v)}x
dx
dv
=
(1−v)
2v
2
⇒\frac{1-v}{2v^{2} } dv=\frac{dx}{x}
2v
2
1−v
dv=
x
dx
Integrating on both sides;
\int \frac{1-v}{2v^{2} } dv=\int \frac{dx}{x}∫
2v
2
1−v
dv=∫
x
dx
By integrating of \int \frac{1-v}{2v^{2} } dv∫
2v
2
1−v
dv the value will be come -\frac{1}{2}log(v)-\frac{1}{2}v−
2
1
log(v)−
2
1
v
So,
-\frac{1}{2}log(v)-\frac{1}{2}v=logx+c−
2
1
log(v)−
2
1
v=logx+c where c=c= constant
⇒log(v)+v=-2logx-2clog(v)+v=−2logx−2c
∴log(\frac{y}{x} )+\frac{y}{x} =-2logx-2clog(
x
y
)+
x
y
=−2logx−2c (∵v=\frac{y}{x}v=
x
y
)
So the homogeneous equation is log(\frac{y}{x} )+\frac{y}{x} =-2logx-2clog(
x
y
)+
x
y
=−2logx−2c