(x2-y2)dx+2xydy=0 how can you solve it
Answers
Answer :
y² + x² = cx
Solution :
Here ,
The given differential equation is ;
(x² - y²)dx + 2xy·dy = 0
The given differntial equation can be rewritten as ;
=> 2xy·dy = (y² - x²)dx
=> dy/dx = (y² - x²)/2xy = f(x,y) {say}
Now ,
=> f(kx,ky) = {(ky)² - (kx)²}/2kx·ky
=> f(kx,ky) = (k²y² - k²x²)/2xy·k²
=> f(kx,ky) = k²(y² - x²)/2xy·k²
=> f(kx,ky) = (y² - x²)/2xy
=> f(kx,ky) = f(x,y)
Since f(kx,ky) = f(x,y) , thus f(x,y) is homogeneous .
Hence ,
dy/dx = (y² - x²)/2xy --------(1)
is a homogeneous differntial equation .
Now ,
Let y = vx ---------(2)
Now ,
Differentiating both sides of eq-(2) with respect to x , we get ;
=> dy/dx = d(vx)/dx
=> dy/dx = v·(dx/dx) + x·(dv/dx)
=> dy/dx = v·1 + x·(dv/dx)
=> dy/dx = v + x·(dv/dx) ---------(3)
Now ,
From eq-(1) and (3) , we get ;
=> v + x·(dv/dx) = (y² - x²)/2xy
=> v + x·(dv/dx) = {(vx)² - x²}/2x·vx
=> v + x·(dv/dx) = (v²x² - x²)/2vx²
=> v + x·(dv/dx) = x²(v² - 1)/2vx²
=> v + x·(dv/dx) = (v² - 1)/2v
=> x·(dv/dx) = (v² - 1)/2v - v
=> x·(dv/dx) = (v² - 1 - 2v²)/2v
=> x·(dv/dx) = (-v² - 1)/2v
=> x·(dv/dx) = -(v² + 1)/2v
=> [2v/(v² + 1)]·dv = -(1/x)·dx ------(4)
Now ,
Integrating both sides of eq-(4) , we get ;
=> ∫[2v/(v² + 1)]·dv = -∫(1/x)·dx
=> ln(v² + 1) = -ln(x) + ln(c)
=> ln(v² + 1) = ln(c/x)
=> v² + 1 = c/x
=> (y/x)² + 1 = c/x [Using eq-(2) , v = y/x]
=> y²/x² + 1 = c/x
=> (y² + x²)/x² = c/x
=> y² + x² = x²(c/x)
=> y² + x² = cx , where c is the integrating constant .
Hence ,
Required solution is : y² + x² = cx
Answer:
ANSWER
Given, (x
2
+y
2
)dx−2xydy=0
⇒(x
2
+y
2
)dx=2xydy
⇒
dx
dy
=
2xy
x
2
+y
2
.... (i)
Let y=vx
Thus,
dx
dy
=v+x
dx
dv
Thus, v+x
dx
dv
=
2x(vx)
x
2
+(vx)
2
⇒v+x
dx
dv
=
2v
1+v
2
⇒x
dx
dv
=
2v
1+v
2
−v
⇒x
dx
dv
=
2v
1+v
2
−2v
2
⇒x
dx
dv
=
2v
1−v
2
⇒
x
dx
=
1−v
2
2v
dv
⇒
x
dx
−
1−v
2
2v
dv=0 .... (ii)
Integrating both sides, we have
logx+log(1−v
2
)=logC
⇒logx(1−v
2
)=logC
⇒x(1−v
2
)=C
⇒x(1−
x
2
y
2
)=C
⇒x(
x
2
x
2
−y
2
)=C
⇒x
2
−y
2
=Cx