Math, asked by jharanagiri, 3 months ago

(x2-y2)dx+2xydy=0 how can you solve it​

Answers

Answered by AlluringNightingale
4

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

Answered by shuklamamta548
2

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

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