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प्रश्न
Solve the differential equation (x2 + y2)dx- 2xydy = 0
उत्तर
(x2 + y2)dx- 2xydy = 0
(x2 + y2) dx = 2xydy
`dy/dx = (x^2 + y^2)/(2xy)`.........(i)
The equation is a homogeneous equation
Let y= vx,
Differentiat ing w.r.t. x, we get,
`dy/dx=v+x(dv)/dx`
`dy/dx=(x^2+y^2)/(2xy) " from "(i)`
`v+x(dv)/dx=(x^2+(vx)^2)/(2x.(vx))`
`v+x(dv)/dx=(1+v^2)/(2v)`
`x(dv)/dx=(1+v^2)/(2v)-v`
`x(dv)/dx=(1+v^2-2v^2)/(2v)`
`x(dv)/dx=(1-v^2)/(2v)`
`(2v)/(1-v^2)dv=1/xdx`.......(ii)
Which is in variables separatable form
∴ Integrating both sides, we get
`int(2v)/(1-v^2)dv=int1/xdx + c_1`
`therefore -log|1-v^2|=log|x|+logc`
`therefore log|x(1-v^2)|=log|c|`
`therefore x(1-v^2)=c`
Resubstituting `v=y/x` we get
`x(1-y^2/x^2)=c`
`x((x^2-y^2)/x^2)=c `
`therefore x^2 - y^2 = cx`, where c is constant
which is the required general solution
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