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प्रश्न
Solve the differential equation `dy/dx=(y+sqrt(x^2+y^2))/x`
उत्तर
`dy/dx=(y+sqrt(x^2+y^2))/x`
Put y = vx
`therefore dy/dx=v+x (dv)/dx`
`therefore (1) becomes, v+x (dv)/dx=(vx+sqrt(x^2+v^2x^2))/x`
`therefore v+x (dv)/dx =v+sqrt(1+v^2)`
`therefore 1/sqrt(1+v^2) dv=1/x dx`
Integrating, we get,
`int 1/sqrt(1+v^2) dv=int 1/x dx+c_1`
`therefore log |v+sqrt(1+v^2)|=log|x|+logc, where c_1=logc`
`therefore (y+sqrt(x^2+y^2))/x=cx`
`(y+sqrt(x^2+y^2))=cx^2 ` is the general solution
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