Question

solve the differential equation (3xy+y^2)dx=(x^2+xy)dy​

Answer 1

Step-by-step explanation:

$(3xy + {y}^{2} )dx = ( {x}^{2} + xy)dy \\ \\ \frac{dy}{dx} = \frac{3xy + {y}^{2} }{ {x}^{2} + xy} \\ \\ it \: \: is \: the \: \: homogenious \: \: differential \: \: equation \\ maximum \: \: power \: \: 2 \\ \\ let \: \: y \: = vx \\ \frac{dy}{dx} = v + x \frac{dv}{dx} \\ \\ v + x \frac{dv}{dx} = \frac{3x(vx) + (vx) {}^{2} }{ {x}^{2} + x(vx)} \\ \\ v + x \frac{dv}{dx} = \frac{ {x}^{2}(3v + {v}^{2}) }{ {x}^{2} (1 + v)} \\ \\ v + \frac{dv}{dx} = \frac{3v + {v}^{2} }{1 + v} \\ \\ \frac{dv}{dx} = \frac{3v + {v}^{2} }{1 + v} - v \\ \\ \frac{dv}{dx} = \frac{3v + {v}^{2} - v - {v}^{2} }{1 + v} \\ \\ \frac{dv}{dx} =( \frac{2v}{1 +v } )\\ \\( \frac{1 + v}{v} ) dv = dx \\ \\ \int( \frac{1}{v} + 1)dv =\int( dx )\\ \\ log(v) + v = x + c \\ \\ log( \frac{y}{x} ) + \frac{y}{x} = x + c$

Required ans....

Thank you.....