Let the solution curve of the differential equation [x ~d y=(x^2+y^2+y) d x, x0], interse…

Question

Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

Accepted Answer

$${{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx$$

$$ \Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}$$

$$ \Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}}  + {y \over x}$$

Let $${y \over x} = v$$

$$ \Rightarrow v + x{{dv} \over {dx}} = \sqrt {1 + {v^2}}  + v$$

$$ \Rightarrow {{dv} \over {\sqrt {1 + {v^2}} }} = {{dx} \over x}$$

OR $$\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$$

at $$x = 1,\,y = 0$$

$$ \Rightarrow C = 0$$

$${y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}}  = x$$

At $x = 2$,

$${y \over 2} + \sqrt {1 + {{{y^2}} \over 4}}  = 2$$

$$ \Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y$$

OR $$y = {3 \over 2}$$

Solutions · 1