Question

Find an implicit expression for the solution y of the IVP (6x^5 − xy) + (−x^2 +xy^2)y'=0
with initial condition y(0) = 1.

Answer 1

Step-by-step explanation:

We have,

$(6 {x}^{5} - xy) + ( - {x}^{2} + x {y}^{2} )y^{ \prime} = 0$

$\implies y^{ \prime} = \frac{ (xy - 6 {x}^{5} )} {( - {x}^{2} + x {y}^{2} ) }$

$\implies \frac{dy}{dx} = \frac{ (y - 6 {x}^{4} )} {( {y}^{2} - x ) }$

$\implies {y}^{2} dy - xdy = ydx - 6 {x}^{4} dx$

$\implies {y}^{2} dy + 6 {x}^{4} dx = ydx + xdy$

$\implies {y}^{2} dy + 6 {x}^{4} dx = d(xy)$

$\implies \int{y}^{2} dy + \int6 {x}^{4} dx = \int \: d(xy)$

$\implies\frac{ {y}^{3} }{3} + 6 \frac{ {x}^{5} }{5} = xy + c$

since y(0) =1, so c=1/3

$\implies\frac{ {y}^{3} }{3} + \frac{ 6{x}^{5} }{5} = xy + \frac{1}{3}$