Question

solve the ODE 3y^2 dy/dx + xy^3=3​

Answer 1

Step-by-step explanation:

We have,

$3 {y}^{2} \frac{dy}{dx} + x {y}^{3} = x$

Order of differential equation is 1

$\implies3 {y}^{2} \frac{dy}{dx} = x(1 - {y}^{3} )$

$\implies \frac{1}{x}. \frac{dy}{dx} = \frac{(1 - {y}^{3}) }{3 {y}^{2} }$

$\implies \frac{3 {y}^{2} }{1 - {y}^{3} } dy = x.dx$

Integrating both sides

$\implies - \int \frac{ - 3{y}^{2} }{1 - {y}^{3} } dy = \int \: x \: dx$

putting 1 - y³ = t => -3y²dy = dt

$\implies - \int \frac{dt}{t} = \int \: x \: dx$

$\implies - ln(t) = \frac{ {x}^{2} }{2} + c$

$\implies ln(1 - {y}^{3} ) = - \frac{ {x}^{2} }{2} + c$

$\implies(1 - {y}^{3} ) = {e}^{ - \frac{ {x}^{2} }{2} + c }$

$\implies(1 - {y}^{3} ) = k {e}^{ - \frac{ {x}^{2} }{2} } \: \: \: (where \: \: k = {e}^{c} )$

$\implies {y}^{3} = 1 - k {e}^{ - \frac{ {x}^{2} }{2} }$

$\implies \: y = \sqrt[3]{1 - k {e}^{ - \frac{ {x}^{2} }{2} } }$