Answer to Question #151727 in Differential Equations for arbaz

Question

solve dy/dx = x^3(y^2 + 1 )

Answer 1

Solution: The Given Ordinary Differential Equation is First order separable ODE form. A first order separable ODE has the form $N(y)~y'=M(x)$ .

Therefore rewrite in the form of a first order separable form ODE

$\frac{1}{1+y^2}~y'=x^3$

Integrate both sides with respect to $x$

$\int\frac{1}{1+y^2}~y'dx=\int x^3dx$

$\int\frac{1}{1+y^2}~y'dx=\int\frac{1}{1+[y(x)]^2}~y'(x)dx=\int\frac{1}{1+y^2}~\frac{dy}{dx}dx=\int\frac{1}{1+y^2}~dy=\int x^3dx$

$\therefore$ considering,

$\int\frac{1}{1+y^2}~dy=\int x^3dx$ [using integration formula's]

$arctan(y)=\frac{x^4}{4}+c~~~~or ~~~~~tan^{-1}y=\frac{x^4}{4}+c$

$\therefore y=tan(\frac{x^4}{4}+c)$

This is general solution of the given differential equation.