Solve initial value problem $\frac{dy}{dx}=x^2(1+y), y(0)=3$

Question

We have

$\int\frac{1}{1+y}dy=\int{x^2dx}$,

$\ln|1+y|=\frac{x^3}3+C$

$C=\ln4$

The solution from my textbook is $y=4e^{\frac{x^3}3}-1$, which implies that $1+y$ is always positive. Why is it positive?

Answer 1

The solution involving the arbitrary constant is $$\ln|1+y|=\frac{x^3}{3}+C.$$ Put $y(0)=3,$ we get $C=\ln (4).$ So, we have the solution to the given IVP as $$\ln|1+y|=\frac{x^3}{3}+\ln(4).$$ Hence, $|1+y|=e^{\frac{x^3}{3}+\ln (4)}.$ Or we can write, $$|1+y|=4e^{\frac{x^3}{3}}.$$ So, $1+y=4e^{\frac{x^3}{3}}$ or $1+y=-4e^{\frac{x^3}{3}}.$
But latter does not satisfy the given initial condition. Because, in that $y(0)=-5.$ So,we have the solution to the given IVP as $1+y=4e^{\frac{x^3}{3}}.$ Now it is easy to see that $1+y>0$ as exponential function $e^x$ is always positive, whatever real value $x$ might take.

Answer 2

You are right that the immediate conclusion is just that $$ 4e^{\tfrac{x^3}{3}}=|y+1|=\begin{cases}y+1&y\geq-1\\-y-1&y\leq-1\end{cases} $$ However, we know from the initial condition that $y(0)=3>-1$ so the real question is:

Why does the solution $y$ stay above $-1$?

Since we know that $y=4e^{\tfrac{x^3}{3}}-1$ (initially at least in a small neighbourhood around $x=0$) we see that if we walk left or right along the trajectory $y$ will remain always in the upper branch since $y=4e^{\tfrac{x^3}{3}}>-1$.

Answer 3

Continuing,

$$1+y=e^{C+\frac{x^3}{3}}=C_1e^{\frac{x^3}{3}}$$

Initial value condition gives $C_1=1+3=4$

$1+y$ has sign of arbitrary constant $C_1=4$ because the exponential is always positive.