In problem 17 –26, solve the initial value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{y}\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$

A first-order ordinary differential equation having the form $\frac{dy}{dx}=f(x,y),y\left(x_0\right)=y_0$is referred to as an initial value problem.

The solution of an initial value problem is a particular curve that satisfies the given differential equation and passes through the point $(x_0,y_0)$.

The particular solution is determined from the general solution. By plugging the initial condition into the general condition, the numerical value of the integrating constant is determined. It results a particular curve belonging to the family of curves.

The given equation is

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

After separating the variables, equation (1) can be written as

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

Integrate both sides of equation (2). It results,

$$\begin{gathered} \int \frac{dy}{1-y}=\int x^{3}dx \\ \log \left(\frac{1}{1-y}\right)=\frac{x^4}{4}+\log C[C=IntegratingCons\tan t] \\ \log \left(\frac{1}{C(1-y)}\right)=\frac{x^4}{4} \\ y=1-C{e}^{-\frac{x^4}{4}} \end{gathered}$$

Now, apply the initial condition by plugging in the general solution. It follows,

3=1-C

C=-2

Hence, the solution of the given initial value problem is $y=1+2e^{-\frac{x^4}{4}}.$ .