Question

Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. In Problems 39-42 find an explicit solution of the given initial-value problem. Use a graphing utility to plot the graph of each solution. Compare each solution curve in a neighborhood of $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^2\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The solution is$\mathrm{y}=-\frac{1}{\mathrm{x}+\mathrm{c}}+1$ .

Step-by-step solution

Definition

A first-order differential equation of the form $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$is said to be separable or to have separable variables.

Integrate

Consider the initial value problem,$\frac{dy}{dx}={(y-1)}^2,y\left(0\right)=1.$

Separate the variables$\frac{dy}{{(y-1)}^2}=dx\mathrm{.....}\text{(1)}$

Apply integral of the above equation.

$\begin{array}{c}\int \frac{\mathrm{dy}}{{(\mathrm{y}-1)}^2}=\int \mathrm{d}\mathrm{x}\\ -\frac{1}{\mathrm{y}-1}=\mathrm{x}+\mathrm{c}\\ \mathrm{y}-1=-\frac{1}{\mathrm{x}+\mathrm{c}}\\ \mathrm{y}=-\frac{1}{\mathrm{x}+\mathrm{c}}+1\end{array}$

As the value $c$cannot be find by the initial condition $y\left(0\right)=1$.

Thus,$y=1$is a singular solution as left-hand side of (1) is undefined at $y=1$.

The singular solution $y=1$satisfied the initial value problem.

Graph

Sketch the graph of the singular solution$y=1$ is shown below:

Curve

Using a graphing utility, all the solution curves for the given differential equation are plotted near the neighborhood of $\left(0,1\right)$as shown below. We see that all the solutions approach the singular solution$y=1$ in the neighborhood.

Observe that from the figure, the solution in the interval $\left[0,1\right]$approaches to the singular solution at $y=1$.