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$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^2\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The solution is $y=1+\frac{1}{10}\tan \left(\frac{x}{10}\right)$.

Step-by-step solution

Definition

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

Integrate

Consider the initial value problem (IVP)$\frac{dy}{dx}={(y-1)}^2+\mathrm{0.01.}$

Separate the variables$\frac{dy}{{(y-1)}^2+0.01}=dx$

Integrate on both sides.

$\begin{array}{c}\int \frac{\mathrm{dy}}{{(\mathrm{y}-1)}^2+0.01}=\int \mathrm{d}\\ \mathrm{x}\frac{1}{0.1}{\tan }^{-1}\left(\frac{\mathrm{y}-1}{0.1}\right)=\mathrm{x}+\mathrm{C}\\ 10{\tan }^{-1}\left(10\right(\mathrm{y}-1\left)\right)=\mathrm{x}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Use}\frac{1}{0.1}=10\\ {\tan }^{-1}\left(10\right(\mathrm{y}-1\left)\right)=\frac{\mathrm{x}+\mathrm{C}}{10}\\ 10(\mathrm{y}-1)=\tan \left(\frac{\mathrm{x}+\mathrm{C}}{10}\right)\\ \mathrm{y}=1+\frac{1}{10}\tan \left(\frac{\mathrm{x}+\mathrm{C}}{10}\right)\end{array}$

Apply initial condition

The initial condition is $y\left(0\right)=1$.

Substitute $x=0,y=1$in $y=1+\frac{1}{10}\tan \left(\frac{x+C}{10}\right)$.

$\begin{array}{c}y\left(0\right)=1+\frac{1}{10}\tan \left(\frac{x+C}{10}\right)\\ 1=1+\frac{1}{10}\tan \left(\frac{0+C}{10}\right)\\ 0=\frac{1}{10}\tan \left(\frac{C}{10}\right)\\ \tan \left(\frac{C}{10}\right)=0\\ \frac{C}{10}={\tan }^{-1}\left(0\right)\\ c=0\end{array}$

Solution

Substitute$C=0$ in $y=1+\frac{1}{10}\tan \left(\frac{x+C}{10}\right)$.

Thus, the solution becomes as follows:

$\begin{array}{c}y=1+\frac{1}{10}\tan \left(\frac{x+C}{10}\right)\\ y=1+\frac{1}{10}\tan \left(\frac{x+0}{10}\right)\\ y=1+\frac{1}{10}\tan \left(\frac{x}{10}\right)\end{array}$

Therefore, the solution of the IVP is $y=1+\frac{1}{10}\tan \left(\frac{x}{10}\right)$.