Chapter 12, Problem 12.1P

(a)

To determine

The solution for $x_1\left(t\right)$ by separating variables.

Answer to Problem 12.1P

The solution for $x_1\left(t\right)$ by separating variables is $x_1\left(t\right)={\left(t+k\right)}^2+1$ for any constant $k$.

Explanation of Solution

Given nonlinear first-order equation,

$\begin{array}{l}\dot{x}=2\sqrt{x-1}\\ \frac{dx}{dt}=2\sqrt{x-1}\end{array}$

Separating same variables in one side of the equation

$\frac{dx}{\sqrt{x-1}}=2dt$

Integrate the above equation and solving

$\begin{array}{l}{\displaystyle \int \frac{dx}{\sqrt{x-1}}}={\displaystyle \int 2dt}\\ \frac{{\left(x-1\right)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2t+C\end{array}$

Here, $C$ is the integration constant.

$\begin{array}{c}\frac{{\left(x-1\right)}^{\frac{1}{2}}}{\frac{1}{2}}=2t+C\\ 2\sqrt{x-1}=2t+C\\ \sqrt{x-1}=t+\frac{C}{2}\end{array}$

Squaring on both sides

$x-1={\left(t+\frac{C}{2}\right)}^2$

Let, $k=\frac{C}{2}$ another constant.

$\begin{array}{c}x-1={\left(t+k\right)}^2\\ x={\left(t+k\right)}^2+1\end{array}$

Conclusion:

Therefore, the solution for $x_1\left(t\right)$ is ${\left(t+k\right)}^2+1$ for any constant $k$.

(b)

To determine

Show that there is another solution for $\dot{x}=2\sqrt{x-1}$ is $x_2\left(t\right)=1$.

Answer to Problem 12.1P

The another solution for $\dot{x}=2\sqrt{x-1}$ is $x_2\left(t\right)=1$.

Explanation of Solution

Assume $x_2\left(t\right)=1$ is the another solution of the nonlinear first order differential equation, then it will satisfy the nonlinear first order differential equation.

Substitute $1$ for $x$ in $\dot{x}=2\sqrt{x-1}$ and solve

$\begin{array}{c}\dot{x}=2\sqrt{x-1}\\ =2\sqrt{1-1}\\ =0\end{array}$

Since, $x_2\left(t\right)=1$ satisfies the nonlinear first order differential equation $\dot{x}=2\sqrt{x-1}$. It is true that the another solution for $\dot{x}=2\sqrt{x-1}$ is $x_2\left(t\right)=1$.

Conclusion:

The another solution for $\dot{x}=2\sqrt{x-1}$ is $x_2\left(t\right)=1$.

(c)

To determine

Show that although $x_1\left(t\right)$ and $x_2\left(t\right)$ are solutions for $\dot{x}=2\sqrt{x-1}$, neither $A{x}_1\left(t\right)$, nor $B{x}_2\left(t\right)$, nor $x_1\left(t\right)+x_2\left(t\right)$ are solutions.

Answer to Problem 12.1P

It is shown that $A{x}_1\left(t\right)$, $B{x}_2\left(t\right)$, and $x_1\left(t\right)+x_2\left(t\right)$ are not the solutions for $\dot{x}=2\sqrt{x-1}$.

Explanation of Solution

From part (a), the solution for $x_1\left(t\right)$ is ${\left(t+k\right)}^2+1$.

Now, consider $x=A{x}_1\left(t\right)$.

Substitute ${\left(t+k\right)}^2+1$ for $x_1\left(t\right)$ in $x=A{x}_1\left(t\right)$

$\begin{array}{l}x=A{x}_1\left(t\right)\\ x=A\left[{\left(t+k\right)}^2+1\right]\end{array}$

Differentiate the above equation with respect to $t$

$\begin{array}{c}\dot{x}=\frac{d}{dt}A\left[{\left(t+k\right)}^2+1\right]\\ =A\left[2\left(t+k\right)\right]\\ =2A\left[\left(t+k\right)\right]\end{array}$

Now, substitute $A\left[{\left(t+k\right)}^2+1\right]$ for $x$ in $\dot{x}=2\sqrt{x-1}$

$\begin{array}{c}\dot{x}=2\sqrt{A\left[{\left(t+k\right)}^2+1\right]-1}\\ =2\sqrt{A{\left(t+k\right)}^2+A-1}\end{array}$

Substitute the above equation in $\dot{x}=2A\left[\left(t+k\right)\right]$

$\begin{array}{c}2\sqrt{A{\left(t+k\right)}^2+A-1}=2A\left(t+k\right)\\ \sqrt{A{\left(t+k\right)}^2+A-1}=A\left(t+k\right)\end{array}$

Thus, $A{x}_1\left(t\right)$ does not satisfy the nonlinear first order equation $\dot{x}=2\sqrt{x-1}$. $A{x}_1\left(t\right)$ is not the solution of the equation $\dot{x}=2\sqrt{x-1}$.

From part (b), the solution for $x_2\left(t\right)$ is $1$.

Now, consider $x=B{x}_2\left(t\right)$.

Substitute $1$ for $x_2\left(t\right)$ in $x=B{x}_2\left(t\right)$

$\begin{array}{l}x=B\left(1\right)\\ x=B\end{array}$

Differentiate the above equation with respect to $t$

$\begin{array}{c}\dot{x}=\frac{dB}{dt}\\ =0\end{array}$

Now, substitute $B$ for $x$ in $\dot{x}=2\sqrt{x-1}$

$\dot{x}=2\sqrt{B-1}$

Substitute the above equation in $\dot{x}=0$

$\begin{array}{c}2\sqrt{B-1}=0\\ B=1\end{array}$

Thus, $B{x}_2\left(t\right)$ does not satisfy the nonlinear first order equation $\dot{x}=2\sqrt{x-1}$. $B{x}_2\left(t\right)$ is not the solution of the equation $\dot{x}=2\sqrt{x-1}$.

Now, consider $x=x_1\left(t\right)+x_2\left(t\right)$.

Substitute ${\left(t+k\right)}^2+1$ for $x_1\left(t\right)$ and $1$ for $x_2\left(t\right)$ in $x=x_1\left(t\right)+x_2\left(t\right)$

$\begin{array}{c}x={\left(t+k\right)}^2+1+1\\ ={\left(t+k\right)}^2+2\end{array}$

Differentiate the above equation with respect to $t$

$\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left({\left(t+k\right)}^2+2\right)\\ \dot{x}=2\left(t+k\right)\end{array}$

Now, substitute ${\left(t+k\right)}^2+2$ for $x$ in $\dot{x}=2\sqrt{x-1}$

$\begin{array}{c}\dot{x}=2\sqrt{{\left(t+k\right)}^2+2-1}\\ =2\sqrt{{\left(t+k\right)}^2+1}\end{array}$

Substitute the above equation in $\dot{x}=2\left(t+k\right)$

$\begin{array}{c}2\sqrt{{\left(t+k\right)}^2+1}=2\left(t+k\right)\\ \sqrt{{\left(t+k\right)}^2+1}=\left(t+k\right)\end{array}$

Thus, $x_1\left(t\right)+x_2\left(t\right)$ does not satisfy the nonlinear first order equation $\dot{x}=2\sqrt{x-1}$. $x_1\left(t\right)+x_2\left(t\right)$ is not the solution of the equation $\dot{x}=2\sqrt{x-1}$.

Conclusion:

It is shown that $A{x}_1\left(t\right)$, $B{x}_2\left(t\right)$, and $x_1\left(t\right)+x_2\left(t\right)$ are not the solutions for $\dot{x}=2\sqrt{x-1}$.