Question

Every autonomous first-order equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{y}\mathbf{\right)}$is separable. Find explicit solutions, ${\mathbf{y}}_1\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{y}}_2\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{y}}_3\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and${\mathbf{y}}_4\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of the differential equation$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{y}\mathbf{-}{\mathbf{y}}^3$that satisfy, in turn, the initial conditions, ${\mathbf{y}}_1\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{y}}_2\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{1}{2}\mathbf{,}{\mathbf{y}}_3\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{1}{2}$and${\mathbf{y}}_4\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$. Use a graphing utility to plot the graphs of each solution. Compare these graphs with those predicted in Problem 19 of Exercises 2.1. Give the exact interval of definition for each solution.

Short answer

$\therefore {\mathrm{y}}_1\left(\mathrm{x}\right)=\frac{2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}};\mathrm{x}>\ln \left(\frac{\sqrt{3}}{2}\right)$

$y_2\left(x\right)$is defined, for all real numbers.

${\mathrm{y}}_3\left(\mathrm{x}\right)=\frac{-{\mathrm{e}}^{\mathrm{x}}}{\sqrt{3+{\mathrm{e}}^{2\mathrm{x}}}},-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$

${\mathrm{y}}_4\left(\mathrm{x}\right)=\frac{-2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}},\text{where}\mathrm{x}>\ln \left(\frac{\sqrt{3}}{2}\right).$

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.

Separate differential equation

On separating of variables, we have

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{y}-{\mathrm{y}}^3}=\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{y}(1-\mathrm{y})(1+\mathrm{y})}=\mathrm{dx}\end{array}$

Using partial fractions, we get$\left[\frac{1}{y}+\frac{\frac{1}{2}}{1-y}-\frac{\frac{1}{2}}{1+y}\right]dy=dx$

On integration, we get$\ln \left|y\right|-\frac{1}{2}\ln |1-y|-\frac{1}{2}\ln |1+y|=x+c\cdots \left(1\right)$

Find constant

When$y>1;\left|y\right|=y$

$\begin{array}{l}|1-y|=y-1\\ |1+y|=1+y\end{array}$

We obtain,$\mathrm{lny}-\frac{1}{2}\ln (\mathrm{y}-1)-\frac{1}{2}\ln (1+\mathrm{y})=\mathrm{x}+\mathrm{c}$

Or$\ln \frac{\mathrm{y}}{\sqrt{\mathrm{y}-1}\sqrt{1+\mathrm{y}}}=\mathrm{x}+\mathrm{c}$

$\ln \frac{\mathrm{y}}{\sqrt{{\mathrm{y}}^2-1}}=\mathrm{x}+\mathrm{c}$

When $x=0,y=2>1$, we have$\ln \frac{2}{\sqrt{4-1}}=c$

$\therefore c=\ln \left(\frac{2}{\sqrt{3}}\right)$

Apply initial condition

Put ${\mathrm{y}}_1\left(0\right)=2$in (1) implies $\ln \frac{{\mathrm{y}}_1}{\sqrt{{\mathrm{y}}_1^2-1}}=\mathrm{x}+\ln \left(\frac{2}{\sqrt{3}}\right)$

$\therefore x=\ln \left(\left(\frac{y_1}{\sqrt{y_1^2-1}}\right)\left(\frac{\sqrt{3}}{2}\right)\right)$

$=\ln \frac{\sqrt{3}{\mathrm{y}}_1}{2\sqrt{{\mathrm{y}}_1^2-1}}$

$\begin{array}{c}\therefore {\mathrm{e}}^{\mathrm{x}}=\frac{\sqrt{3}{\mathrm{y}}_1}{2\sqrt{{\mathrm{y}}_1^2-1}}\\ \Rightarrow {\mathrm{e}}^{2\mathrm{x}}=\frac{3{\mathrm{y}}_1^2}{4({\mathrm{y}}_1^2-1)}\end{array}$

Solving

$\begin{array}{l}{\mathrm{e}}^{2\mathrm{x}}(4{\mathrm{y}}_1^2-4)=3{\mathrm{y}}_1^2\\ {\mathrm{y}}_1^2(4{\mathrm{e}}^{2\mathrm{x}}-3)=4{\mathrm{e}}^{2\mathrm{x}}\end{array}$

${\mathrm{y}}_1^2=\frac{4{\mathrm{e}}^{2\mathrm{x}}}{4{\mathrm{e}}^{2\mathrm{x}}-3}$

$\therefore {\mathrm{y}}_1=\frac{2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}}$

(Here positive square root is chosen because of initial condition)

Notice that $y_1\left(x\right)$is defined, when $4e^{2x}-3>0$

${\mathrm{e}}^{2\mathrm{x}}>\frac{3}{4}$

$\begin{array}{c}{\mathrm{e}}^{2\mathrm{x}}>\frac{3}{4}\\ {\mathrm{e}}^{\mathrm{x}}>\frac{\sqrt{3}}{2}\\ \therefore \mathrm{x}>\ln \left(\frac{\sqrt{3}}{2}\right)\end{array}$

$\therefore {\mathrm{y}}_1\left(\mathrm{x}\right)=\frac{2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}};\mathrm{x}>\ln \left(\frac{\sqrt{3}}{2}\right)$

Graph

Graph is given below:

Solve for 0<y<1

When $0<\mathrm{y}<1;\left|\mathrm{y}\right|=\mathrm{y},|1-\mathrm{y}|=1-\mathrm{y}$and$|1+y|=1+y$

So, we obtain$\mathrm{lny}-\frac{1}{2}\ln (1-\mathrm{y})-\frac{1}{2}\ln (1+\mathrm{y})=\mathrm{x}+\mathrm{c}$

Or $\ln \frac{\mathrm{y}}{\sqrt{1-{\mathrm{y}}^2}}=\mathrm{x}+\mathrm{c}$ -------(2)

Let $y_2\left(0\right)=\frac{1}{2}$in $\left(2\right)$, we find $c=\ln \frac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}$

$=\ln \frac{1}{\sqrt{3}}$

Find constant

By replacing the value of $c$in (2), we get

$\begin{array}{c}\ln \frac{y_2}{\sqrt{1-y_2^2}}=x+\ln \frac{1}{\sqrt{3}}\\ \ln \frac{\sqrt{3}{y}_2}{\sqrt{1-y_2^2}}=x\\ \frac{\sqrt{3}{y}_2}{\sqrt{1-y_2^2}}=e^x\\ 3y_2^2=e^{2x}(1-y_2^2)\end{array}$

And $\begin{array}{c}{y}_2^2(3+e^{2x})=e^{2x}\\ y_2=\frac{e^x}{\sqrt{3+e^{2x}}}\left(\text{As}{y}_2\left(0\right)=\frac{1}{2},\text{sopositivesquarerootischosenfor}{y}_2\right)\end{array}$

$y_2\left(x\right)$ is defined, for all real numbers.

Graph

${\mathrm{y}}_2\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{3+{\mathrm{e}}^{2\mathrm{x}}}}\text{,where}-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }.\text{\hspace{0.17em}\hspace{0.17em}graph\hspace{0.17em}\hspace{0.17em}isgiven\hspace{0.17em}\hspace{0.17em}below}:$

Solve for -1<y<0 to find constant

When$-1<y<0,\left|y\right|=-y,|1-y|=1-y$ and$|1+\mathrm{y}|=1+\mathrm{y}$

We obtain,$\ln (-y)-\frac{1}{2}\ln (1-y)-\frac{1}{2}\ln (1+y)=x+c$

$\therefore \ln \frac{-y}{\sqrt{1-y^2}}=x+c$ -------(3)

${\mathrm{y}}_3\left(0\right)=-\frac{1}{2}$implies $n\frac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}=x+c$

$\ln \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=c$

$c=\ln \frac{1}{\sqrt{3}}$

Substitute value of constant

Substituting the above value in (3), we obtain

$\begin{array}{c}\ln \frac{-{\mathrm{y}}_3}{\sqrt{1-{\mathrm{y}}_3^2}}=\mathrm{x}+\ln \frac{1}{\sqrt{3}}\\ \ln \frac{-\sqrt{3}{\mathrm{y}}_3}{\sqrt{1-{\mathrm{y}}_3^2}}=\mathrm{x}\\ \frac{-\sqrt{3}{\mathrm{y}}_3}{\sqrt{1-{\mathrm{y}}_3^2}}={\mathrm{e}}^{\mathrm{x}}\\ 3{\mathrm{y}}_3^2={\mathrm{e}}^{2\mathrm{x}}(1-{\mathrm{y}}_3^2)\\ {\mathrm{y}}_3^2(3+{\mathrm{e}}^{2\mathrm{x}})={\mathrm{e}}^{2\mathrm{x}}\\ {\mathrm{y}}_3=\frac{\pm {\mathrm{e}}^{\mathrm{x}}}{\sqrt{3+{\mathrm{e}}^{2\mathrm{x}}}}\end{array}$

Since, $y_3\left(0\right)=-\frac{1}{2}$we choose negative sign.

Graph

The graph of$y_3\left(x\right)=\frac{-e^x}{\sqrt{3+e^{2x}}},-\infty <x<\infty \text{\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}given\hspace{0.17em}\hspace{0.17em}below}:$

Solve for y<-1to find constant

When,$y<-1$ we have$\left|y\right|=-y,|1+y|=-(1+y)$

Equation $\left(1\right)$reduces to$n(-y)-\frac{1}{2}\ln (1-y)-\frac{1}{2}\ln (-1-y)=x+c$

Or $\ln \frac{-\mathrm{y}}{\sqrt{{\mathrm{y}}^2-1}}=\mathrm{x}+\mathrm{c}$ -----(4)

Let $x=0$and $y=-2$in (4), we get $n\frac{2}{\sqrt{4-1}}=0+c$

$c=\ln \frac{2}{\sqrt{3}}$

Substitute value of constant

Now substitute value of constant in equation (4) to get:

$\begin{array}{c}\ln \frac{-{\mathrm{y}}_4}{\sqrt{{{\mathrm{y}}_4}^2-1}}=\mathrm{x}+\ln \frac{2}{\sqrt{3}}\\ \frac{-\sqrt{3}{\mathrm{y}}_4}{2\sqrt{{{\mathrm{y}}_4}^2-1}}={\mathrm{e}}^{\mathrm{x}}\\ \frac{3{{\mathrm{y}}_4}^2}{4({{\mathrm{y}}_4}^2-1)}={\mathrm{e}}^{2\mathrm{x}}\\ 3{{\mathrm{y}}_4}^2=4{\mathrm{e}}^{2\mathrm{x}}({{\mathrm{y}}_4}^2-1)\\ {{\mathrm{y}}_4}^2(4{\mathrm{e}}^{2\mathrm{x}}-3)=4{\mathrm{e}}^{2\mathrm{x}}\\ {\mathrm{y}}_4\left(\mathrm{x}\right)=\frac{\pm 2{\mathrm{e}}^{2\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}}\end{array}$

Graph

The graph of$\mathrm{y}\left(0\right)=-2,{\mathrm{y}}_4\left(\mathrm{x}\right)=\frac{-2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}}\text{is\hspace{0.17em}\hspace{0.17em}given\hspace{0.17em}\hspace{0.17em}below}:$

Solve

Now, $4e^{2x}-3>0$gives $e^{2x}>\frac{3}{4}$

$\begin{array}{l}{\mathrm{e}}^{\mathrm{x}}>\frac{\sqrt{3}}{2}\text{or}\mathrm{x}>\ln \frac{\sqrt{3}}{2}\\ \therefore {\mathrm{y}}_4\left(\mathrm{x}\right)=\frac{-2{\mathrm{e}}^{\mathrm{x}}}{\sqrt{4{\mathrm{e}}^{2\mathrm{x}}-3}},\text{where}\mathrm{x}>\ln \left(\frac{\sqrt{3}}{2}\right).\end{array}$