Question

(a) Find a solution of the initial-value problem consisting of the differential equation in Example 3 and each of the initial conditions: $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}=\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}=\mathbf{-}\mathbf{2}$, androle="math" localid="1667811686463" $\mathbf{y}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

(b) Find the solution of the differential equation in Example 3 when${\mathbf{lnc}}_1$is used as the constant of integration on the left-hand side in the solution and$\mathbf{4}{\mathbf{lnc}}_1$is replaced by$\mathbf{lnc}$. Then solve the same initial-value problems in part (a).

Short answer

1. The solution is$\mathrm{y}=2\left(\frac{3-{\mathrm{e}}^{4\mathrm{x}-1}}{3+{\mathrm{e}}^{4\mathrm{x}-1}}\right)$.
2. The solution is $y=2\left(\frac{-3e+e^{4x}}{-3e-e^{4x}}\right)$.

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Simplify differential equation(a)

Consider the initial value problem$\frac{dy}{dx}=y^2-4$

And the initial conditions $y\left(0\right)=2,y\left(0\right)=-2$, and $y\left(\frac{1}{4}\right)=1$.

The differential equation can be written as $\frac{dy}{dx}=y^2-4$

$\begin{array}{c}\frac{dy}{y^2-4}=dx\\ \frac{dy}{(y+2)(y-2)}=dx\\ \frac{1}{4}\cdot \frac{(y+2-(y-2\left)\right)dy}{(y+2)(y-2)}=dx\\ \left[\frac{\frac{1}{4}}{y-2}-\frac{\frac{1}{4}}{y+2}\right]dy=dx\end{array}$

Integrate

Take integration on both sides,

$\begin{array}{c}\frac{1}{4}\int \frac{1}{y-2}dy-\frac{1}{4}\int \frac{1}{y+2}dy=\int dx\\ \frac{1}{4}\ln |y-2|-\frac{1}{4}\ln |y+2|=x+c_1\\ \frac{1}{4}\ln \left|\frac{y-2}{y+2}\right|=x+c_1\end{array}$

Multiplying both sides by 4:

$\ln \left|\frac{y-2}{y+2}\right|=4x+c_2$;$c_2=4c_1$

Simplify integral

Taking anti log both sides:

$\begin{array}{c}\left|\frac{y-2}{y+2}\right|=e^{4x+c_2}\\ \frac{y-2}{y+2}=\pm e^{4x+c_2}\\ \frac{y-2}{y+2}=\pm e^{4x}{e}^{c_2}\\ \frac{y-2}{y+2}=c{e}^{4x}\text{Here}\pm e^{c_2}=c\\ y-2=yc{e}^{4x}+2c{e}^{4x}\\ y-yc{e}^{4x}=2+2c{e}^{4x}\\ y(1-c{e}^{4x})=2(1+c{e}^{4x})\\ y=2\left(\frac{1+c{e}^{4x}}{1-c{e}^{4x}}\right)\end{array}$

Find critical point

The differential equation can be written as$\frac{dy}{dx}=y^2-4$

Factor the right hand side of the differential equation$\frac{dy}{dx}=(y-2)(y+2)$

To find the critical points, set $\frac{dy}{dx}=0$.

role="math" localid="1667812888396" $\begin{array}{l}(\mathrm{y}-2)(\mathrm{y}+2)=0\\ \mathrm{y}=2\text{or}\mathrm{y}=-2\end{array}$

Thus, $y\left(x\right)=2$and $y\left(x\right)=-2$are two constant (equilibrium) solutions that satisfy the given initial conditions $y\left(0\right)=2$and $y\left(0\right)=-2$.

Find constant

Substitute $y\left(\frac{1}{4}\right)=1$in $y=2\left(\frac{1+c{e}^{4x}}{1-c{e}^{4x}}\right)$.

$1=2\left[\frac{1+c{e}^{4\left(\frac{1}{4}\right)}}{1-c{e}^{4\left(\frac{1}{4}\right)}}\right]$

Find solution

Substitute the value of $c$in $y=2\left(\frac{1+c{e}^{4x}}{1-c{e}^{4x}}\right)$.

$\begin{array}{l}y=2\left[\frac{1+\left(\frac{-1}{3e}\right)e^{4x}}{1-\left(\frac{-1}{3e}\right)e^{4x}}\right]\\ y=2\left(\frac{3-e^{4x-1}}{3+e^{4x-1}}\right)\end{array}$

Separate differential equation(b)

Consider the differential equation$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{y}}^2-4$

The differential equation can be written as $\frac{dy}{dx}=y^2-4$

$\begin{array}{c}\frac{\mathrm{dy}}{{\mathrm{y}}^2-4}=\mathrm{dx}\\ \frac{\mathrm{dy}}{(\mathrm{y}+2)(\mathrm{y}-2)}=\mathrm{dx}\\ \frac{1}{4}\cdot \frac{(\mathrm{y}+2-(\mathrm{y}-2\left)\right)\mathrm{dy}}{(\mathrm{y}+2)(\mathrm{y}-2)}=\mathrm{dx}\\ \left[\frac{\frac{1}{4}}{\mathrm{y}-2}-\frac{\frac{1}{4}}{\mathrm{y}+2}\right]\mathrm{dy}=\mathrm{dx}\end{array}$

Integrating

Integrating both sides:

Solve for y

$\begin{array}{c}cy-2c=e^{4x}(y+2)\\ y(c-e^{4x})=2(e^{4x}+c)\\ y=2\left(\frac{c+e^{4x}}{c-e^{4x}}\right)\end{array}$

Apply condition

Substitute $y=-2$in $y=2\left(\frac{c+e^{4x}}{c-e^{4x}}\right)$.

$\begin{array}{c}-2=2\left(\frac{c+e^{4x}}{c-e^{4x}}\right)\\ -c+e^{4x}=c+e^{4x}\\ 2c=0\\ c=0\end{array}$

Hence the initial condition $y\left(0\right)=-2$gives$c=0$ and $y\left(x\right)=-2$.

Apply condition

However, the initial condition $y\left(0\right)=2$doesn't correspond to a value of $c$, and it must simply be recognized that $y\left(x\right)=2$is a solution of the initial value problem.

Substitute$x=\frac{1}{4}$ and $y=1$in $y=2\left(\frac{c+e^{4x}}{c-e^{4x}}\right)$.

Thus, a solution of the initial value problem is $y=2\left(\frac{-3e+e^{4x}}{-3e-e^{4x}}\right)$.