Question Statement

Solve the differential equation: $\frac{dy}{dx}=y^2+4$ subject to the initial condition $y(0)=-2$.

---

Background and Explanation

This is a first-order separable differential equation. The solution method involves rearranging the equation so that all terms involving $y$ are on one side and all terms involving $x$ are on the other, then integrating both sides. This requires the standard arctangent integration formula.

---

Solution

We begin with the given differential equation and initial condition: $\frac{dy}{dx}=y^2+4;y(0)=-2$

Step 1: Separate the variables

To separate the variables, divide both sides by $(y^2+4)$ and multiply by $dx$: $\frac{dy}{y^2+4}=dx$

Step 2: Integrate both sides

Recognizing that $4=2^2$, we rewrite and integrate: $\int \frac{dy}{(y{)}^2+(2{)}^2}=\int dx$

Using the integration formula $\int \frac{dy}{y^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{y}{a}\right)+C$: $\frac{1}{2}{\tan }^{-1}\left(\frac{y}{2}\right)=x+c(1)$

Step 3: Apply the initial condition

We are given $y(0)=-2$, which means when $x=0$, $y=-2$. Substituting into equation (1):

$\begin{array}{rl}\frac{1}{2}{\tan }^{-1}\left(\frac{-2}{2}\right)& =0+c\\ \frac{1}{2}{\tan }^{-1}(-1)& =c\\ \frac{1}{2}\left(-\frac{\pi }{4}\right)& =c\\ c& =-\frac{\pi }{8}\end{array}$

Step 4: Write the particular solution

Substituting $c=-\frac{\pi }{8}$ back into equation (1), we obtain the implicit solution:

$\boxed{\frac{1}{2}{\tan }^{-1}\left(\frac{y}{2}\right)=x-\frac{\pi }{8}}$

This can also be written explicitly by solving for $y$: $y=2\tan \left(2x-\frac{\pi }{4}\right)$

---

Key Formulas or Methods Used

• Separation of variables: For $\frac{dy}{dx}=f(x)g(y)$, rearrange to $\frac{dy}{g(y)}=f(x)dx$
• Arctangent integral: $\int \frac{dy}{y^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{y}{a}\right)+C$
• Principal value: ${\tan }^{-1}(-1)=-\frac{\pi }{4}$
• Initial value condition: Substitute given $(x_0,y_0)$ to solve for the constant of integration $c$

---

Summary of Steps

1. Separate variables: Rearrange to get $\frac{dy}{y^2+4}=dx$
2. Integrate: Use the arctangent formula to obtain $\frac{1}{2}{\tan }^{-1}\left(\frac{y}{2}\right)=x+c$
3. Find constant: Substitute $x=0$ and $y=-2$ to calculate $c=-\frac{\pi }{8}$
4. Final solution: $\frac{1}{2}{\tan }^{-1}\left(\frac{y}{2}\right)=x-\frac{\pi }{8}$, or equivalently $y=2\tan \left(2x-\frac{\pi }{4}\right)$