Question

(5 points) Solve the IVP. y' = y²-4, y(0) = 0

Answer 1

`x=-(1)/(y)-(1)/(4)y`

Explanation:

Solve the given initial-value problem.

`y' = y^2-4; \ y(0)=0`

`\hrulefill`

The given differential equation is separable. Solve by doing the following.

`\boxed{\left\begin{array}{ccc}\text{\underline{Separable Differential Equation:}}\\(dy)/(dx) =f(x)g(y)\\\\\rightarrow\int(dy)/(g(y))=\int f(x)dx \end{array}\right }`

`y' = y^2-4; \ \text{Note that} \ y'=(dy)/(dx) \\\\\\\Longrightarrow (dy)/(dx)=y^2-4\\\\\\\Longrightarrow dy=(y^2-4)dx\\\\\\\Longrightarrow (1)/(y^2-4) dy=dx\\\\\\\Longrightarrow \int(1)/(y^2-4) dy=\int dx\\\\\\\Longrightarrow \int(1)/(y^2) dy-(1)/(4)y =x+C\\\\\\\Longrightarrow \int y^(-2) dy-(1)/(4)y =x+C\\\\\\\Longrightarrow -y^(-1)-(1)/(4)y =x+C\\\\\\\Longrightarrow \boxed{-(1)/(y) -(1)/(4)y =x+C}`

Now use the initial condition to find the value of the arbitrary constant, "C"

`-(1)/(y) -(1)/(4)y =x+C; \ y(0)=0\\\\\\\Longrightarrow -(1)/((0)) -(1)/(4)(0) =(0)+C\\\\\\\therefore \boxed{C=0}`

Now we can write the solution as:

`\boxed{\boxed{x=-(1)/(y)-(1)/(4)y }}`