Question

In Exercises \(11-16,\) solve the initial value problem. Confirm your answer by checking that it conforms to the slope field of the differential equation. $$\frac{d y}{d x}=2 x \sqrt{x+2}\( and \)y=0\( when \)x=-1$$

Short answer

The solution to the initial value differential equation is \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \).

Step-by-step solution

Integrate the differential equation

The given equation can be written as \( dy = 2x\sqrt{x+2}dx \). Let's integrate both sides of the equation with respect to \(x\). Doing so will yield the general solution of the differential equation: \( y = \frac{4}{5}(x+2)^{5/2} + C \), where C is the constant of integration.

Apply the initial condition

The initial condition given is \( y = 0 \) when \( x = -1 \). Substituting these values into the general solution, we get: \( 0 = \frac{4}{5}(x+2)^{5/2} + C. \) Solving for \( C \) yields \( C = -4/5 \).

Write down the final specific solution

Substitute the found constant \( C = -4/5 \) back into the general solution, the specific solution conforming to the initial condition becomes: \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \).