Question

Determine whether the following equations are separable. If so, solve the initial value problem. $$2 y y^{\prime}(t)=3 t^{2}, y(0)=9$$

Short answer

#Answer# The particular solution for the initial value problem is \(y(t) = \sqrt{2t^3 + 81}\).

Step-by-step solution

Rewrite the equation in terms of y'

(Write the content here) Divide both sides of the equation by \(2y\) to get the equation in terms of \(y'(t)\): $$y'(t) = \frac{3t^2}{2y}$$

Check for separability

The given equation is $$y'(t) = \frac{3t^2}{2y}$$ We observe that the equation is indeed separable, as we can rewrite it as: $$y'(t)dy = \frac{3t^2}{2y}dt$$

Integrate both sides

Now, integrate both sides of the equation with respect to their corresponding variables: $$\int y'(t) dy = \int \frac{3t^2}{2y} dt$$ Integrating, we have: $$\frac{1}{2}y^2 = t^3 + C$$

Apply the initial condition

Given that \(y(0) = 9\), we can apply this initial condition to find the value of \(C\): $$\frac{1}{2}(9)^2 = (0)^3 + C$$ $$C = \frac{81}{2}$$

Find the particular solution

Now, substitute the value of \(C\) back into the equation and solve for \(y(t)\): $$\frac{1}{2}y^2 = t^3 + \frac{81}{2}$$ Multiply both sides by 2: $$y^2 = 2t^3 + 81$$ Take the square root of both sides: $$y(t) = \pm\sqrt{2t^3 + 81}$$ Since \(y(0) = 9\) and \(9 = \sqrt{81}\), the particular solution is the positive square root: $$y(t) = \sqrt{2t^3 + 81}$$

Key concepts

Initial value problem

In mathematics, particularly in differential equations, an initial value problem involves finding a specific solution to a differential equation that satisfies an initial condition. This initial condition is a starting point that ensures we obtain the unique solution that passes through a specified point on its curve.

For the problem at hand, we are given the differential equation \(2 y y^{\prime}(t)=3 t^{2}\) along with the initial condition \(y(0)=9\). This tells us that when \(t=0\), the value of \(y\) should be \(9\).

Initial conditions are crucial because without them, a differential equation can have infinitely many solutions. By providing an initial condition, we are able to narrow down to one specific, unique solution that fits both the differential equation and this initial requirement. This is a common task in solving real-world problems where the conditions at a specific time are known, and we need to predict behavior over time.

Integration techniques

Integration is an essential part of solving differential equations. Separable differential equations, like the one in our problem, involve rearranging and integrating both sides of the equation to find the solution.

In the provided problem, after confirming the equation is separable, we rewrote it as \(y'(t)dy = \frac{3t^2}{2y}dt\). The idea is to integrate both sides:

• On the left side, integrate in terms of \(y\).
• On the right side, integrate in terms of \(t\).

After integrating, we obtained the equation \(\frac{1}{2}y^2 = t^3 + C\).

The integration process introduces a constant \(C\), known as the constant of integration, which is determined by the initial value condition. This constant holds the key to finding the precise particular solution to the problem at hand. Remember that integration reverses differentiation, which means finding the original function from its rate of change.

Particular solution

A particular solution is a specific solution to a differential equation that meets the criteria of the initial conditions given.

Once we have integrated both sides and found the general solution, \(\frac{1}{2}y^2 = t^3 + C\), the next step is to apply the initial condition \(y(0)=9\). By substituting this into the equation, we find the value of \(C\):

• Substitute \(t = 0\) and \(y = 9\) into the equation.
• Calculate \(C\) which turns out to be \(C = \frac{81}{2}\).

Now, substituting \(C\) back into the equation gives us \(\frac{1}{2}y^2 = t^3 + \frac{81}{2}\).

Finally, solving for \(y(t)\) gives us \(y(t) = \sqrt{2t^3 + 81}\), which is the particular solution. Since \(y(0) = 9\), we select the positive root, ensuring the solution meets the initial condition.