Question

Solving initial value problems 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

Question: Determine whether the given initial-value problem is separable. If it is separable, solve the initial-value problem: $$2 y y^\prime(t) = 3t^2$$, and \(y(0) = 9\) Answer: The given initial value problem is separable, and the solution is \(y(t)=\sqrt{\frac{9}{2}t^3 + 81}\).

Step-by-step solution

Check if the equation is separable

To check if the given equation is separable, let's rewrite the equation in terms of t and y as follows: $$2 y y^\prime(t) = 3t^2$$ We want to reorganize it so that we have a function as a product of two separate functions of 'y' and 't' : $$\frac{2 y}{3t^2}y^{\prime}(t) = 1 $$ Now that it's multiplied by a function of 'y' and 't,' it's separable.

Integrate both sides

Now that the equation is separable, we will integrate both sides to find the function 'y(t)': $$\int\frac{2 y}{3t^2}y^{\prime}(t) dt= \int 1 dt$$ We can rewrite the left side of the equation to clarify the integration: $$\int \frac{2 y(y)}{3t^2} dy = \int 1 dt$$ Next, we perform the integration: $$\frac{1}{3}y^2(t)=\frac{3}{2} t^3 + C$$

Apply the initial condition

Now let's apply the initial condition, \(y(0) = 9\) to find the constant: $$\frac{1}{3}*(9)^2 = \frac{3}{2}(0)^3 + C$$ $$81 = 0 + 3C$$ $$C = 27$$

Write the final solution for y(t)

Let's replace the constant in the solution, and rewrite y(t) in terms of t: $$\frac{1}{3}y^2(t) = \frac{3}{2}t^3 + 27$$ $$y^2(t) = \frac{9}{2}t^3 + 81$$ Now take the square root of both sides to solve for y(t): $$y(t) = \pm\sqrt{\frac{9}{2}t^3 + 81}$$ Since we know that \(y(0)=9\), we can reject the negative square root: $$y(t)=\sqrt{\frac{9}{2}t^3 + 81}$$

Key concepts

Initial Value Problems

An initial value problem in the realm of differential equations is an equation that requires the solution to pass through a specific point, called an initial condition. In the example of the exercise provided, the differential equation
\[2 y y^{\textprime}(t)=3 t^{2}, y(0)=9\]
poses an initial condition where the value of the function \(y(t)\) is 9 when \(t\) is 0. This information is crucial, as it allows us to determine the specific solution to the differential equation that not only meets the general condition but also this very specific initial condition.

When dealing with initial value problems, it's important to solve for the general solution of the differential equation first and then apply the initial conditions to find the particular solution that fits the specific situation. Incorporating the initial condition, in this case, enables us to find the value of the integration constant, which makes the solution unique and applicable to the given condition.

Integrating Factors

An integrating factor is a function that is typically used to simplify non-separable differential equations, making them easier to solve. While the differential equation in our exercise was separable and didn't require an integrating factor, it's important to understand how they can be valuable in other cases.

An integrating factor is selected in such a way that it multiplies each term of the differential equation and transforms it into an exact equation. Exact equations have the property that they can be straightforwardly integrated with respect to one of the variables, usually after some manipulation or simplification. The choice of an integrating factor is not random and generally depends on the form of the original differential equation.

For instance, if you have a linear first-order differential equation of the form \(y^{\textprime} + p(x)y = q(x)\), an integrating factor can be expressed as \(e^{\int p(x) dx}\). Multiplying through by this factor would allow the left-hand side of the equation to be written as the derivative of a product of this factor and the function \(y\).

Integration of Differential Equations

Integration is a fundamental process in solving differential equations, as it allows us to move from the derivative of a function to the function itself. In the step-by-step solution for our separable differential equation, we have
\[\frac{2 y}{3t^2}y^{\textprime}(t) = 1 \]
by separating the variables related to \(y\) and \(t\), we can integrate both sides to find the function \(y(t)\).

This step involves finding the antiderivative of each side with respect to its respective variable, resulting in a general solution that includes an arbitrary constant. The integration process simplifies the equation, but often, as is the case with real-world problems, an additional step of applying initial conditions or boundary conditions is required to find this constant and arrive at a specific solution.