Question

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{t}{y}, y(1)=2$$

Short answer

Answer: Yes, the differential equation is separable. The solution for the initial value problem is $$y(t)=\sqrt{t^2+3}$$.

Step-by-step solution

Check if the equation is separable

We need to determine if the equation $$y^{\prime}(t)=\frac{t}{y}$$ can be written as the product of a function of t and a function of y. In this case, it is indeed separable because we can rewrite the equation as $$y^{\prime}(t) = t \cdot \frac{1}{y}$$, which is a function of t (t) multiplied by a function of y (1/y).

Separate the variables

Now that we know the equation is separable, we can rewrite the equation in the separated form: $$y^{\prime}(t) = t \cdot \frac{1}{y}$$ becomes $$y dy = t dt$$, with the variable y on the left side and t on the right side.

Integrate both sides

We can now integrate both sides of the equation: $$\int y dy = \int t dt$$. The left side integrates to $$\frac{y^2}{2} + C_1$$ and the right side integrates to $$\frac{t^2}{2} + C_2$$, where C_1 and C_2 are constants of integration.

Combine constants

We can combine the constants of integration into a single constant C by writing $$\frac{y^2}{2} = \frac{t^2}{2} + C$$.

Apply initial condition

Next, we apply the initial condition y(1) = 2: $$\frac{(2)^2}{2} = \frac{(1)^2}{2} + C$$. This simplifies to: $$2=C+\frac{1}{2}$$, from which we get $$C=\frac{3}{2}$$. The equation is now in the form $$\frac{y^2}{2} = \frac{t^2}{2} + \frac{3}{2}$$.

Solve for y(t)

Finally, we need to solve for y(t). First, multiply both sides of the equation by 2: $$y^2=t^2+3$$, and then take the square root of both sides: $$y(t)=\pm \sqrt{t^2+3}$$. Since y(1) = 2, we use the positive square root, i.e., $$y(t)=\sqrt{t^2+3}$$. This is the solution for the initial value problem.

Key concepts

Initial Value Problems

An Initial Value Problem (IVP) is a differential equation that comes with a specific value for the unknown function at a given point. This means, aside from having to solve the differential equation, we also have to find the specific solution that satisfies the condition provided. In our exercise, the IVP is given by:

• The differential equation: \(y'(t) = \frac{t}{y}\).
• The initial condition: \(y(1) = 2\).

To solve the IVP, our goal is to find the function \(y(t)\) such that it not only solves the differential equation but also fulfills the initial condition. Solving an IVP often requires a series of steps including separating variables, integrating, and applying the initial condition to determine any constants involved.

Integration

Integration is a fundamental concept in calculus used to find the antiderivative of a function. In the context of differential equations, integration is the step where we find the general solution of the separated variables. Once we have separated the equation as in:

• \(y dy = t dt\)

We integrate both sides. This involves:

• Finding the antiderivative of \(y\) on the left, which results in \(\frac{y^2}{2} + C_1\).
• Finding the antiderivative of \(t\) on the right, resulting in \(\frac{t^2}{2} + C_2\).

Understanding integration allows us to move from the separated equation to a general solution that includes constants of integration, which can then be further adjusted to meet the initial condition of the IVP.

Variable Separation

The method of separating variables involves rearranging a differential equation so that each variable appears on a different side of the equation. It is essential in solving separable differential equations because it allows us to deal with each variable independently. In this problem, the differential equation \(y'(t) = \frac{t}{y}\) is transformed using variable separation. This is done by rewriting it as \(y dy = t dt\), where all the \(y\) terms are on one side and the \(t\) terms on the other. This simplification is key to setting up the problem for integration, allowing us to easily apply the antiderivatives to each side separately. The successful separation of variables is vital as it enables the next steps in solving the differential equation.

Constant of Integration

The constant of integration represents an unknown constant that appears after integrating a function. When we integrate both sides of a differential equation that involved separated variables, we introduce constants of integration on each side, denoted as \(C_1\) and \(C_2\). However, because the solution's constants can be combined, we often represent them as a single constant, \(C\). Having identified \(C\) in \(\frac{y^2}{2} = \frac{t^2}{2} + C\), we use the initial condition \(y(1) = 2\) to solve for \(C\), ensuring the solution satisfies the particular initial value problem. In this case:

• Applying the condition gives: \(2 = C + \frac{1}{2}\)
• Solving for \(C\) results in: \(C = \frac{3}{2}\)

This adjusted constant then helps us get the unique solution \(y(t) = \sqrt{t^2 + 3}\), tailored explicitly to the problem's initial condition.