Question

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=t y^{2} ; \quad y(t)=\frac{2}{2-t^{2}}\)

Short answer

Question: Verify if the given solution \(y(t) = \frac{2}{2-t^2}\) satisfies the given differential equation \(y' = ty^2\) and the initial condition \(y(0)=1\). Answer: Yes, the given solution \(y(t) = \frac{2}{2-t^2}\) satisfies the given differential equation \(y' = ty^2\) and the initial condition \(y(0)=1\).

Step-by-step solution

Calculate the derivative of the given solution

To calculate the derivative of the given solution \(y(t) = \frac{2}{2-t^2}\), we will first find its derivative with respect to \(t\), \(y'(t)\). An easy way to find the derivative is to use the power rule and chain rule. Rewrite the equation as \(y(t) = 2(2-t^2)^{-1}\) and find the derivative using the chain rule: \(y'(t) = \frac{d}{dt}\left[2(2-t^2)^{-1}\right] = 2(-1)(2-t^2)^{-2}(-2t) = 4t(2-t^2)^{-2}\)

Check if the derivative satisfies the given differential equation

In this step, we will verify if the given solution satisfies the given differential equation \(y' = ty^2\). Given \(y(t) = \frac{2}{2-t^2}\) and \(y'(t) = 4t(2-t^2)^{-2}\): Plugging in the values of \(y\) and \(y'\) into the given differential equation: \(4t(2-t^2)^{-2} = t \left(\frac{2}{2-t^2}\right)^2\) Simplifying the equation: \(4t(2-t^2)^{-2} = t\left(\frac{4}{(2-t^2)^2}\right)\) Both sides are equal, so the given solution does satisfy the given differential equation.

Check if the given solution satisfies the initial condition

Now, let's check if the given solution satisfies the initial condition \(y(0) = 1\). We will plug \(t = 0\) into the given solution, \(y(t)\): \(y(0) = \frac{2}{2-(0)^2} = \frac{2}{2} = 1\) Thus, the given solution does satisfy the initial condition, and we have confirmed that the given solution is correct for the given differential equation and initial condition.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in various fields such as physics, engineering, economics, and biology. These equations often describe how quantities change over time. For instance, they can model population growth, heat conduction, or motion dynamics.

In the context of the problem provided, the differential equation is given as \(y^{\prime} = t y^{2}\). This is a first-order differential equation. Solving this type of equation means finding a function \(y(t)\) that satisfies the equation for all values of \(t\) in the domain. The provided solution \(y(t) = \frac{2}{2-t^2}\) demonstrates how the differential equation can be evaluated for a specific function.

To solve these equations, various methods such as separation of variables, integration, and substitution are often used. The solution process typically involves calculating appropriate derivatives and integrating to find the general solution or a specific solution that fits given conditions.

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of a composition of functions. It helps to connect the derivative of a composite function to the derivatives of its inner and outer functions.

In our problem's solution, the chain rule is employed to find the derivative of \(y(t) = 2(2-t^2)^{-1}\). The expression \((2-t^2)^{-1}\) is treated as an inner function, while \(2 \cdot\) is the outer function. Applying the chain rule gives the derivative:

• First, differentiate the outer function \(2u^{-1}\) with respect to its input \(u\).
• Next, multiply by the derivative of the inner function \(2-t^2\).

This approach results in \(y'(t) = 4t(2-t^2)^{-2}\), showing how derivatives of more complex expressions can be systematically managed using the chain rule.

Initial Conditions

Initial conditions are essential when solving differential equations because they help determine a unique solution from a family of potential solutions. They are specific values provided for the function or its derivatives at a certain point, ensuring that the solution fits the particular situation described by the problem.

In our example, the initial condition is given as \(y(0) = 1\). This means that when \(t=0\), the value of \(y(t)\) must equal 1. Using this condition, we test the provided solution \(y(t) = \frac{2}{2-t^2}\) by substituting \(t = 0\):

• Substitute \(t = 0\) into the equation, leading to \(y(0) = \frac{2}{2-0^2} = 1\).

Thus, the initial condition directly confirms that the given solution is appropriate for the particular instance of the differential equation.

Solution Verification

Solution verification is the process of checking whether a given solution truly satisfies both the differential equation and any initial conditions. This step is critical to confirm that the work is correct and the findings are applicable.

In our context, we have to check if the solution \(y(t) = \frac{2}{2-t^2}\) satisfies the differential equation \(y' = ty^2\). This involves calculating the derivative using the chain rule and substituting both \(y(t)\) and \(y'(t)\) back into the differential equation.

• Evaluate the left side of the equation using the found derivative: \(4t(2-t^2)^{-2}\).
• Evaluate the right side using \(y(t)\): \(t \left(\frac{2}{2-t^2}\right)^2 = t\left(\frac{4}{(2-t^2)^2}\right)\).

Matching both sides confirms that the solution is correct. Verifying the initial condition similarly ensures that the solution holds true for specific initial values, completing the verification process.