Question

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=t^{2}+y ; \quad y(t)=3 e^{t}-\left(t^{2}+2 t+2\right)\)

Short answer

Question: Verify whether the given solution \(y(t)=3 e^{t}-\left(t^{2}+2 t+2\right)\) is the correct solution to the given differential equation \(y^{\prime}=t^{2}+y\), with the initial condition \(y(0)=1\). Answer: The given solution, \(y(t)=3 e^{t}-\left(t^{2}+2 t+2\right)\), is the correct solution to the given differential equation, \(y^{\prime}=t^{2}+y\), with the initial condition \(y(0)=1\).

Step-by-step solution

Identify the given differential equation and its solution with the initial condition

The given first-order differential equation is \(y^{\prime}=t^{2}+y\). The initial condition is \(y(0)=1\). We are given its solution as \(y(t)=3 e^{t}-\left(t^{2}+2 t+2\right)\).

Find the derivative of the given solution

In order to confirm if the given solution is correct, we first need to find its derivative, which represents \(y^{\prime}\). Differentiate the given solution, \(y(t)\), with respect to \(t\) using the rules of differentiation: $$ \frac{d}{dt}\left(3 e^{t} - (t^{2}+2 t+2)\right) = 3 e^{t} - 2t - 2. $$

Substitute the obtained derivative and the given solution into the differential equation

Now, we need to substitute the derivatives we found in the previous step and the given solution into the original differential equation to check if it holds true. The differential equation is \(y^{\prime}=t^{2}+y\). Substituting the obtained derivative and the given solution into the equation: $$ 3 e^{t} - 2t - 2 = t^{2} + \left(3 e^{t} - \left(t^{2}+2 t+2\right)\right) $$

Simplify the equation and check if the identity holds

Now, we will simplify the equation to see if it becomes an identity (which would confirm the given solution is correct). $$ 3 e^{t} - 2t - 2 = t^{2} + 3 e^{t} - t^{2} - 2t - 2 $$ After simplifying, we observe that the equation holds true, as the left side is equal to the right side.

Verify the initial condition

Lastly, we will check if the given solution satisfies the initial condition: \(y(0)=1\). Plug in the value of \(t=0\) into the given solution: $$ y(0) = 3 e^{0} - (0^{2} + 2 \cdot 0 + 2) = 3 \cdot 1 - 2 = 1 $$ The initial condition is satisfied, as \(y(0) = 1\). Hence, the given solution, \(y(t)=3 e^{t}-\left(t^{2}+2 t+2\right)\), is the correct solution to the given differential equation, \(y^{\prime}=t^{2}+y\), with the initial condition \(y(0)=1\).

Key concepts

Initial Value Problem

An initial value problem involves solving a differential equation with a given initial condition. This means we have a differential equation, and to find the specific solution, we need an extra condition that tells us the value of the function at a particular point. In our example, the differential equation is \(y^{\prime}=t^{2}+y\) and the initial condition is \(y(0)=1\). This initial condition helps narrow down the solution to a specific function that not only satisfies the differential equation but also passes through the point \((0, 1)\).
A great analogy is thinking of the differential equation as a general map and the initial condition as a particular starting point, guiding us exactly where to go on that map.

Solution Verification

Solution verification is the process of confirming that a proposed solution actually satisfies the given differential equation and its initial condition. To verify, we start by determining if the given function, \(y(t) = 3 e^{t} - (t^{2} + 2t + 2)\), works with the differential equation \(y^{\prime}=t^{2}+y\).
We do this by differentiating the function to find \(y^{\prime}\), and checking if it equals \(t^{2} + y\) when substituted back into the equation. In this case, upon differentiation and substitution, the left side and right side of the equation become identical, thereby proving that the function is indeed a solution.
Lastly, we verify that \(y(t)\) also satisfies the initial condition by substituting the starting point \(t = 0\) into \(y(t)\). If \(y(0) = 1\) holds true, the solution is verified.

Differentiation

Differentiation is a fundamental concept used to calculate the rate of change of a function with respect to a variable. In the context of solving differential equations, differentiation is used to find the derivative of the proposed solution. For the function \(y(t) = 3 e^{t} - (t^{2}+2t+2)\), we apply the rules of differentiation to calculate \(y^{\prime}\), which leads us to \(3 e^{t} - 2t - 2\).
This calculation involves differentiating each component of the function separately:

• The derivative of \(3 e^{t}\) is \(3 e^{t}\) because the derivative of the exponential function \(e^{t}\) is itself.
• The derivative of \(-t^{2}\) is \(-2t\).
• The derivative of \(-2t\) is \(-2\).
• The derivative of a constant \(-2\) is \(0\).

This allows us to substitute \(y^{\prime}\) back into the differential equation and validates that the solution satisfies the equation itself.

First-Order Differential Equation

A first-order differential equation is one that involves the first derivative of an unknown function and the independent variable, but no higher order derivatives. For first-order equations, the key task is to find a function \(y(t)\) whose derivative matches a given combination of functions of \(t\) and \(y(t)\).
The differential equation \(y^{\prime}=t^{2}+y\) is linear and has its highest derivative as \(y^{\prime}\), which makes it a first-order differential equation. Solving it involves methods tailored for first-order equations, such as separation of variables, integrating factors, or, as shown in this example, checking that a given solution satisfies the equation and the initial condition.
First-order differential equations appear everywhere in scientific fields because they often model real-world phenomena where the rate of change is dependent on present conditions or states.