Question

\(y^{\prime}+t^{2} y=3\), with \(y(0)=-2\), on \([0,5]\).

Short answer

Integrating factor: \( e^{\frac{t^3}{3}} \), solution involves non-elementary integral.

Step-by-step solution

Identify the Type of Differential Equation

The given equation is a first-order linear non-homogeneous ordinary differential equation (ODE) of the form: \[ y' + p(t) y = g(t) \] where \( p(t) = t^2 \) and \( g(t) = 3 \).

Find the Integrating Factor

The integrating factor, \( \,\mu(t)\,\), is found using the formula: \[ \mu(t) = e^{\int p(t) \, dt} = e^{\int t^2 \, dt} = e^{\frac{t^3}{3}}. \]

Multiply Through by the Integrating Factor

Multiply the entire differential equation by \( \,\mu(t)\,\) to make the left side an exact derivative: \[ e^{\frac{t^3}{3}} y' + e^{\frac{t^3}{3}} t^2 y = 3e^{\frac{t^3}{3}}. \]

Write as an Exact Differential

Recognize that the left side can be rewritten as a derivative: \[ \frac{d}{dt}\left(e^{\frac{t^3}{3}} y \right) = 3e^{\frac{t^3}{3}}. \]

Integrate Both Sides

Integrate both sides with respect to \(t\): \[ \int \frac{d}{dt}\left(e^{\frac{t^3}{3}} y \right) \, dt = \int 3e^{\frac{t^3}{3}} \, dt. \] This results in:\[ e^{\frac{t^3}{3}} y = \int 3e^{\frac{t^3}{3}} \, dt. \] Solving the right integral is more involved, requiring knowledge of exponential integrals (or stating it's a standard result):\[ e^{\frac{t^3}{3}} y = E(t) + C. \]

Apply Initial Condition

To find the constant \( C \), use the initial condition \( y(0) = -2 \). Substitute into the integrated function after solving the integral for specific functions of this form:\[ e^{0}(-2) = E(0) + C \] implies that \( C = -2 - E(0) \).

Solve for \( y(t) \)

Rearrange the integrated equation to solve for \( y(t) \): \\[ y(t) = e^{-\frac{t^3}{3}}(E(t) + C). \] Substitute the expression for \( C \) back into the equation:\[ y(t) = e^{-\frac{t^3}{3}}(E(t) -2 - E(0)). \]

Simplify the Solution

Recognize the solution includes an integrating factor term: \[ y(t) = e^{-rac{t^3}{3}}(F(t) - 2) \] where \( F(t) \) considers past evaluated integrals for exponential functions. Calculate specific \( F(t) \) values if needed, or use numerical methods for practical time points if required.

Key concepts

Integrating Factor

When you encounter a first-order linear ordinary differential equation, an integrating factor is a powerful tool to simplify the equation. The general form is given as \( y' + p(t) y = g(t) \). Here, \( p(t) \) represents a function of \( t \), and \( g(t) \) is the non-homogeneous term.
An integrating factor, \( \mu(t) \), is found using the formula \( \mu(t) = e^{\int p(t) \, dt} \). This factor transforms the original differential equation into a form where the left-hand side becomes the exact derivative of a product of two functions.

• Calculate the integrating factor: Integrate \( p(t) \) to determine \( \mu(t) \).
• Multiply the entire differential equation by \( \mu(t) \) to make the left side an exact differential.

For example, if \( p(t) = t^2 \), the integrating factor becomes \( e^{\frac{t^3}{3}} \). This simplifies solving the differential equation by transforming it into its corresponding integrals.

Initial Conditions

Initial conditions are crucial in determining a particular solution amidst many possible solutions to a differential equation. An initial condition takes the form \( y(t_0) = y_0 \), providing a specific value for the solution at a certain point \( t = t_0 \).

• Use the given initial condition to solve for any constant of integration that arises when solving the equation.
• Substitute the initial condition into the solution to find the specific constant value, making the solution unique.

In our example, the initial condition is \( y(0) = -2 \). This initial point is used to find the constant \( C \) in the integrated solution, thus providing a unique solution to the differential equation based on the condition.

Exact Differential

An exact differential in the context of differential equations allows us to recognize that an expression on one side of the equation is the derivative of a specific function. Transforming a differential equation into a form where the left side is an exact differential makes it solvable through direct integration.

• An exact differential takes the form \( \frac{d}{dt}( \mu(t)y ) \), where \( \mu(t) \) is the integrating factor.
• Integrating both sides after achieving an exact differential simplifies solving the equation.

In the provided example, multiplying the entire differential equation by the integrating factor allowed us to rewrite it as \( \frac{d}{dt}(e^{\frac{t^3}{3}} y) = 3e^{\frac{t^3}{3}} \). This showcases how recognizing an exact differential leads to straightforward integration.

Exponential Integrals

Exponential integrals can appear when dealing with exponential functions within differential equations. These integrals involve terms of the form \( e^{f(t)} \), which can be solved or simplified using standard methods or known results from tables of integrals.
While exponential integrals might seem daunting, they can often be expressed in terms of elementary functions, especially when the function within the exponent is linear or quadratic.

• Identify the form of the integral required when addressing terms like \( 3e^{\frac{t^3}{3}} \).
• Use integration techniques or known results to solve the integral.

In our solution, integrating \( 3e^{\frac{t^3}{3}} \, dt \) might require consulting integral tables or using numerical methods for more complicated expressions. This integration step is essential in obtaining a specific form for the solution and finding any involved constants.