Question

Find the solution to the initial value problem. $$ y^{\prime}=3 y-x+6 x^{2}, y(0)=-1 $$

Short answer

The solution to the initial value problem is \( y(x) = e^{3x}(F(x) + C) \), where \( F(x) \) results from integrating the right side and \( C \) is found using the initial condition.

Step-by-step solution

Identify the Type of Differential Equation

This is a first-order linear differential equation of the form \( y' = a(x)y + b(x) \), where \( a(x) = -3 \) and \( b(x) = -x + 6x^2 \).

Find the Integrating Factor

The integrating factor is given by \( \mu(x) = e^{\int -3\, dx} = e^{-3x} \).

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{-3x} \), resulting in \( e^{-3x} y' - 3e^{-3x} y = (-x + 6x^2)e^{-3x} \).

Write Left Side as Derivative

The left-hand side can be written as the derivative of the product: \( \frac{d}{dx}(e^{-3x}y) = (-x + 6x^2)e^{-3x} \).

Integrate Both Sides

Integrate both sides of the equation with respect to \( x \): \[ \int \frac{d}{dx}(e^{-3x}y) \, dx = \int (-x + 6x^2)e^{-3x} \, dx \]. The left side becomes \( e^{-3x}y \). You will need integration by parts or a suitable substitution for the right side.

Solve the Integral on the Right

Use integration by parts or a table of integrals to solve \( \int (-x + 6x^2)e^{-3x} \, dx \). For simpler integration, separate the expression: 1. \(-x e^{-3x}\)2. \(6x^2 e^{-3x}\)

Apply Initial Condition

After solving the integrals, you should have \( e^{-3x} y = F(x) + C \). Apply the initial condition \( y(0) = -1 \) to find \( C \).

Solve for \( y \)

Multiply both sides by \( e^{3x} \) to solve for \( y \). This results in \( y = e^{3x}(F(x) + C) \). Substitute \( C \) from the initial condition to find the particular solution.

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves the rate of change of a variable and the variable itself. These equations can usually be expressed in the form \( y' = a(x)y + b(x) \). Here, \( y' \) denotes the derivative of \( y \) with respect to \( x \), and \( a(x) \) and \( b(x) \) are given functions of \( x \).

In our exercise, we have \( y' = 3y - x + 6x^2 \), identifying it as a first-order linear differential equation. This classification underlines that the equation involves the first derivative of \( y \), making the understanding of its behaviour dependent on \( y \) alone without involving higher derivatives.

• It is crucial to correctly identify \( a(x) \) and \( b(x) \) in the given equation for effective problem-solving.
• In our case, rearranging gives \( y' = 3y - x + 6x^2 \).
• Comparing with the standard form, \( a(x) = -3 \) and \( b(x) = -x + 6x^2 \).

Understanding these basic concepts will help simplify the problem into solvable parts.

Integrating Factor

To find the solution of a first-order linear differential equation, we often use an integrating factor. This is a strategic function we multiply through the equation to enable us to rewrite it in a simpler, integrable form.

In our case, the integrating factor \( \mu(x) \) is obtained from the function \( a(x) \). Calculate it using the formula: \[ \mu(x) = e^{ \int a(x) \, dx } \].
This calculation leads to \( \mu(x) = e^{-3x} \).

• Important to note is that the choice of integrating factor depends on \( a(x) \), which in this problem is a constant \(-3\).
• Multiplying the entire differential equation by this integrating factor helps us rewrite the left-hand side as a derivative of a product, simplifying the integration process that follows.

Thus, the integrating factor method transforms the equation into a form where both sides can be integrated with respect to \( x \).

Integration by Parts

When faced with complex integrals, such as those involving products of polynomials and exponential functions, we often rely on a technique known as integration by parts. This technique helps us solve integrals that are not straightforward.

In our differential equation problem, to solve \( \int (-x + 6x^2)e^{-3x} \, dx \), we would consider splitting it into simpler, separate integrals: \(-xe^{-3x}\) and \(6x^2e^{-3x}\).

The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \].

• We choose \( u \) and \( dv \) such that the resulting integral \( \int v \, du \) is simpler to integrate.
• For instance, in \( -xe^{-3x} \), let \( u = -x \) making \( dv = e^{-3x} \, dx \).
• This technique reduces a complex integral into more manageable parts, facilitating the evaluation.

Understanding integration by parts is a valuable skill in calculus, allowing us to tackle integrals that combine different functions.

Initial Condition

An initial condition in a differential equation provides the necessary information to find a particular solution from a family of solutions. By setting a specific value for \( y \) at a known \( x \) value, we can find unique solutions that satisfy the equation and condition.

In our case, the initial condition is \( y(0) = -1 \).

• This condition allows us to determine the constant of integration \( C \), which arises from integrating the differential equation.
• After integrating both sides, apply this condition by substituting \( x = 0 \) to find \( C \).

For example, if after integration, we find \( e^{-3x}y = F(x) + C \), substituting \( x = 0 \) and \( y = -1 \) enables us to solve for \( C \).
By then plugging \( C \) back into the expression, we obtain a specific solution. Thus, initial conditions are crucial for transforming a general solution into a particular one that exactly meets the problem's requirements.