Question

Solve the initial-value problem $y^{\prime }-2y=4x+3y(0)=-2$.

Short answer

The particular solution is $y=-2x-\frac{5}{2}+\frac{1}{2}{e}^{2x}$.

Step-by-step solution

Identify the Type of Differential Equation

The given equation is a first-order linear ordinary differential equation of the form $y^{\prime }-2y=4x+3$. We'll use an integrating factor to solve it.

Calculate the Integrating Factor

An integrating factor $\mu (x)$ for an equation of the form $y^{\prime }+P(x)y=Q(x)$ is given by $\mu (x)=e^{\int P(x)dx}$. For our equation, $P(x)=-2$, so$$\mu (x)=e^{-2x}.$$

Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation $y^{\prime }-2y=4x+3$ by the integrating factor $e^{-2x}$:$$e^{-2x}{y}^{\prime }-2e^{-2x}y=(4x+3)e^{-2x}.$$

Recognize the Left Side as a Derivative

Notice that the left side of the equation $e^{-2x}{y}^{\prime }-2e^{-2x}y$ can be written as the derivative of a product:$$\frac{d}{dx}(e^{-2x}y).$$

Integrate Both Sides

Integrate both sides with respect to $x$:$$\int d(e^{-2x}y)=\int (4x+3)e^{-2x}dx.$$This simplifies to $$e^{-2x}y=\int (4x+3)e^{-2x}dx.$$

Solve the Right Side Integral

The integral $\int (4x+3)e^{-2x}dx$ requires integration by parts. Let $u=4x+3$ and $dv=e^{-2x}dx$. Use integration by parts:$$\int udv=uv-\int vdu,$$where $du=4dx$ and $v=-\frac{1}{2}{e}^{-2x}$. Perform the integration to find the expression.

Calculate Specific Integrals

Calculate:$$\int 4x{e}^{-2x}dx=-2x{e}^{-2x}-\int -2e^{-2x}dx=-2x{e}^{-2x}-e^{-2x},$$$$\int 3e^{-2x}dx=-\frac{3}{2}{e}^{-2x}.$$Thus, $$-2x{e}^{-2x}-e^{-2x}-\frac{3}{2}{e}^{-2x}=-2x{e}^{-2x}-\frac{5}{2}{e}^{-2x}.$$

Write the General Solution

Combine the results:$$e^{-2x}y=-2x{e}^{-2x}-\frac{5}{2}{e}^{-2x}+C,$$which yields the general solution,$$y=-2x-\frac{5}{2}+C{e}^{2x}.$$

Apply Initial Condition

Use the initial condition $y(0)=-2$ to find $C$. Substitute $x=0$:$$-2=-2(0)-\frac{5}{2}+C{e}^0⟹C=\frac{1}{2}.$$

Write the Particular Solution

Substitute the value of $C$ back into the general solution:$$y=-2x-\frac{5}{2}+\frac{1}{2}{e}^{2x}.$$

Key concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation that contains a function of one independent variable and its derivatives. It's a key concept in calculus and analysis, used to describe a wide range of phenomena in physics, engineering, biology, and more. In this exercise, the given equation is a first-order linear ODE:
$y^{\prime }-2y=4x+3$
Here, the function $y$ depends on the variable $x$, and $y^{\prime }$ denotes the derivative of $y$ with respect to $x$. Identifying the type of ODE is crucial as it dictates the methods available for finding a solution. For first-order linear ODEs like this one, the integrating factor method is a common means to find a solution.

Integrating Factor Method

The integrating factor method is a powerful technique for solving linear first-order ordinary differential equations. The method involves multiplying the entire ODE by a strategic function, known as the integrating factor, which simplifies the equation. The general form of such an equation is:
$y^{\prime }+P(x)y=Q(x)$
For the given problem, $P(x)=-2$, and the integrating factor $\mu (x)$ is calculated as:
$$\mu (x)=e^{\int P(x)dx}=e^{-2x}$$
Multiplying the entire equation by this integrating factor transforms the left side of the equation into the derivative of a product, making it straightforward to integrate. By doing this, the problem simplifies significantly and allows us to solve the differential equation more easily.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It's particularly useful when dealing with integrals that arise from using the integrating factor method. For the given problem, after multiplying the differential equation by the integrating factor, we arrive at the integral:
$$\int (4x+3)e^{-2x}dx$$
To solve this, we apply integration by parts, which can be recalled by the formula:
$$\int udv=uv-\int vdu$$
Here, we choose:

• $u=4x+3$
• $dv=e^{-2x}dx$

This gives us $du=4dx$ and $v=-\frac{1}{2}{e}^{-2x}$. Substituting these into the formula and performing the integration yields the necessary expression to continue solving the ODE.

Particular Solution

The particular solution to a differential equation is the solution that satisfies both the differential equation and any given initial conditions. In this problem, once we have the general solution:
$$y=-2x-\frac{5}{2}+C{e}^{2x}$$
We apply the initial condition $y(0)=-2$ to find the constant $C$. Substituting $x=0$ into the general solution gives:
$$-2=-\frac{5}{2}+C\cdot 1$$
Which leads to solving for $C$ as $\frac{1}{2}$. Inserting $C$ back into the general solution provides the particular solution:
$$y=-2x-\frac{5}{2}+\frac{1}{2}{e}^{2x}$$
This particular solution is unique to the given initial condition and is the final answer to the initial value problem.