Chapter 9: Problem 7
Find the general solution of the following equations. $$y^{\prime}(x)+2 y=-4$$
Short Answer
Expert verified
The general solution is given by:
$$ y(x) = Ce^{-2x} - 2, $$
where \(C\) is an arbitrary constant.
Step by step solution
01
Identify the type of ODE
This is a linear first-order ODE, as we can write it in the following form:
$$ y'(x) + P(x)y(x) = Q(x), $$
where \(P(x) = 2\) and \(Q(x) = -4.\)
02
Solve the homogeneous equation
To solve the homogeneous equation, we need to find the complementary function, which satisfies this equation:
$$y'(x) + 2y(x) = 0.$$
This is a separable ODE, and we can solve it by integrating:
$$\frac{y'(x)}{y(x)} = -2.$$
Now, integrate both sides with respect to \(x\):
$$\int \frac{dy}{y} = -2 \int dx.$$
This gives us:
$$\ln|y(x)| = -2x + C_1.$$
Taking the exponential of both sides, we get the complementary function:
$$y_c(x) = Ce^{-2x},$$
where \(C = e^{C_1}\) is an arbitrary constant.
03
Solve the particular solution
Now, we must find a particular solution for the original equation. Since the right side is a constant function, we assume that our particular solution is of the form:
$$y_p(x) = A,$$
where \(A\) is a constant to be determined.
Differentiating \(y_p(x)\) with respect to \(x\), we get:
$$y_p'(x) = 0.$$
Now, substitute \(y_p(x)\) and \(y_p'(x)\) into the original equation:
$$0 + 2A = -4.$$
Solving for \(A\), we find that \(A = -2\). Therefore, the particular solution is:
$$y_p(x) = -2.$$
04
Combine the homogeneous and particular solutions
Finally, to find the general solution of the ODE, we combine the homogeneous and particular solutions:
$$y(x) = y_c(x) + y_p(x) = Ce^{-2x} - 2.$$
This is the general solution of the given ODE.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear First-Order ODE
A linear first-order ordinary differential equation (ODE) is one of the simplest forms of differential equations. It's called 'first-order' because it involves only the first derivative of the function being studied. A differential equation of this type looks like this:
- \( y'(x) + P(x)y(x) = Q(x) \)
- \( y' \) signifies the rate of change of \( y \).
- \( P(x)y \) and \( Q(x) \) represent expressions involving the function itself and known terms.
Particular Solution
A particular solution is a specific solution to a differential equation that satisfies the equation but does not include the arbitrary constants found in a general solution. For the equation \( y'(x) + 2y = -4 \), we searched for a particular solution by assuming a constant function form:
- \( y_p(x) = A \)
- \( y_p(x) = -2 \)
Homogeneous Equation
The homogeneous equation is derived from a given linear first-order ODE by setting the non-homogeneous part (\( Q(x)\)) to zero. For our equation
To solve a homogeneous equation, you can often use separation of variables, meaning rewriting it such that all terms involving the function \( y(x) \) are on one side and terms involving \( x \) on the other:
- \( y'(x) + 2y(x) = 0 \)
To solve a homogeneous equation, you can often use separation of variables, meaning rewriting it such that all terms involving the function \( y(x) \) are on one side and terms involving \( x \) on the other:
- \( \frac{y'}{y} = -2 \)
- \( y_c(x) = Ce^{-2x} \)
General Solution
The general solution of a differential equation combines both the homogeneous (complementary function) and the particular solution. This combination captures all possible solutions, incorporating any arbitrary constants. For our ODE, the general solution is:
This form allows for flexibility in adapting the solution to various initial conditions. When initial conditions are provided (values of \( y \) at specific \( x \)), we can solve for \( C \).
The general solution is fundamental for understanding the complete behavioral spectrum of the system described by the differential equation, encapsulating all potential scenarios that the system might experience.
- \( y(x) = Ce^{-2x} - 2 \)
This form allows for flexibility in adapting the solution to various initial conditions. When initial conditions are provided (values of \( y \) at specific \( x \)), we can solve for \( C \).
The general solution is fundamental for understanding the complete behavioral spectrum of the system described by the differential equation, encapsulating all potential scenarios that the system might experience.