Question

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

Short answer

Question: Find the general solution of the first-order linear differential equation given as \(y^{\prime}(t) = ky + b\), with the constraint that \(ky+b < 0\). Answer: \(y(t) = Ce^{kt} - \frac{b}{k}\), where \(C\) is the constant of integration.

Step-by-step solution

Identify the given equation type

The given differential equation is a first-order linear equation with the form \(y^{\prime}(t) = ky + b\).

Solve the differential equation

To solve this equation, we can use an integrating factor. The integrating factor is given by \(e^{\int k dt} = e^{kt}\). Multiply both sides of the equation by the integrating factor: $$e^{kt}y^{\prime}(t) = ke^{kt}y + be^{kt}.$$ Now, notice that the left-hand side of this equation is the derivative of the product \(y(t) e^{kt}\): $$\frac{d}{dt}(y(t)e^{kt}) = ke^{kt}y + be^{kt}.$$ Integrate both sides with respect to \(t\): $$\int\frac{d}{dt}(y(t)e^{kt})dt = \int (ke^{kt}y + be^{kt}) dt.$$ The left side we can rewrite as $$y(t)e^{kt} = \int (ke^{kt}y + be^{kt}) dt + C,$$ where \(C\) is the constant of integration.

Solve for y(t)

Now, we solve for \(y(t)\) by dividing both sides by \(e^{kt}\): $$y(t) = e^{-kt}\left(\int (ke^{kt}y + be^{kt}) dt + C\right).$$ Given the constraint that \(ky + b < 0\), the differential equation to integrate is \(ke^{kt}(y-\frac{b}{k})\). Thus, $$y(t) = e^{-kt}\left(\int ke^{kt}\left(y-\frac{b}{k}\right) dt + C\right).$$ Notice that \(\int ke^{kt}\left(y-\frac{b}{k}\right) dt = e^{kt}(y-\frac{b}{k}) + D\), where \(D\) is another constant. Therefore, we have $$y(t) = e^{-kt}\left(e^{kt}\left(y-\frac{b}{k}\right) + C\right),$$ which can be simplified to $$y(t) = Ce^{kt} - \frac{b}{k}.$$

Verify the solution

We can check our solution by calculating the first derivative of our proposed solution and comparing it with the given equation. The first derivative is: $$y^{\prime}(t) = Cke^{kt}.$$ Now, substitute the proposed solution into the given equation: $$Cke^{kt} = k\left(Ce^{kt}-\frac{b}{k}\right) + b.$$ Distribute the \(k\): $$Cke^{kt} = kCe^{kt}-b+b,$$ which confirms that the proposed solution, \(y(t) = Ce^{kt} - \frac{b}{k}\), is indeed a general solution to the given first-order linear differential equation.

Key concepts

Integrating Factor

The method of the integrating factor is an effective technique for solving first-order linear differential equations. It involves multiplying the equation by an integrating factor that is generally expressed as an exponential function of the independent variable. This factor is carefully chosen so that the equation becomes an exact derivative, which makes it easier to integrate.

In this exercise, the integrating factor is found by the expression \(e^{\int k dt} = e^{kt}\), which simplifies the original equation. Multiplication by this factor enables the left-hand side of the differential equation to be expressed as the derivative of \(y(t)e^{kt}\), thus paving the way for straightforward integration. Using the integrating factor effectively transforms the problem into a form that can be integrated directly to find the solution.

Separation of Variables

Separation of variables is a classic strategy for solving differential equations where each variable can be isolated on one side of the equation. However, this method may not be directly applicable for all equations, especially when they are not naturally separable.

In the given exercise, we have a non-separable first-order linear differential equation, but it's worth mentioning separation of variables as a general method. Typically applicable to equations of the form \(\frac{dy}{dx}=g(x)h(y)\), it requires rearranging the terms so that each side of the equation contains only one variable and its derivatives before integrating. Despite not using this method for our particular problem, it is a valuable tool to have when faced with differential equations that can be manipulated into this format.

General Solution of Differential Equation

The general solution of a differential equation represents all possible solutions that satisfy the equation, encompassing any specific solutions that may exist. The general solution includes a constant of integration which accounts for initial conditions or boundary values that are not specified within the differential equation itself.

In our case, the general solution is \(y(t) = Ce^{kt} - \frac{b}{k}\). This formula includes the arbitrary constant \(C\), reflecting the infinite number of solutions that could satisfy the given first-order linear equation. Each unique value of \(C\) corresponds to a particular solution curve on the plane, highlighting how the general solution captures a family of possible behaviors of the dynamic system described by the differential equation.

Constant of Integration

The constant of integration is a fundamental aspect of indefinite integration. Each time we integrate a function with respect to its variable, we add a constant to the result because anti-differentiation can only determine a function up to an added constant.

In the step-by-step solution we discussed, the constant of integration \(C\) appears naturally as a result of the integration process. When solving differential equations, \(C\) is essential for representing the most general form of the solution. It is determined by initial conditions or other constraints on the problem, which for this exercise, were not provided, so the general solution includes this constant. The importance of the constant of integration cannot be understated, as it allows for tailored solutions that adhere to real-world scenarios and conditions.