Question

Explain how to solve a separable differential equation of the form \(g(t) y^{\prime}(t)=h(t)\).

Short answer

Question: Solve the separable differential equation \(g(t)y'(t) = h(t)\) and obtain its general solution. Answer: The general solution for the given separable differential equation is \(y(t) = K \cdot e^{\int \frac{h(t)}{g(t)} dt}\), where \(K\) is an arbitrary constant.

Step-by-step solution

Rewrite the equation to separate the variables

Divide both sides of the equation by \(g(t)\), and we get: \[y'(t) = \frac{h(t)}{g(t)}\] Now, we want to have all the terms involving \(y(t)\) on one side of the equation and all the terms involving \(t\) on the other side. To do this, we can rewrite the equation as: \[\frac{dy(t)}{dt} = \frac{h(t)}{g(t)}\]

Divide both sides by \(y\)

To separate the variables, we want to have \(y\) and \(t\) on opposite sides of the equation. We can achieve this by dividing both sides by \(y\): \[\frac{1}{y(t)}\frac{dy(t)}{dt} = \frac{h(t)}{g(t)}\]

Integrate both sides

Now that we have separated the variables, we can integrate both sides of the equation with respect to \(t\). We get: \[\int \frac{1}{y(t)}\frac{dy(t)}{dt} dt = \int \frac{h(t)}{g(t)} dt\] Using the substitution method, we can simplify the integral on the left-hand side by setting \(u = y(t)\) and \(du = dy(t)\): \[\int \frac{1}{u} du = \int \frac{h(t)}{g(t)} dt\]

Solve the integrals

Solve the integrals on both sides of the equation: \[ln(|y(t)|) +C_1 = \int \frac{h(t)}{g(t)} dt + C_2\] To simplify and combine the constants, we can rewrite the equation as: \[ln(|y(t)|) = \int \frac{h(t)}{g(t)} dt + C\] where \(C = C_2 - C_1\).

Solve for \(y(t)\)

To solve for \(y(t)\), we can take the exponent of both sides of the equation: \[|y(t)| = e^{\int \frac{h(t)}{g(t)} dt + C}\] Since \(C\) is also an arbitrary constant, we can combine it with the constant of integration in the exponent: \[|y(t)| = e^{C} \cdot e^{\int \frac{h(t)}{g(t)} dt}\] Now, remove the absolute value by introducing another arbitrary constant, \(K\): \[y(t) = K \cdot e^{\int \frac{h(t)}{g(t)} dt}\] We have now obtained the general solution for a separable differential equation of the given form.

Key concepts

Integrating Factors

An integrating factor is a function used to multiply both sides of a differential equation to facilitate integration, often transforming a non-exact equation into an exact one. While the example at hand deals with separable differential equations directly, it's worth knowing how integrating factors might play a role in more complex scenarios. For instance, if one encounters a linear first-order differential equation of the form \(y' + p(t)y = g(t)\), an integrating factor, typically \(e^{\int p(t) dt}\), is used to enable the product rule of integration. Once multiplied, the left-hand side of the equation becomes the derivative of the product of the integrating factor and \(y(t)\), making it possible to integrate both sides with ease.

Variable Separation

Variable separation is a technique for solving first-order separable differential equations where variables related to \(y\) are isolated on one side and those related to \(t\) on the other. As seen in the solution steps, separate the functions involving \(y\) and \(t\), and then integrate both sides. For the given exercise, this step is crucial as it separates variables to prepare for individual integration. Remember, this method can only be applied when a differential equation is indeed separable, which is the case for equations of the form \(g(t) y'(t) = h(t)\).

Exponential Function Integration

Exponential function integration is particularly useful in solving differential equations involving growth and decay processes. In our exercise, after separating variables and integrating, we introduce the natural logarithm and exponentiate to solve for \(y(t)\). The relationship between logarithms and exponentials simplifies the process of solving for the variable \(y\). Integrating the exponential function typically yields an expression involving the natural logarithm, as shown in step 4 of the solution, and exponentiating both sides of the equation effectively 'undoes' the logarithm, bringing us closer to the solution for \(y(t)\).

Arbitrary Constants

In the process of integrating differential equations, we introduce arbitrary constants (\(C_1\), \(C_2\), etc.), which are essential in forming the general solution. These constants arise because indefinite integrals are not unique; adding a constant to any function's integral will yield another valid integral. After integration in the exercise, we combine constants from both sides to form a single arbitrary constant \(C\). Finally, by exponentiating and recognizing that \(e^C\) is also arbitrary (as it could be any positive number), we introduce a new arbitrary constant \(K\), encapsulating the exponential of any added constant and catering to the positive or negative values of \(y(t)\). Understanding the role of arbitrary constants is crucial as they ultimately determine the particular solution to a differential equation given initial conditions.