Question

use the method of Problem 35 to solve the given differential equation. $$ y^{\prime}-2 y=t^{2} e^{2 t} $$

Short answer

Based on the solution above, answer the following question: **Question:** Find the general solution of the given first-order linear differential equation: \(y^{\prime}-2 y=t^{2} e^{2 t}\). **Answer:** The general solution to the given differential equation is \(y(t) = e^{2t} \left(\frac{t^3}{3} + C\right)\), where C is the integration constant.

Step-by-step solution

Identify the Integrating Factor

The given equation is $$ y^{\prime}-2 y=t^{2} e^{2 t} $$ It can be written in the standard form of a linear first-order differential equation as $$ y^{\prime} - 2y = t^2 e^{2t}. $$ Comparing this to the general form of a linear first-order differential equation, which is $$ y^{\prime} + p(t) y = f(t) , $$ we can see that $$ p(t) = -2 \text{ and } f(t) = t^2 e^{2t}. $$ Now let's find the integrating factor, which is given by $$ \mu(t) = e^{\int p(t) dt} . $$ Substituting \(p(t) = -2\), we have $$ \mu(t) = e^{\int -2 dt} = e^{-2t}. $$

Multiply each side by the Integrating Factor

Next, multiply both sides of the given differential equation by the integrating factor \(\mu(t) = e^{-2t}\) to obtain $$ e^{-2t}(y^{\prime} - 2y) = (t^{2} e^{2 t}) e^{-2t} . $$

Rewrite the Left Side as the Product Rule

The left side of the equation can be rewritten using the product rule in the form: $$ \frac{d}{dt}(e^{-2t} y) = t^2. $$ This is because the product rule states that $$ \frac{d}{dt}(u \cdot v) = u' \cdot v + u \cdot v', $$ where \(u = e^{-2t}\) and \(v = y\). So, we have $$ \frac{d}{dt}(e^{-2t} y) = (-2 e^{-2t}) y + (e^{-2t}) y' = e^{-2t}(y^{\prime} - 2y). $$

Integrate both sides

Now we integrate both sides of the equation: $$ \int \frac{d}{dt}(e^{-2t} y) dt = \int t^2 dt. $$ We have $$ e^{-2t} y = \frac{t^3}{3} + C , $$ where C is the integration constant.

Isolate y(t)

Finally, we solve for \(y(t)\) by dividing by \(e^{-2t}\): $$ y(t) = e^{2t} \left(\frac{t^3}{3} + C\right). $$ This is the general solution to the given differential equation.

Key concepts

Integrating Factor Method

The integrating factor method is a powerful tool for solving linear first-order differential equations of the form \( y' + p(t)y = f(t) \).To apply this method, one must first identify an integrating factor \( \(\mu(t)\) = e^{\int p(t) dt} \).This special function is designed to turn the original equation into an exact equation, making it easier to solve. The process includes multiplying both sides of the equation by this integrating factor, which often simplifies the left side into a form that can be recognized as the derivative of a product of two functions, allowing for direct integration. This method is particularly useful because it can be applied to a wide variety of problems with consistently reliable results.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental in expressing the relation between varying quantities and their rates of change, serving as the cornerstone of many scientific and engineering disciplines. In particular, linear first-order differential equations have the form \( y' + p(t)y = f(t) \),where \( y' \) denotes the derivative of the function \( y \) with respect to \( t \), \( p(t) \) is a coefficient function, and \( f(t) \) is a known function of \( t \). These equations can model various phenomena, from population dynamics to electrical circuits, and are solved using analytical methods like the integrating factor method, separation of variables, or numerical techniques.

Product Rule

The product rule is a fundamental theorem in calculus used to find the derivative of the product of two functions. It states that if \( u(t) \) and \( v(t) \) are both differentiable functions, then the derivative of their product \( u(t)v(t) \) is given by:\[ \frac{d}{dt}(u(t) \( \cdot \) v(t)) = u'(t) v(t) + u(t) v'(t) \].The product rule is pivotal when simplifying differential equations using the integrating factor method. By recognizing that the left side of the modified equation is the derivative of a product, the equation can be integrated directly, rather than having to expand and integrate term by term.

Integration

Integration is the process of finding the integral of a function, which is the inverse operation to differentiation. It plays a crucial role in solving differential equations by providing a way to reverse the process of differentiation and find the original function or a general solution. The integral of a function can represent many concepts such as area under a curve, accumulated totals, or in the case of differential equations, the general solution to the equation. There are two types of integrals: indefinite integrals, which include a constant of integration \( C \), and definite integrals, which are evaluated over a specific interval. The method of integration used to solve differential equations usually involves finding an antiderivative of the given function, allowing for the determination of the original function being sought.