Question

find the solution of the given initial value problem. $$ t^{3} y^{\prime}+4 t^{2} y=e^{-t}, \quad y(-1)=0 $$

Short answer

Question: Determine the function y(t) that solves the given initial value problem. Differential equation: \(ty^{\prime} + 4y = te^{(-t)}\) Initial condition: \(y(-1) = 0\) Answer: \(y(t) = \frac{-t e^{-t} + e^{-t} - 2e}{t^4}\)

Step-by-step solution

Write the linear differential equation in standard form

To obtain the standard form of the given differential equation, we need to write it as: $$ y^{\prime} + P(t) y = Q(t) $$ In our case, we divide both sides of the given equation by \(t^3\) to isolate \(y^\prime\): $$ y^{\prime}+\frac{4}{t}y=\frac{e^{-t}}{t^3} $$ So, we have \(P(t) = \frac{4}{t}\), and \(Q(t) = \frac{e^{-t}}{t^3}\).

Determine the integrating factor

The integrating factor, denoted by \(μ(t)\), is given by: $$ μ(t) = e^{\int P(t) dt} $$ Here, \(P(t) = \frac{4}{t}\). So, we have: $$ μ(t)=e^{\int \frac{4}{t} dt} = e^{4\ln|t|}=t^4 $$

Multiply the standard form equation by the integrating factor

We will multiply both sides of the equation \(y^{\prime}+\frac{4}{t}y=\frac{e^{-t}}{t^3}\) by the integrating factor \(t^4\): $$ t^4y^{\prime} + 4t^3 y = t e^{-t} $$ Now, we can notice that the left-hand side of the equation is the derivative of a product: $$ \frac{d}{dt}(t^4y) = t e^{-t} $$

Integrate both sides of the equation

Integrate both sides of the equation with respect to t: $$ \int \frac{d}{dt}(t^4y) dt = \int t e^{-t} dt $$ The integration yields: $$ t^4y = \int t e^{-t} dt + C $$ To integrate the right side, we use integration by parts. Let \(u=t\), and \(dv=e^{-t} dt\). Then, \(du = dt\), and \(v=-e^{-t}\). Therefore, $$ \int t e^{-t} dt = -t e^{-t} - \int (-e^{-t}) dt = -t e^{-t} + e^{-t} + k $$ Substituting this back into our equation, we get: $$ t^4y = -t e^{-t} + e^{-t} + k $$ Now we can solve for y(t): $$ y(t) = \frac{-t e^{-t} + e^{-t} + k}{t^4} $$

Solve for y(t) using the initial condition

We now substitute the initial condition y(-1) = 0 to find the value of k: $$ 0 = \frac{-(-1) e^{-(-1)} + e^{-(-1)} + k}{(-1)^4} $$ Solving for k, we obtain \(k=-2e\). Finally, we substitute the value of k back into the equation for y(t): $$ y(t) = \frac{-t e^{-t} + e^{-t} - 2e}{t^4} $$ This is the solution for the given initial value problem.

Key concepts

Linear Differential Equation

Linear differential equations are foundational in differential equations and are defined by their linearity in terms of the function and its derivatives. When we look at an equation like \(t^{3} y^{\text{'} }+4 t^{2} y=e^{-t}\), we are looking at a first-order linear differential equation. Once reformulated to standard form \(y^{\text{'} } + P(t) y = Q(t)\), where P(t) and Q(t) are both functions of \(t\), it becomes clear that the solution involves finding an expression for \(y\) that satisfies this equation for any value of \(t\).
The crucial first step is transforming the given equation into the standard linear form. This structure allows us to utilize specific techniques tailor-made for linear equations, such as finding an integrating factor, which will subsequently enable us to solve the equation in a systematic manner. The beauty of linear differential equations lies in their predictability and the straightforward methods through which we can approach them, thus making these types of problems rewarding for students who enjoy a clear step-by-step process.

Integrating Factor

The integrating factor is a technique used when solving linear differential equations and is a key ally in transforming an unwieldy expression into one ready for integration. The integrating factor, usually denoted as \(\mu(t)\), is derived from an equation's function \(P(t)\).
In the exercise, we find the integrating factor using the formula \(\mu(t) = e^{\int P(t) dt}\), where \(P(t)\) is identified from the standard form of the equation. After calculating, we get \(\mu(t)=t^4\), a power function that, when multiplied with the original differential equation, turns the left-hand side into the derivative of the product of \(\mu(t)\) and the function \(y\). This is where the magic happens: the once separable terms are now linked, facilitating the next step—integration. An integrating factor simplifies the equation in exactly the right way, enabling an elegant path to the solution.

Integration by Parts

Integration by parts is a powerful tool often used when the function we want to integrate is a product of two simpler functions. It derives from the product rule for differentiation and is based on the formula \(\int u dv = uv - \int v du\).
In our problem, when faced with the task of integrating \(t e^{-t}\), we recognize that it is the product of two functions, making integration by parts the ideal choice. By identifying \(u=t\) (which becomes easier to integrate when differentiated) and \(dv=e^{-t} dt\) (which upon integration yields a manageable expression), we allow ourselves to turn a complex integral into a pair of simpler ones. The elegant dance of choosing \(u\) and \(dv\), differentiating, and integrating, then substituting back into our formula, leads us to our desired result—a neater integral that brings us closer to finding our function \(y(t)\). This technique not only serves to solve the integral at hand but also instills an appreciation for the interplay between differentiation and integration.