Question

The solution of the initial value problem \(t y^{\prime}+4 y=\alpha t^{2}, y(1)=-\frac{1}{3}\) is known to exist on \(-\infty

Short answer

Answer: The value of the constant α is -2.

Step-by-step solution

Solve the homogeneous equation

First, we solve the homogeneous equation \((t y^{\prime}+4 y=0)\). This is a first-order linear differential equation which can be rewritten in standard form as: \(y^{\prime}+\frac{4}{t}y=0\) We know that the integrating factor is given by: \(IF(t) = e^{\int\left(\frac{4}{t}\right)dt} = e^{4\ln (t)} = t^{4}\) Now, we multiply the standard form of the equation by the integrating factor: \(t^{4} y^{\prime}+4 t^{3}y=0\) Now, notice that the LHS is equal to \((t^{4}y)^{\prime}\). Thus, we get the equation: \((t^{4}y)^{\prime}=0\) Now, we integrate both sides of the equation: \(\int(t^{4}y)^{\prime} dt=\int 0 dt\) \(t^{4}y = C\) \(y = \frac{C}{t^{4}}\) The general form of the homogeneous solution is found.

Find a particular solution for the inhomogeneous equation

Now we solve the inhomogeneous equation: \(t y^{\prime}+4 y=\alpha t^{2}\) We can make a guess of the particular solution in the form of \(y_p = A t^B\). Then, \(y_p^{\prime} = A B t^{B-1}\) Substituting these into the inhomogeneous equation: \(t(A B t^{B-1})+4(A t^B)=\alpha t^{2}\) If we try to match the powers of both sides of the equation, we get \(B=2\) and \(A B + 4 A = \alpha\). Since we have \(B=2\), the particular solution \(y_p\) that we get is: \(y_p = A t^{2}\)

Find the general solution and apply the initial condition

We know that the general solution is given by the sum of the homogeneous solution and particular solution. Therefore, \(y(t) = y_h(t) + y_p(t) = \frac{C}{t^{4}} + A t^{2}\) Now, we apply the initial condition \(y(1)=-\frac{1}{3}\): \(-\frac{1}{3} = \frac{C}{1^4} + A(1^2)\) \(-\frac{1}{3} = C + A\)

Determine α

We have the relationship between A and C, and we know that \(A B + 4 A = \alpha\). Therefore, \(\alpha = 2 A + 4 A = 6 A\) And since \(-\frac{1}{3} = C + A\), we get \(A = -\frac{1}{3} - C\) SubSTITUTE A into the previous equation to find α: \(\alpha = 6 \left(-\frac{1}{3} - C\right) = -2 - 6 C\) Since the problem states that the solution exists for \(-\infty<t<\infty\) and we have \(y(t) = \frac{C}{t^{4}} + A t^{2}\), we notice that the term with C, which involves \(t^{4}\) in the denominator, vanishes as \(t\) approaches infinity. Therefore, C should be equal to zero to have a solution that exists on \(-\infty<t<\infty\). So, \(C=0\) Now we have A and can find the value for α: \(\alpha = -2 - 6(0) = -2\) Thus, the value of the constant α is equal to -2.

Key concepts

Differential Equations

Differential equations are a type of equation that involves functions and their derivatives. They play a fundamental role in mathematics, physics, engineering, and many other disciplines, as they describe how a particular quantity changes over time or space. The initial value problem is a common type of differential equation where one is given an equation along with an initial condition, which specifies the value of the unknown function at a particular point. In our exercise example, we tackled a first-order linear differential equation and used the initial condition to find a specific solution. This is commonly encountered by students who might find the algebraic rearrangement and integration challenging. However, breaking down the process into systematic steps can simplify understanding and solving these equations.

Integrating Factor

An integrating factor is a function that is used to simplify the solving process of linear differential equations. The purpose of an integrating factor is to create a new equation where the left-hand side becomes the derivative of a product of functions, which can then be integrated more straightforwardly. In our problem, the integrating factor was found using the coefficient of y from the homogeneous equation. It can often be confusing to determine what this factor is and how it's used, but by following the formula for the integrating factor, \(e^{\int P(t)dt}\), where P(t) is the coefficient of y, students can apply this method to a range of problems to find the homogeneous solution more easily.

Homogeneous Solution

The homogeneous solution of a differential equation corresponds to the solution of the equation when the right-hand side is set to zero, effectively representing the system's natural behavior without external forces. This solution is crucial for understanding the system's structure and is often denoted as \(y_h(t)\). In the provided exercise, after applying the integrating factor and integrating, we found the general form of the homogeneous solution as \(y_h(t) = \frac{C}{t^{4}}\). While students might struggle to interpret the significance of the homogeneous solution, it's essential to consider it as the 'base' or 'complementary' part of the overall general solution.

Particular Solution

In contrast to the homogeneous solution, the particular solution, denoted as \(y_p(t)\), specifically addresses the non-homogeneous part of the differential equation: the part with non-zero right-hand side. Finding a particular solution often involves making an educated guess based on the form of the non-homogeneous function. In our exercise, we guessed \(y_p = A t^B\), which after differentiation and substitution into the original equation, led us to find the constants A and B. The particular solution is tailored to the external forces or input and, when combined with the homogeneous solution, forms the general solution to the initial value problem. This step might require creative intuition and practice in recognizing patterns but remains a vital component in solving differential equations.