Question

try to transform the given equation into one with constant coefficients by the method of Problem 34. If this is possible, find the general solution of the given equation. $$ y^{\prime \prime}+3 t y^{\prime}+t^{2} y=0, \quad-\infty

Short answer

Based on the given problem, find the general solution of the given second-order linear ODE with variable coefficients. Given ODE: $$y''(t) + 3ty'(t) + t^2y(t) = 0$$ General solution: $$y(t) = (C_{1}t + C_{2}) e^{-\frac{3}{4}t^2}$$ where \(C_{1}\) and \(C_{2}\) are arbitrary constants.

Step-by-step solution

Apply the transformation method

Let's apply a transformation on the dependent variable as follows: $$ y(t) = u(t) e^{\int r(t) dt}, $$ where \(r(t)\) is an unknown function to be determined. Now, we need to find the derivatives of \(y(t)\) and substitute them back into the given ODE.

Compute first and second derivatives of y(t)

Let's compute the first and second derivatives of \(y(t)\): $$ y'(t) = u'(t) e^{\int r(t) dt} + u(t) r(t) e^{\int r(t) dt}, $$ and $$ y''(t) = u''(t) e^{\int r(t) dt} + 2 u'(t) r(t) e^{\int r(t) dt} + u(t) r'(t) e^{\int r(t) dt} + u(t) r^2(t) e^{\int r(t) dt}. $$

Substitute the derivatives in the given ODE

Now, let's substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) in the given ODE and simplify: $$ u''(t) + (2 r(t) + 3t)u'(t) + (r'(t) + r^{2}(t) + t^{2})u(t) = 0. $$

Make the ODE constant coefficient

To transform the ODE into one with constant coefficients, set the coefficients of \(u(t)\) and \(u'(t)\) to be constant: 1. \(2r(t) + 3t = 0\) for the coefficient of \(u'(t)\). Solving for \(r(t)\), we get: $$ r(t) = -\frac{3}{2}t. $$ 2. For the coefficient of \(u(t)\), we have: $$ r'(t) + r^2(t) + t^2 = 0. $$ Now, we calculate \(r'(t)\): $$ r'(t) = -\frac{3}{2}. $$ So the expression becomes: $$ -\frac{3}{2} + \left(-\frac{3}{2}t\right)^2 + t^2 = 0. $$ Simplifying, we have: $$ \frac{9}{4}t^2 - \frac{3}{2} + t^2 = 0, $$ which is always true for any value of \(t\). That means we have successfully transformed the given ODE with variable coefficients into an ODE with constant coefficients: $$ u''(t) = 0. $$

Solve the transformed ODE

Now, let's solve the transformed ODE: $$ u''(t) = 0, $$ Integrating twice, we get: $$ u'(t) = C_{1}, $$ $$ u(t) = C_{1}t + C_{2}. $$

Transform back to the original variable

To find the general solution of the given equation, we need to transform back to the original variable \(y(t)\) using the formula: $$ y(t) = u(t)e^{\int r(t) dt}. $$ Now we calculate \(\int r(t) dt\): $$ \int r(t) dt = \int -\frac{3}{2}t dt = -\frac{3}{4}t^2 + K, $$ where \(K\) is an arbitrary constant. Since \(e^K\) can be absorbed into the constants \(C_1\) and \(C_2\), we can ignore it for the purpose of finding \(y(t)\). So, $$ y(t) = (C_{1}t + C_{2}) e^{-\frac{3}{4}t^2}, $$ which is the general solution of the given equation for \(-\infty < t < \infty\).

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions of one independent variable and their derivatives. They are a fundamental tool in calculus and are used to describe various phenomena in engineering, physics, economics, and other disciplines. An ODE provides a way to model the relationship between a variable, its rate of change, and other influencing factors.

In the context of this exercise, we are dealing with a second-order ODE, which involves the second derivative of the unknown function. The equation is \( y'' + 3ty' + t^2y = 0 \). Here, the main variable is \( y \), which depends on \( t \), and the equation describes how \( y \) changes as \( t \) changes. It's important to understand the role of each term, where \( y'' \) represents acceleration or curvature, \( 3ty' \) indicates a velocity component affected by time \( t \), and \( t^2y \) is a position term modified by time.

The goal is to find a function \( y(t) \) that satisfies this equation under all circumstances in the given interval.

Transformation of Variables

To simplify the complexity of an ordinary differential equation like the one in the exercise, we often use the method of transformation of variables. This technique involves changing variables to convert a complicated ODE into a simpler one, often with constant coefficients.

For the given ODE, a transformation is applied by substituting \( y(t) = u(t)e^{\int r(t) dt} \), where \( u(t) \) is a new function, and \( r(t) \) is a function determined during the transformation process. The purpose of this substitution is to eliminate the dependence on \( t \) within the coefficients, resulting in an ODE with constant coefficients, which are generally easier to solve.

This method requires us to compute derivatives of the transformed variable \( y(t) \) and substitute back into the original equation. By choosing \( r(t) \) cleverly, you can simplify the parameters in the equation, notably reducing them to constants, as shown in the step-by-step solution.

Constant Coefficients

Differential equations with constant coefficients have terms whose coefficients are constants rather than functions of the independent variable. This type of differential equation is generally more straightforward to solve because the coefficients do not vary with changes in the variable, leading to more predictable behavior and solutions.

In our exercise, after the application of the transformation of variables, the resulting differential equation is \( u''(t) = 0 \). This is a prime example of a differential equation with constant coefficients because the terms \( u''(t) \), \( u'(t) \), and \( u(t) \) are not multiplied by \( t \) or any function of \( t \).

Solving ODEs with constant coefficients involves straightforward integration, making the problem-solving process easier to manage. This simplification is what makes the transformation method particularly powerful in handling more complex differential equations.

General Solution

The general solution of a differential equation is a function that encompasses all possible solutions of the equation. It typically contains arbitrary constants, which can be adjusted to fit specific initial or boundary conditions. This allows the solution to be tailored to particular applications or scenarios.

For the given exercise, the general solution was derived through several transformations and simplifications. By integrating the transformed equation, we obtained \( u(t) = C_1t + C_2 \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions if given.

To express the solution in terms of the original variable \( y(t) \), we revert the transformation: \( y(t) = (C_1t + C_2)e^{-\frac{3}{4}t^2} \). This general solution covers every potential scenario that the original ODE might describe over the entire domain from \(-\infty \) to \( \infty \).

• The presence of \( e^{-\frac{3}{4}t^2} \) suggests a damping or decaying effect on the solution.
• The terms \( C_1t \) and \( C_2 \) indicate linear growth or decline depending on the specific initial conditions.

Understanding how to derive and interpret the general solution is crucial for applying differential equations to real-world problems.