Question

Use the method outlined in Problem 28 to solve the given differential equation. $$ t y^{\prime \prime}-(1+t) y^{\prime}+y=t^{2} e^{2 t}, \quad t>0 ; \quad y_{1}(t)=1+t \quad(\text { see Problem } 15) $$

Short answer

Question: Using the reduction of order and variation of parameters methods, find the general solution to the following nonhomogeneous differential equation: $$ ty''-(1+t)y'+y= t^2e^{2t}, \quad t>0; $$ with a known solution \(y_1(t) = 1+t\). Answer: The general solution to the given nonhomogeneous differential equation is: $$ y(t) = c_1(1+t) + c_2(-\frac{1}{3}t^{-1} + Ce^{2t} + C_1)(1+t). $$

Step-by-step solution

Find the second linearly independent solution, y_2(t)

Using the reduction of order method, we assume that our second linearly independent solution, \(y_2(t)\), is in the form of: $$ y_2(t) = u(t)(1+t), $$ where \(u(t)\) is a function to be determined. We also need the first and second derivatives of \(y_2(t)\): $$ y_2'(t) = u'(t)(1+t) + u(t), $$ $$ y_2''(t) = u''(t)(1+t) + 2u'(t) $$

Substitute y_2(t) into the given differential equation

Now, we substitute \(y_2(t)\) and its derivatives, \(y_2'(t)\) and \(y_2''(t)\), into the given differential equation: $$ t(u''(t)(1+t) + 2u'(t)) - (1 + t)(u'(t)(1+t) + u(t)) + u(t)(1+t) = t^2e^{2t} $$ We simplify the equation and solve for \(u(t)\).

Simplify the equation and solve for u(t)

Simplify the equation: $$ tu''(t)(1+t) + 2tu'(t) - u'(t)(1+t)^2 - u(t)(1+t) + u(t)(1+t) = t^2e^{2t} $$ Cancel out the terms and we get: $$ tu''(t)(1+t) + 2tu'(t) - u'(t)(1+t)^2 = t^2e^{2t} $$ Now we simplify the equation further by dividing through by \(t(1+t)\): $$ u''(t) + \frac{2-u'(t)(1+t)}{t} = \frac{te^{2t}}{1+t} $$

Apply an integrating factor to solve for u'(t)

Next, we will apply an integrating factor to the equation for \(u'(t)\). Taking \(M(t) = 2-u'(t)(1+t)\) and \(N(t) = te^{2t}/(1+t)\), we find the integrating factor, \(\mu(t)\). The integrating factor can be found as follows: $$ \mu(t) = e^{\int P(t) dt} = e^{\int \frac{2}{t} dt} = e^{2 \ln(t)} = t^2 $$ Now, we multiply the equation by the integrating factor, \(\mu(t) = t^2\): $$ t^2u''(t) + 2t - t^3u'(t)(1+t) = t^3e^{2t} $$ Notice that the left-hand side of the equation is now an exact differential: $$ \frac{d}{dt}(tu^{2}(t)(1-u'(t)(1+t))) = t^3e^{2t} $$

Integrate both sides to obtain u'(t)

Now, we integrate both sides with respect to \(t\): $$ \int \frac{d}{dt}(tu^{2}(t)(1-u'(t)(1+t))) dt = \int t^3 e^{2t} dt $$ After integrating and isolating \(u'(t)\), we will have: $$ u'(t) = \frac{1}{3} t^{-2} - \frac{1}{9} te^{-2t} + C $$

Integrate u'(t) to find u(t)

Next, we integrate \(u'(t)\) to obtain: $$ u(t) = \int\left(\frac{1}{3} t^{-2} - \frac{1}{9} te^{-2t}\right) dt + C_1 $$ After evaluating this integral, we have: $$ u(t) = -\frac{1}{3}t^{-1} + Ce^{2t} + C_1 $$

Find the second linearly independent solution, y_2(t)

Now that we have found \(u(t)\), we can find the second linearly independent solution, \(y_2(t)\), in terms of \(u(t)\): $$ y_2(t) = u(t)(1+t) = (-\frac{1}{3}t^{-1} + Ce^{2t} + C_1)(1+t) $$

Apply the method of variation of parameters

With both linearly independent solutions, \(y_1(t)=1+t\) and \(y_2(t)=(-\frac{1}{3}t^{-1} + Ce^{2t} + C_1)(1+t)\), we can now apply the method of variation of parameters to find the general solution to the nonhomogeneous differential equation. Our final solution will be a linear combination of both linearly independent solutions, \(y(t)=c_1y_1(t)+c_2y_2(t)\).

Key concepts

Reduction of Order

Reduction of order is a technique used to find a second linearly independent solution to a second-order homogeneous differential equation when one solution, say y_1(t), is already known. Imagine you're solving a puzzle and you have one piece already in place. Reduction of order helps you figure out the shape of the missing piece using the one you have.

The general form of the second solution, y_2(t), is assumed to be the product of an unknown function u(t) and the known solution, like this: $$ y_2(t) = u(t)y_1(t). $$This approach effectively reduces the order of the differential equation by using the known solution, allowing us to find u(t) that determines y_2(t). After substituting y_2(t) and its derivatives into the original equation, we solve for u(t) to find the missing piece of the puzzle.

Integrating Factor

An integrating factor is a special multiplier that turns a non-exact differential equation into an exact one, making it easier to solve. It's like finding the right key that will unlock a door. This function, usually denoted as μ(t), is determined by an expression involving the original differential equation.

For example, if we have a first-order linear differential equation:

• $$ P(t)y' + Q(t)y = g(t) $$

, the integrating factor is obtained by:

• $$ μ(t) = e^{tt t t default( t t default( t default(t)) dt}). $$

By multiplying the entire equation by this factor, the left-hand side often becomes the derivative of a product of μ(t) and y(t), leading to a simpler equation that can be integrated to find y(t).

Variation of Parameters

Variation of parameters is a method to solve nonhomogeneous differential equations. It's akin to custom fitting a suit; you take a general solution for the homogeneous equation and adjust it to fit the nonhomogeneous case. When you have two linearly independent solutions, y_1(t) and y_2(t), of the associated homogeneous equation, these become the parameters you can vary.

The general solution of the nonhomogeneous equation is:

• $$ y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t) $$

,where c_1 and c_2 are constants, and y_p(t) is a particular solution. You adjust c_1 and c_2 to find a solution that satisfies the original nonhomogeneous equation.

Nonhomogeneous Differential Equation

Nonhomogeneous differential equations are those that include terms independent of the function and its derivatives being sought. In the world of differential equations, these are the problems that deal with external forces or influences. They can be represented as:

• $$ a_2(t)y'' + a_1(t)y' + a_0(t)y = g(t), $$

where g(t) is a non-zero function representing the nonhomogeneity.

The power of the nonhomogeneous equation lies in its ability to model more complex and realistic scenarios, such as damped oscillations with external forces or circuits with varying currents.

Linearly Independent Solutions

Linearly independent solutions are the backbone of solving homogeneous linear differential equations. Consider them as distinct ingredients in a recipe that cannot be made from one another. A set of solutions is linearly independent if no solution in the set can be written as a linear combination of the others.

They are essential because they form the 'basis' for the solution space to a differential equation. For second-order equations, we need two linearly independent solutions. The Wronskian determinant is a useful tool to test for independence. If the Wronskian is non-zero at some point on the interval of interest, the functions are linearly independent. For our two solutions, y_1(t) and y_2(t), we're seeking, this would mean:

• $$ W(y_1, y_2)(t) = t t default(y_1(t))y_2'(t) - y_1'(t)y_2(t) $$

,is not equal to zero.