Question

For the given differential equation, $$ y^{\prime \prime}-2 y^{\prime}+2 y=e^{-t} \sin 2 t+2 t+t e^{-t} \sin t $$

Short answer

Question: Solve the given second-order linear non-homogeneous differential equation: \(y''(t) - 2y'(t) + 2y(t) = e^{-t} + te^{-t} + e^{-t}\sin 2t\) Answer: The general solution to the given differential equation is: \(y(t) = e^{1t}(A\cos{t} + B\sin{t}) + (A_1t + A_2)e^{-t} + (B_1t + B_2) + (C_1t + C_2)e^{-t}\sin{t} + (D_1t + D_2)e^{-t}\cos{t}\), where the constants \(A\), \(B\), \(A_1\), \(A_2\), \(B_1\), \(B_2\), \(C_1\), \(C_2\), \(D_1\), and \(D_2\) can be determined by any given initial conditions.

Step-by-step solution

Solve the homogeneous equation

To solve the homogeneous equation, we first set the right side of the given equation to zero: $$ y''(t) - 2y'(t) + 2y(t) = 0 $$ This is a second-order linear homogeneous differential equation with constant coefficients. We will look for a solution of the form \(y_h(t) = e^{rt}\), where \(r\) is a constant. Taking the derivatives and substituting into the equation, we get: $$ r^2e^{rt}- 2re^{rt} + 2e^{rt} = 0 $$ Factoring out \(e^{rt}\) and setting the remaining part to zero, we now have the characteristic equation: $$ r^2 - 2r + 2 = 0 $$

Solve the characteristic equation

To find the values of \(r\), we will solve the quadratic equation \(r^2 - 2r + 2 = 0\). Since it does not factor, we will use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2-4(1)(2)}}{2(1)} $$ $$ r = 1 \pm i $$ Since the roots are complex with real part 1 and imaginary part 1, the general solution to the homogeneous equation is of the form: $$ y_h(t) = e^{1t}(A\cos{t} + B\sin{t}) $$ where \(A\) and \(B\) are constants.

Find a particular solution

We will now find a particular solution to the given non-homogeneous equation. We notice that the right side of the equation contains terms with \(e^{-t}\), \(t\), and \(e^{-t}\sin 2t\). For the linear operator with constant coefficients, we will set up a particular solution of the form: $$ y_p(t) = (A_1t + A_2)e^{-t} + (B_1t + B_2) + (C_1t + C_2)e^{-t}\sin{t} + (D_1t + D_2)e^{-t}\cos{t} $$ Taking the first and second derivatives of \(y_p(t)\), we substitute them back into the given differential equation. Then, match the coefficients of like terms on both sides of the equation to get a linear system of equations. Finally, solve for the constants \(A_1\), \(A_2\), \(B_1\), \(B_2\), \(C_1\), \(C_2\), \(D_1\), and \(D_2\).

Combine the complementary and particular solutions

We found the complementary solution \(y_h(t)\) from the homogeneous equation and the particular solution \(y_p(t)\) from the given non-homogeneous equation. Therefore, the general solution to the given differential equation will be the sum of these two solutions: $$ y(t) = y_h(t) + y_p(t) = e^{1t}(A\cos{t} + B\sin{t}) + (A_1t + A_2)e^{-t} + (B_1t + B_2) + (C_1t + C_2)e^{-t}\sin{t} + (D_1t + D_2)e^{-t}\cos{t} $$ This solution will satisfy the original differential equation for any values of the constants \(A\), \(B\), \(A_1\), \(A_2\), \(B_1\), \(B_2\), \(C_1\), \(C_2\), \(D_1\), and \(D_2\), which can be determined by any given initial conditions.

Key concepts

Homogeneous Linear Differential Equation

When we encounter a differential equation like \(y^{\textstyle\backprime \backprime}-2 y^{\textstyle\backprime}+2 y=0\), we're dealing with a homogeneous linear differential equation. Here's what makes it so:

• It's linear because the equation involves the function \(y\) and its derivatives without being multiplied or divided by each other.
• The term homogeneous comes into play since each term includes the function or one of its derivatives, and no free-standing numbers or non-zero functions appear on the rhs (right-hand side) of the equation.

Handling such an equation usually involves finding the general solution, characterizing the behavior of any solution before initial conditions are considered. The method often starts by seeking solutions that have a certain structure, like \(y_h(t) = e^{rt}\), which, when substituted into the homogeneous equation, simplifies our search to find the so-called roots of the equation.

Characteristic Equation

Once we have our structured solution attempt like \(e^{rt}\), the next step involves a crucial part called the characteristic equation. Derived from substituting \(y_h(t)\) into the homogeneous linear differential equation, it helps us pinpoint the specific values of \(r\) that will make \(y_h(t)\) a valid solution.

For the equation \(y^{\textstyle\backprime \backprime}-2 y^{\textstyle\backprime}+2 y=0\), our characteristic equation takes the form: \(r^2 - 2r + 2 = 0\). Solving it may lead us to real roots, repeated roots, or as in this case, complex roots. The nature of these roots dictates the form of our homogeneous solution and essentially sets the backbone of how we address the entire differential equation.

Particular Solution

While the general solution of the homogeneous equation covers a vast array of possibilities, the particular solution, \(y_p(t)\), brings us closer to solving non-homogeneous differential equations. It represents a specific solution that satisfies the original, non-homogeneous equation.

To find it, we compose a function that resembles the non-homogeneous part, often choosing terms that would not appear in the homogeneous solution to avoid overlap. For the exercise at hand involving terms like \(e^{-t} \text{and}\ t\), the chosen \(y_p(t)\) includes these elements, adjusted with undetermined coefficients to be solved later. This process, named the method of undetermined coefficients, allows us to craft a tailored function that, along with the homogeneous solution, forms the complete answer to the equation.

Complex Roots

The presence of complex roots in the characteristic equation, such as \(1 \text{pm} i\), brings about a unique twist to our solution. Rather than working directly with complex numbers, we use Euler's formula which connects complex exponentials to sine and cosine functions.

With our complex roots, the homogeneous solution takes a form that includes the real part of the root as the exponent's coefficient and the imaginary part as the argument for the cosine and sine: \(y_h(t) = e^{1t}(A\text{cos}{t} + B\text{sin}{t})\). This representation captures the oscillatory nature that complex roots imply, and it's an essential underpinning to understand when dealing with differential equations bearing such roots.

Method of Undetermined Coefficients

The method of undetermined coefficients is a strategic approach used to find the particular solution for non-homogeneous linear differential equations. It involves creating a trial solution with arbitrary constants, structured to mimic the form of the non-homogeneous part of the equation. For example, \(y_p(t)\) in the original exercise was crafted to reflect the components of the rhs like \(e^{-t}\) and \(t\).

Once we have our trial solution, we differentiate, substitute into the given equation, and equate the coefficients of similar terms from both sides—thereby creating a system of linear equations that, when solved, determine the values of our undetermined coefficients, such as \(A_1\), \(A_2\), and so on. This process effectively locks in the particular solution, ensuring it works seamlessly with the homogeneous one to satisfy the differential equation comprehensively.