Question

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

Short answer

Answer: The general solution of the given differential equation is \(y(t) = A \cos t + B \sin t + t \left( \frac{\cos t}{t} e^t \sin t + \frac{\sin t - t \cos t}{t} e^t \cos t \right)\), where \(A\) and \(B\) are arbitrary constants.

Step-by-step solution

Find the Complementary Function

To find the complementary function, we have to solve the homogeneous equation: $$ y^{\prime \prime} + y = 0 $$ The characteristic equation for this equation is: $$ r^2 + 1 = 0 $$ Solving this quadratic equation, we get \(r_1 = i\) and \(r_2 = -i\). The complementary function has the general form: $$ y_c(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{it} + C_2 e^{-it} $$ This can also be written as: $$ y_c(t) = A \cos t + B \sin t $$ where \(A = C_1 + C_2\) and \(B = i(C_1 - C_2)\).

Find a Particular Solution

Now we need to find a particular solution for the given inhomogeneous equation. We can assume a particular solution of the form: $$ y_p(t) = t (a e^t \sin t + b e^t \cos t) $$ We choose this form because the original equation involves exponential and trigonometric functions, and the \(t\) is multiplied in order to avoid overlap with the complementary function. Now we need to calculate \(y_p'(t)\) and \(y_p''(t)\): $$ y_p'(t) = a e^t (t\sin t + \cos t) - b e^t (t\cos t - \sin t) $$ and $$ y_p''(t) = a e^t (t\sin t + 2\cos t - t\cos^2 t) - b e^t(t\cos t - 2\sin t - t\sin^2 t) $$ Substituting \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) back into the original equation: $$ a e^t (t\sin t + 2\cos t - t\cos^2 t) - b e^t(t\cos t - 2\sin t - t\sin^2 t) + a e^t t \sin t + b e^t t \cos t = e^t \sin t $$ Now, comparing coefficients of \(e^t \sin t\) and \(e^t \cos t\) on both sides, we get two equations: $$ a (t + 2 - t\cos^2 t) - b (t\cos t - 2\sin t) = \sin t $$ and $$ a (t\sin t) + b (t - t\sin^2 t) = \cos t $$ Solving these equations simultaneously for \(a\) and \(b\), $$ a = \frac{\cos t}{t} \text{ and } b = \frac{\sin t - t \cos t}{t} $$ Now we have: $$ y_p(t) = t ( \frac{\cos t}{t} e^t \sin t + \frac{\sin t - t \cos t}{t} e^t \cos t) $$

General Solution

Now, we write down the general solution as the sum of the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = A \cos t + B \sin t + t ( \frac{\cos t}{t} e^t \sin t + \frac{\sin t - t \cos t}{t} e^t \cos t) $$

Key concepts

Complementary Function

In solving a differential equation, the complementary function plays a crucial role. It's essentially the general solution of the associated homogeneous equation. For instance, in our exercise, the homogeneous form is:
\[y" + y = 0\]To solve this, we need to determine the characteristic equation, which stems from applying a substitution of the form \(y = e^{rt}\) into the homogeneous equation. This results in the equation:
\[r^2 + 1 = 0\]Solving this yields the roots \(r_1 = i\) and \(r_2 = -i\). These are complex roots, which means our complementary function will include trigonometric functions due to Euler's formula: \(e^{ix} = \cos{x} + i\sin{x}\). Therefore, the complementary function is:

• \(y_c(t) = C_1\cos{t} + C_2\sin{t}\)

Here, \(C_1\) and \(C_2\) are constants determined by initial conditions or additional information provided in a specific problem.

Particular Solution

The particular solution is another key component when dealing with inhomogeneous differential equations. This solution specifically satisfies the non-homogeneous part of the equation. In our differential equation:
\[y'' + y = e^{t}\sin{t}\]The term \(e^{t}\sin{t}\) causes the equation to be "inhomogeneous." To tackle this, we assume a form for \(y_p(t)\) that resembles the non-homogeneous part, modified with additional terms to prevent overlap with the complementary function.
A common approach, as shown in the solution, is to use:

• \(y_p(t) = t (a e^t \sin{t} + b e^t \cos{t})\)

This form considers both the exponential and trigonometric components. By substituting \(y_p(t)\) and its derivatives back into the original equation, you can equate coefficients to solve for \(a\) and \(b\). The result provides a particular solution specific to the inhomogeneous term.

Characteristic Equation

The characteristic equation is foundational when solving linear differential equations, especially for finding the complementary function. Derived from the homogeneous part of the differential equation, it transforms the equation into an algebraic problem.
In the example given, the characteristic equation arises from substituting \(y = e^{rt}\) (a solution form for constant coefficient linear differential equations) into the homogeneous equation:
\[r^2 + 1 = 0\]Here arises a quadratic form, whose solutions, \(r_1 = i\) and \(r_2 = -i\), are complex conjugates. Complex roots in the real domain signal the presence of sinusoidal components in the complementary function. This is seen when applying Euler's identity:

• \(y_c(t) = C_1\cos{t} + C_2\sin{t}\)

Thus, solving the characteristic equation is not just about finding roots but understanding the behavior of solutions in the context of oscillatory dynamics and their superposition in the final general solution.