Question

Find the general solution of the given differential equation. $$ 2 y^{\prime \prime}-3 y^{\prime}+y=0 $$

Short answer

Answer: The general solution of the given differential equation is \(y(t) = c_1 + c_2 e^{-2t} - \frac{1}{4}\sin(2t) + \frac{1}{8}\cos(2t)\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Find the complementary function

We first find the complementary function by solving the homogeneous equation associated with the given inhomogeneous equation: $$ y^{\prime\prime}+2y^{\prime}=0 $$ To solve this, assume a solution in the form of \(y=e^{rt}\). Substituting this into the equation gives: $$ r^2e^{rt} + 2re^{rt} = 0 $$ Divide by \(e^{rt}\) to get the characteristic equation: $$ r^2 + 2r = 0 $$ Factor the left side of the equation: $$ r(r + 2) = 0 $$ So, the roots are \(r_1 = 0\) and \(r_2 = -2\). Thus, the complementary function is given by the sum of the linearly independent solutions: $$ y_c(t) = c_1 e^{0t} + c_2 e^{-2t} = c_1 + c_2 e^{-2t} $$ where \(c_1\) and \(c_2\) are constants.

Find a particular solution

Next, we find a particular solution for the given inhomogeneous equation using the method of undetermined coefficients. Given the right-hand side consists of a constant term and a sinusoidal term, we assume a particular solution of the form: $$ y_p(t) = A + B\sin(2t) + C\cos(2t) $$ Now we find the first and second derivatives of \(y_p(t)\): $$ y_p^{\prime}(t) = 2B\cos(2t) - 2C\sin(2t) $$ $$ y_p^{\prime\prime}(t) = -4B\sin(2t) - 4C\cos(2t) $$ Substitute \(y_p(t)\), \(y_p^{\prime}(t)\), and \(y_p^{\prime\prime}(t)\) into the given inhomogeneous differential equation: $$ (-4B\sin(2t) - 4C\cos(2t)) + 2(2B\cos(2t) - 2C\sin(2t)) = 3 + 4\sin(2t) $$ By comparing the coefficients of the sinusoidal terms, we obtain: $$ -4B - 4C = 3 \quad (1) $$ $$ 4C - 4B = 4 \quad (2) $$ Solving the linear system, we find \(A = 0\), \(B = -\frac{1}{4}\), and \(C = \frac{1}{8}\). Thus, the particular solution is: $$ y_p(t) = -\frac{1}{4}\sin(2t) + \frac{1}{8}\cos(2t) $$

Combine complementary function and particular solution

Finally, we combine the complementary function \(y_c(t)\) and the particular solution \(y_p(t)\) to form the general solution: $$ y(t) = y_c(t) + y_p(t) = c_1 + c_2 e^{-2t} - \frac{1}{4}\sin(2t) + \frac{1}{8}\cos(2t) $$ This is the general solution of the given differential equation.

Key concepts

Complementary Function

When you're dealing with differential equations, finding the complementary function is a crucial step. To do this, first focus on the homogeneous version of your differential equation. This means you look at the equation without the non-homogeneous part.
For our example, the homogeneous equation would be:

• \(y'' + 2y' = 0\)

Solving this requires assuming the solution has the form \(y = e^{rt}\), where \(r\) is a constant to be determined. Plugging this into the homogeneous equation gives the characteristic equation, a polynomial whose roots help us find the complementary function. Here, this simplifies to:

• \(r^2 + 2r = 0\)

The roots of this equation (

• \(r_1 = 0\), \(r_2 = -2\)

) provide the fundamental solutions: constant functions and exponential functions. Therefore, the complementary function is:

• \(y_c(t) = c_1 + c_2 e^{-2t}\)

where \(c_1\) and \(c_2\) are arbitrary constants reflecting the general nature of the solution.

Particular Solution

Finding a particular solution involves addressing the inhomogeneous part of your differential equation. This part involves coefficients that aren't solved by the homogeneous equation alone. In our example, the non-homogeneous part is \(3 + 4\sin{2t}\).
We use the method of undetermined coefficients here. Assume a form for the particular solution based on the right-hand side of the equation. For trigonometric functions like sine and cosine, you can propose:

• \(y_p(t) = A + B\sin{2t} + C\cos{2t}\)

Next, find the derivatives \(y'_p(t)\) and \(y''_p(t)\). Upon substituting these into the differential equation, compare coefficients from both sides to determine \(A, B, C\). In this example, solving for \(A = 0\), \(B = -\frac{1}{4}\), and \(C = \frac{1}{8}\) gave us:

• \(y_p(t) = -\frac{1}{4}\sin{2t} + \frac{1}{8}\cos{2t}\)

This gives us a specific solution to the non-homogeneous part, but still needs to be combined with the complementary solution for the full equation.

Undetermined Coefficients

The method of undetermined coefficients is like baking a cake where you have a basic recipe but adjust the ingredients until it tastes just right. It's a systematic guesswork applied to differential equations.
For equations with functions like exponentials, polynomials, or trigonometric functions on the right-hand side, this method presumes a form for the particular solution mirroring this. For instance:

• If the right side is a polynomial, guess a polynomial of that degree.
• If it involves \(\sin\) or \(\cos\), use trigonometric terms with coefficients \(B\) and \(C\).

The trick is consistency: this assumed solution is differentiated and substituted back into the original differential equation. Then, balance the equation by adjusting the coefficients until the left-hand side equals the right-hand side. This method only works well if your differential equation fits certain forms, meaning it's quite specialized yet powerful when applicable.
In our example, undetermined coefficients led us through aligning terms systematically, comparing them, and solving the eventual system of equations efficiently, proffering values of \(A\), \(B\), and \(C\) for a complete solution.