Math / Calculus

QuestionFind the the solution of the differential equations
1. $y^{\prime \prime}+y=\cot (x)$

Studdy Solution

STEP 1

Assumptions
1. The differential equation is $y^{\prime \prime} + y = \cot(x)$ .
2. We are looking for the general solution of this second-order non-homogeneous differential equation.
3. The solution will be composed of the general solution to the corresponding homogeneous equation and a particular solution to the non-homogeneous equation.

STEP 2

First, solve the corresponding homogeneous equation: $y^{\prime \prime} + y = 0$

STEP 3

The characteristic equation for the homogeneous differential equation is: $r^2 + 1 = 0$

STEP 4

Solve the characteristic equation: $r^2 = -1$ $r = \pm i$

STEP 5

The general solution to the homogeneous equation is: $y_h = C_1 \cos(x) + C_2 \sin(x)$ where $C_1$ and $C_2$ are arbitrary constants.

STEP 6

Next, find a particular solution $y_p$ to the non-homogeneous equation $y^{\prime \prime} + y = \cot(x)$ .

STEP 7

Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$ , we will use the method of undetermined coefficients or variation of parameters. In this case, variation of parameters is more suitable due to the nature of $\cot(x)$ .

STEP 8

Assume a particular solution of the form: $y_p = u_1(x) \cos(x) + u_2(x) \sin(x)$ where $u_1(x)$ and $u_2(x)$ are functions to be determined.

STEP 9

Compute the first derivatives of $y_p$ : $y_p' = u_1' \cos(x) - u_1 \sin(x) + u_2' \sin(x) + u_2 \cos(x)$

STEP 10

Compute the second derivatives of $y_p$ : $y_p'' = (u_1'' \cos(x) - 2u_1' \sin(x) - u_1 \cos(x)) + (u_2'' \sin(x) + 2u_2' \cos(x) - u_2 \sin(x))$

STEP 11

Substitute $y_p$ and its derivatives into the original differential equation: $y_p'' + y_p = \cot(x)$

STEP 12

Simplify the equation by combining like terms: $(u_1'' \cos(x) - 2u_1' \sin(x) - u_1 \cos(x)) + (u_2'' \sin(x) + 2u_2' \cos(x) - u_2 \sin(x)) + u_1 \cos(x) + u_2 \sin(x) = \cot(x)$

STEP 13

Group terms involving $\cos(x)$ and $\sin(x)$ : $(u_1'' \cos(x) - 2u_1' \sin(x) + u_2'' \sin(x) + 2u_2' \cos(x)) = \cot(x)$

STEP 14

To simplify, set up the system of equations: $u_1'' \cos(x) + u_2'' \sin(x) = 0$ $-2u_1' \sin(x) + 2u_2' \cos(x) = \cot(x)$

STEP 15

Solve the first equation for $u_1''$ and $u_2''$ : $u_1'' \cos(x) + u_2'' \sin(x) = 0$

STEP 16

Solve the second equation for $u_1'$ and $u_2'$ : $-2u_1' \sin(x) + 2u_2' \cos(x) = \cot(x)$

STEP 17

Divide the second equation by 2: $-u_1' \sin(x) + u_2' \cos(x) = \frac{1}{2} \cot(x)$

STEP 18

Use the Wronskian $W = \cos^2(x) + \sin^2(x) = 1$ to solve for $u_1'$ and $u_2'$ : $u_1' = -\frac{\sin(x) \cot(x)}{2} = -\frac{\cos(x)}{2}$ $u_2' = \frac{\cos(x) \cot(x)}{2} = \frac{1}{2}$

STEP 19

Integrate $u_1'$ and $u_2'$ to find $u_1(x)$ and $u_2(x)$ : $u_1(x) = -\frac{1}{2} \int \cos(x) \, dx = -\frac{1}{2} \sin(x)$ $u_2(x) = \frac{1}{2} \int 1 \, dx = \frac{x}{2}$

STEP 20

Substitute $u_1(x)$ and $u_2(x)$ back into the particular solution: $y_p = u_1(x) \cos(x) + u_2(x) \sin(x)$ $y_p = -\frac{1}{2} \sin(x) \cos(x) + \frac{x}{2} \sin(x)$

STEP 21

Simplify the particular solution: $y_p = \frac{x}{2} \sin(x) - \frac{1}{4} \sin(2x)$

STEP 22

Combine the general solution of the homogeneous equation and the particular solution to get the general solution of the original equation: $y = y_h + y_p$ $y = C_1 \cos(x) + C_2 \sin(x) + \frac{x}{2} \sin(x) - \frac{1}{4} \sin(2x)$

STEP 23

The general solution to the differential equation $y^{\prime \prime} + y = \cot(x)$ is: $y = C_1 \cos(x) + C_2 \sin(x) + \frac{x}{2} \sin(x) - \frac{1}{4} \sin(2x)$

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