Question

solve the equation by the method of undetermined coefficients.
$y - 2y = \sin x$

Explanation:

Step1: Solve the homogeneous equation

First, solve the homogeneous differential equation \( y'' - 2y' = 0 \). The characteristic equation is \( r^2 - 2r = 0 \), which factors as \( r(r - 2)=0 \). So the roots are \( r_1 = 0 \) and \( r_2 = 2 \). The general solution of the homogeneous equation, \( y_h \), is \( y_h = C_1 + C_2 e^{2x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Step2: Find a particular solution

We need to find a particular solution \( y_p \) of the non - homogeneous equation \( y'' - 2y'=\sin x \). Since the non - homogeneous term is \( \sin x \), we assume a particular solution of the form \( y_p = A\cos x + B\sin x \), where \( A \) and \( B \) are undetermined coefficients.

First, find the first and second derivatives of \( y_p \):

• The first derivative \( y_p'=-A\sin x + B\cos x \)
• The second derivative \( y_p''=-A\cos x - B\sin x \)

Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the non - homogeneous equation \( y'' - 2y'=\sin x \):

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Now, we equate the coefficients of \( \cos x \) and \( \sin x \) on both sides of the equation.

For the coefficient of \( \cos x \): \( -A - 2B = 0 \)

For the coefficient of \( \sin x \): \( 2A - B = 1 \)

We solve the system of equations:
From the first equation \( -A - 2B = 0 \), we can express \( A=-2B \).

Substitute \( A = - 2B \) into the second equation \( 2A - B = 1 \):

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Then, substitute \( B = -\frac{1}{5} \) into \( A=-2B \), we get \( A=-2\times(-\frac{1}{5})=\frac{2}{5} \)

So the particular solution \( y_p=\frac{2}{5}\cos x-\frac{1}{5}\sin x \)

Step3: General solution of the non - homogeneous equation

The general solution of the non - homogeneous differential equation \( y'' - 2y'=\sin x \) is the sum of the general solution of the homogeneous equation and a particular solution of the non - homogeneous equation.

So \( y = y_h + y_p=C_1 + C_2 e^{2x}+\frac{2}{5}\cos x-\frac{1}{5}\sin x \)

Answer:

The general solution of the differential equation \( y'' - 2y'=\sin x \) is \( y = C_1 + C_2 e^{2x}+\frac{2}{5}\cos x-\frac{1}{5}\sin x \), where \( C_1 \) and \( C_2 \) are arbitrary constants.