Solve \( \left(D^{2}-4 D+3\right) y=\sin 3 x \cos 2 x \)

We start with the equation \[ (D^2-4D+3)y=\sin 3x\cos 2x. \] This is equivalent to \[ y''-4y'+3y=\sin3x\cos2x. \] ### Step 1. Rewrite the Right‐Hand Side Using the trigonometric identity \[ \sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)], \] with \(A=3x\) and \(B=2x\), we have \[ \sin3x\cos2x=\frac{1}{2}\bigl[\sin(3x+2x)+\sin(3x-2x)\bigr]=\frac{1}{2}\bigl[\sin5x+\sin x\bigr]. \] The differential equation becomes \[ y''-4y'+3y=\frac{1}{2}\sin5x+\frac{1}{2}\sin x. \] ### Step 2. Solve the Homogeneous Equation The homogeneous equation is \[ y''-4y'+3y=0. \] Its characteristic equation is \[ r^2-4r+3=0. \] Factoring, \[ (r-1)(r-3)=0, \] so the roots are \(r=1\) and \(r=3\). Therefore, the general solution of the homogeneous equation is \[ y_h = C_1e^x+C_2e^{3x}, \] where \(C_1\) and \(C_2\) are arbitrary constants. ### Step 3. Find a Particular Solution We look for a particular solution of the form \[ y_p=A\sin5x+B\cos5x+C\sin x+D\cos x, \] where \(A\), \(B\), \(C\), and \(D\) are constants to be determined. #### 3.1 Compute Derivatives Differentiate \(y_p\): \[ y_p = A\sin5x+B\cos5x+C\sin x+D\cos x, \] \[ y_p' = 5A\cos5x-5B\sin5x+C\cos x-D\sin x, \] \[ y_p'' = -25A\sin5x-25B\cos5x-C\sin x-D\cos x. \] #### 3.2 Substitute into the Differential Operator Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the left-hand side \(y''-4y'+3y\): \[ y_p''-4y_p'+3y_p = \bigl[-25A\sin5x-25B\cos5x-C\sin x-D\cos x\bigr] \] \[ -4\bigl[5A\cos5x-5B\sin5x+C\cos x-D\sin x\bigr] +3\bigl[A\sin5x+B\cos5x+C\sin x+D\cos x\bigr]. \] Group the coefficients for like terms. **For \(\sin5x\):** \[ -25A + 20B + 3A = -22A+20B. \] **For \(\cos5x\):** \[ -25B -20A + 3B = -20A-22B. \] **For \(\sin x\):** \[ -C + 4D + 3C = 2C+4D. \] **For \(\cos x\):** \[ -D -4C + 3D = -4C+2D. \] Thus, the left-hand side becomes \[ y_p''-4y_p'+3y_p = \bigl(-22A+20B\bigr)\sin5x+\bigl(-20A-22B\bigr)\cos5x+\bigl(2C+4D\bigr)\sin x+\bigl(-4C+2D\bigr)\cos x. \] This must equal the right-hand side, \[ \frac{1}{2}\sin5x + \frac{1}{2}\sin x. \] This gives us the system of equations: \[ -22A+20B=\frac{1}{2}, \quad -20A-22B=0, \quad 2C+4D=\frac{1}{2}, \quad -4C+2D=0. \] #### 3.3 Solve for \(A\) and \(B\) From \[ -20A-22B=0 \quad\Longrightarrow\quad 20A+22B=0 \quad\Longrightarrow\quad 10A+11B=0, \] so \[ B=-\frac{10}{11}A. \] Substitute into the first equation: \[ -22A+20\Bigl(-\frac{10}{11}A\Bigr)= -22A-\frac{200}{11}A=-\frac{242}{11}A=\frac{1}{2}. \] Thus, \[ A=-\frac{1}{2}\cdot\frac{11}{242}=-\frac{11}{484}=-\frac{1}{44} \quad (\text{since } \frac{11}{484}=\frac{1}{44}). \] Then, \[ B=-\frac{10}{11}\Bigl(-\frac{1}{44}\Bigr)=\frac{10}{484}=\frac{5}{242}. \] #### 3.4 Solve for \(C\) and \(D\) From \[ -4C+2D=0 \quad\Longrightarrow\quad 2D=4C \quad\Longrightarrow\quad D=2C. \] Substitute into \(2C+4D=\frac{1}{2}\): \[ 2C+4(2C)=2C+8C=10C=\frac{1}{2} \quad\Longrightarrow\quad C=\frac{1}{20}. \] Then, \[ D=2C=\frac{1}{10}. \] Thus, the particular solution is \[ y_p=-\frac{1}{44}\sin5x+\frac{5}{242}\cos5x+\frac{1}{20}\sin x+\frac{1}{10}\cos x. \] ### Step 4. Write the General Solution The general solution to the differential equation is \[ y = y_h+y_p = C_1e^x+C_2e^{3x} -\frac{1}{44}\sin5x+\frac{5}{242}\cos5x +\frac{1}{20}\sin x+\frac{1}{10}\cos x. \]