3. \( \quad \frac{d^{2} y}{d x^{2}}+2 y=x^{3}+x^{2}+e^{-2 x}+\cos 3 x \)

Let's solve the differential equation \( \frac{d^{2} y}{d x^{2}}+2 y=x^{3}+x^{2}+e^{-2 x}+\cos 3 x \). First, we need to find the complementary solution \( y_c \) by solving the associated homogeneous equation: \[ \frac{d^{2} y}{d x^{2}} + 2y = 0. \] The characteristic equation is: \[ r^{2} + 2 = 0 \Rightarrow r = \pm i\sqrt{2}. \] Thus, the complementary solution is: \[ y_c = C_1 \cos(\sqrt{2} x) + C_2 \sin(\sqrt{2} x), \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions. Next, we find a particular solution \( y_p \) for the non-homogeneous part \( x^{3}+x^{2}+e^{-2 x}+\cos 3 x \). We can use the method of undetermined coefficients. 1. For \( x^{3}+x^{2} \), we can try a polynomial of the form \( Ax^3 + Bx^2 + Cx + D \). 2. For \( e^{-2x} \), we try \( Ke^{-2x} \). 3. For \( \cos(3x) \), we try \( A\cos(3x) + B\sin(3x) \). Combining these, the trial solution is: \[ y_p = Ax^3 + Bx^2 + Cx + D + Ke^{-2x} + A\cos(3x) + B\sin(3x). \] Next, we substitute \( y_p \) into the original equation and equate coefficients for the different powers of \( x \), \( e^{-2x} \), and the trigonometric terms. This will give us a system of equations to solve for \( A, B, C, D, K \). Once we compute those coefficients and solve for them, the general solution will then be: \[ y = y_c + y_p = C_1 \cos(\sqrt{2} x) + C_2 \sin(\sqrt{2} x) + y_p, \] where \( y_p \) will include the specific polynomial, exponential, and harmonic components from our particular solution process.