91Ó°ÊÓ

First, we need to find the complementary function by solving the homogeneous equation: $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 $$ Write down the characteristic equation: $$ r^2 + 2r - 3 = 0 $$ Now, we need to find the roots of the characteristic equation: $$ r_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ Substitute with values: $$ r_{1,2} = \frac{-2\pm\sqrt{2^2-4(1)(-3)}}{2(1)} $$ After simplification, we get: $$ r_{1,2} = -1 \pm 2 $$ So, the roots are: $$ r_1 = 1, \quad r_2 = -3 $$ Hence, the complementary function is given by: $$ y_c = c_1 e^{x} + c_2 e^{-3x} $$

Now, we will find a particular solution for the given non-homogeneous equation: $$ y^{\prime \prime}+2 y^{\prime}-3 y=-16 x e^{x} $$ We will guess a particular solution of the form: $$ y_p = (Ax^2+Bx+C)e^x $$ Let's find the first derivative, \(y_p'\): $$ y_p' = (2Ax+B)xe^x + (Ax^2+Bx+C)e^x $$ And now the second derivative, \(y_p''\): $$ y_p'' = (2A)xe^x + (2Ax+B)e^x + (2Ax+B)xe^x + (Ax^2+Bx+C)e^x $$ Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the non-homogeneous equation: $$ (2A)xe^x + (2Ax+B)e^x + (2Ax+B)xe^x + (Ax^2+Bx+C)e^x + 2(2Ax+B)xe^x + 2(Ax^2+Bx+C)e^x - 3(Ax^2+Bx+C)e^x = -16xe^x $$ Simplify and collect the coefficients for \(e^x\): $$ (2A + 4A - 3A)x^2+(2B + 2B - 3B)x + (2A+2B+2C)x + (Ax^2 + Bx + C)e^x = -16xe^x $$ $$ x^2e^x - 14xe^x + (2A+2B+2C)e^x = -16xe^x $$ Comparing the coefficients, we get the following system of equations: $$ A = 0 \\ -14 = -16 + 2B \\ 2A+2B+2C = 0 $$ Solving the above system gives: $$ A=0, \quad B=1, \quad C=-1 $$ So, our particular solution is: $$ y_p = (x-1)e^x $$

We have found the complementary function, \(y_c\), and a particular solution, \(y_p\), for the given non-homogeneous equation. The general solution is given by the sum of these two: $$ y(x) = y_c + y_p = (c_1 e^{x} + c_2 e^{-3x}) + (x-1)e^x $$ Therefore, the general solution is: $$ y(x) = c_1 e^{x} + c_2 e^{-3x} + (x-1)e^x $$