91Ó°ÊÓ

To find the complementary function, we need to solve the corresponding homogeneous equation: $$y^{\prime \prime}+4 y=0.$$ This is a linear differential equation with constant coefficients, so we use the characteristic equation to find the solution: $$r^2 + 4 = 0.$$ Solving for r, we get complex roots: $$r = \pm 2i.$$ Thus, our complementary function will be in the form of: $$y_c(x)=c_1 \cos{2x} + c_2\sin{2x}.$$

Our given nonhomogeneous equation was: $$y^{\prime \prime}+4 y=7 e^{x}.$$ To annihilate \(7e^x\), we will integrate it by applying the following annihilator: $$D - 1,$$ where \(D\) is the differentiation operator. So, applying it to \(7e^x\), $$(D - 1)(7e^x) = 7e^x - 7e^x = 0,$$ we get a homogeneous equation. Now apply the same annihilator to the given differential equation: $$(D - 1)(y^{\prime \prime}+4 y - 7e^x) = y^{\prime \prime \prime} + 4y^{\prime} - y^{\prime \prime} - 4y$$

Now we have the following annihilating equation to solve: $$y^{\prime \prime \prime} + 4y^{\prime} - y^{\prime \prime} - 4y=0.$$ To solve this differential equation, we follow the same procedure used to find the complementary function of the given equation. The characteristic equation is obtained as: $$r^3 + 4r - r^2 - 4 = 0.$$ Factoring out \(r\), we get \((r)(r^2 - r + 4) = 0\). Solving it gives us the roots: \(0, 1 \pm 2i\). Hence, the general solution for annihilating equation is: $$y(x) = A + B\cos{2x} + C\sin{2x}.$$

Combining the complementary and particular solutions, we get: $$y(x) = A + c_1 \cos{2x} + c_2\sin{2x}.$$ We can see that the functions corresponding to \(c_1 \cos{2x}\) and \(c_2\sin{2x}\) will be annihilated by the given equation since they are the complementary functions. So, the general solution will be: $$y(x) = A + c_1 \cos{2x} + c_2\sin{2x}.$$

We need to plug the general solution into the given differential equation and find the particular solution. $$y^{\prime \prime}+4 y=7 e^{x}.$$ Plugging in our general solution, we get: $$(-4A + 4c_1) \cos{2x} + (-4A + 4c_2)\sin{2x} = 7 e^{x}.$$ However, this equation is not valid since the left side and right side have different functions. To make it valid, the coefficients of \(\cos{2x}\) and \(\sin{2x}\) on the left side must be equal to zero. Thus, $$4A - 4c_1= 0 \Rightarrow A = c_1,$$ $$4A - 4c_2 = 0 \Rightarrow A = c_2.$$ So, our general solution becomes: $$y(x) = A(1 + \cos{2x} +\sin{2x}).$$ Now, comparing this general solution to the given equation, we can eliminate both \(A\cos{2x}\) and \(A\sin{2x}\) terms as they are complementary parts. Hence, the particular solution is: $$y_p(x) = A.$$