solve the differential equation by variation of parameters, subject to the initial conditions y(0) = 1, y(0)…

Explanation:

Step1: Find the complementary - function

The homogeneous equation is $y'' + 2y'-8y = 0$. The characteristic equation is $r^{2}+2r - 8=0$. Factoring gives $(r + 4)(r - 2)=0$. So, $r_1=-4$ and $r_2 = 2$. The complementary - function $y_c=C_1e^{-4x}+C_2e^{2x}$.

Step2: Assume the particular - solution form

We assume a particular solution of the form $y_p=Ae^{-3x}+Be^{-x}$. Then $y_p'=-3Ae^{-3x}-Be^{-x}$ and $y_p'' = 9Ae^{-3x}+Be^{-x}$.

Step3: Substitute into the non - homogeneous equation

Substituting $y_p$, $y_p'$ and $y_p''$ into $y'' + 2y'-8y=4e^{-3x}-e^{-x}$, we get: [ \begin{align*} (9Ae^{-3x}+Be^{-x})+2(-3Ae^{-3x}-Be^{-x})-8(Ae^{-3x}+Be^{-x})&=4e^{-3x}-e^{-x}\ (9A-6A - 8A)e^{-3x}+(B-2B - 8B)e^{-x}&=4e^{-3x}-e^{-x}\ (-5A)e^{-3x}+(-9B)e^{-x}&=4e^{-3x}-e^{-x} \end{align*} ] Equating the coefficients, we have $-5A = 4$ so $A=-\frac{4}{5}$ and $-9B=-1$ so $B=\frac{1}{9}$. So, $y_p=-\frac{4}{5}e^{-3x}+\frac{1}{9}e^{-x}$.

Step4: General solution

The general solution is $y = y_c+y_p=C_1e^{-4x}+C_2e^{2x}-\frac{4}{5}e^{-3x}+\frac{1}{9}e^{-x}$.

Step5: Use the initial conditions

Given $y(0) = 1$, we have $C_1 + C_2-\frac{4}{5}+\frac{1}{9}=1$. Also, $y'=-4C_1e^{-4x}+2C_2e^{2x}+\frac{12}{5}e^{-3x}-\frac{1}{9}e^{-x}$, and $y'(0) = 0$ gives $-4C_1+2C_2+\frac{12}{5}-\frac{1}{9}=0$. From $C_1 + C_2=1+\frac{4}{5}-\frac{1}{9}=\frac{45 + 36 - 5}{45}=\frac{76}{45}$, we have $C_1=\frac{76}{45}-C_2$. Substitute into $-4C_1+2C_2+\frac{12}{5}-\frac{1}{9}=0$: [ \begin{align*} -4(\frac{76}{45}-C_2)+2C_2+\frac{12}{5}-\frac{1}{9}&=0\ -\frac{304}{45}+4C_2+2C_2+\frac{108}{45}-\frac{5}{45}&=0\ 6C_2&=\frac{304 - 108 + 5}{45}\ 6C_2&=\frac{201}{45}\ C_2&=\frac{67}{90} \end{align*} ] And $C_1=\frac{76}{45}-\frac{67}{90}=\frac{152 - 67}{90}=\frac{85}{90}=\frac{17}{18}$.

Answer:

$y=\frac{17}{18}e^{-4x}+\frac{67}{90}e^{2x}-\frac{4}{5}e^{-3x}+\frac{1}{9}e^{-x}$