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 the roots are $r_1=-4$ and $r_2 = 2$. The complementary - function $y_c(x)=C_1e^{-4x}+C_2e^{2x}$.

Step2: Use the method of variation of parameters

We assume a particular solution of the form $y_p(x)=u_1(x)e^{-4x}+u_2(x)e^{2x}$. We have the following system of equations for $u_1'(x)$ and $u_2'(x)$: $\begin{cases}u_1'e^{-4x}+u_2'e^{2x}=0\-4u_1'e^{-4x}+2u_2'e^{2x}=5e^{-2x}-e^{-x}\end{cases}$ From the first equation, $u_2'=-u_1'e^{-6x}$. Substitute into the second equation: $-4u_1'e^{-4x}+2(-u_1'e^{-6x})e^{2x}=5e^{-2x}-e^{-x}$ $-4u_1'e^{-4x}-2u_1'e^{-4x}=5e^{-2x}-e^{-x}$ $-6u_1'e^{-4x}=5e^{-2x}-e^{-x}$ $u_1'=-\frac{5}{6}e^{2x}+\frac{1}{6}e^{3x}$ Integrating, $u_1(x)=-\frac{5}{12}e^{2x}+\frac{1}{18}e^{3x}+k_1$ From $u_2'=-u_1'e^{-6x}$, we have $u_2'=\frac{5}{6}e^{-4x}-\frac{1}{6}e^{-3x}$ Integrating, $u_2(x)=-\frac{5}{24}e^{-4x}+\frac{1}{18}e^{-3x}+k_2$ The particular solution $y_p(x)=u_1(x)e^{-4x}+u_2(x)e^{2x}$ $y_p(x)=(-\frac{5}{12}e^{2x}+\frac{1}{18}e^{3x})e^{-4x}+(-\frac{5}{24}e^{-4x}+\frac{1}{18}e^{-3x})e^{2x}$ $y_p(x)=-\frac{5}{12}e^{-2x}+\frac{1}{18}e^{-x}-\frac{5}{24}e^{-2x}+\frac{1}{18}e^{-x}$ $y_p(x)=-\frac{5}{8}e^{-2x}+\frac{1}{9}e^{-x}$ The general solution is $y(x)=y_c(x)+y_p(x)=C_1e^{-4x}+C_2e^{2x}-\frac{5}{8}e^{-2x}+\frac{1}{9}e^{-x}$

Step3: Apply the initial conditions

Given $y(0) = 1$, we have $C_1 + C_2-\frac{5}{8}+\frac{1}{9}=1$ $C_1 + C_2=1+\frac{5}{8}-\frac{1}{9}=\frac{72 + 45-8}{72}=\frac{109}{72}$ $y'(x)=-4C_1e^{-4x}+2C_2e^{2x}+\frac{5}{4}e^{-2x}-\frac{1}{9}e^{-x}$ Given $y'(0) = 0$, we have $-4C_1 + 2C_2+\frac{5}{4}-\frac{1}{9}=0$ $-4C_1 + 2C_2=-\frac{5}{4}+\frac{1}{9}=-\frac{45 - 4}{36}=-\frac{41}{36}$ Solving the system $\begin{cases}C_1 + C_2=\frac{109}{72}\-4C_1 + 2C_2=-\frac{41}{36}\end{cases}$ Multiply the first equation by 2: $2C_1+2C_2=\frac{109}{36}$ Subtract the second equation from it: $(2C_1 + 2C_2)-(-4C_1 + 2C_2)=\frac{109}{36}+\frac{41}{36}$ $6C_1=\frac{109 + 41}{36}=\frac{150}{36}=\frac{25}{6}$ $C_1=\frac{25}{36}$ $C_2=\frac{109}{72}-\frac{25}{36}=\frac{109 - 50}{72}=\frac{59}{72}$

Answer:

$y(x)=\frac{25}{36}e^{-4x}+\frac{59}{72}e^{2x}-\frac{5}{8}e^{-2x}+\frac{1}{9}e^{-x}$