Question

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

$y + 2y - 8y=4e^{-3x}-e^{-x}$

$y(x)=$

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 $y_c=C_1e^{-4x}+C_2e^{2x}$.

Step2: Find the particular - solution using variation of parameters

Let $y_1 = e^{-4x}$ and $y_2=e^{2x}$. The Wronskian $W=

$$\begin{vmatrix}e^{-4x}&e^{2x}\\-4e^{-4x}&2e^{2x}\end{vmatrix}$$

=2e^{-2x}+4e^{-2x}=6e^{-2x}$.
The non - homogeneous part is $f(x)=4e^{-3x}-e^{-x}$.
For $f_1(x)=4e^{-3x}$, we find $u_1'=-\frac{y_2f_1(x)}{W}=-\frac{e^{2x}\times4e^{-3x}}{6e^{-2x}}=-\frac{2}{3}e^{x}$, so $u_1=-\frac{2}{3}e^{x}$.
For $f_2(x)=-e^{-x}$, we find $u_2'=\frac{y_1f_2(x)}{W}=\frac{e^{-4x}\times(-e^{-x})}{6e^{-2x}}=-\frac{1}{6}e^{-3x}$, so $u_2=\frac{1}{18}e^{-3x}$.
The particular solution $y_p=u_1y_1 + u_2y_2=-\frac{2}{3}e^{x}\times e^{-4x}+\frac{1}{18}e^{-3x}\times e^{2x}=-\frac{2}{3}e^{-3x}+\frac{1}{18}e^{-x}$.

Step3: General solution

The general solution is $y = y_c + y_p=C_1e^{-4x}+C_2e^{2x}-\frac{2}{3}e^{-3x}+\frac{1}{18}e^{-x}$.

Step4: Apply initial conditions

Given $y(0) = 1$, we have $C_1 + C_2-\frac{2}{3}+\frac{1}{18}=1$.
Given $y'=-4C_1e^{-4x}+2C_2e^{2x}+2e^{-3x}-\frac{1}{18}e^{-x}$, and $y'(0) = 0$, we have $-4C_1+2C_2 + 2-\frac{1}{18}=0$.
Solving the system of equations

$$\begin{cases}C_1 + C_2=1+\frac{2}{3}-\frac{1}{18}=\frac{18 + 12 - 1}{18}=\frac{29}{18}\\-4C_1+2C_2=\frac{1}{18}-2=-\frac{35}{18}\end{cases}$$

Multiply the first equation by 2: $2C_1+2C_2=\frac{29}{9}$. Subtract the second equation from it: $(2C_1 + 2C_2)-(-4C_1+2C_2)=\frac{29}{9}+\frac{35}{18}$.
$6C_1=\frac{58 + 35}{18}=\frac{93}{18}=\frac{31}{6}$, so $C_1=\frac{31}{36}$.
Substitute $C_1$ into $C_1 + C_2=\frac{29}{18}$, we get $C_2=\frac{29}{18}-\frac{31}{36}=\frac{58 - 31}{36}=\frac{27}{36}=\frac{3}{4}$.

Answer:

$y=\frac{31}{36}e^{-4x}+\frac{3}{4}e^{2x}-\frac{2}{3}e^{-3x}+\frac{1}{18}e^{-x}$