solve the differential equation by variation of parameters. 5y - 10y + 10y = e^x sec(x)

Explanation:

Step1: Find the complementary - function

The homogeneous equation is $5y''-10y'+10y = 0$. The characteristic equation is $5r^{2}-10r + 10=0$, or $r^{2}-2r + 2=0$. Using the quadratic formula $r=\frac{2\pm\sqrt{4 - 8}}{2}=1\pm i$. So the complementary - function $y_c(x)=e^{x}(C_1\cos(x)+C_2\sin(x))$.

Step2: Assume the particular - solution form

We use the method of variation of parameters. Assume $y_p(x)=u_1(x)e^{x}\cos(x)+u_2(x)e^{x}\sin(x)$. We have $y_p'=u_1'(e^{x}\cos(x)-e^{x}\sin(x))+u_2'(e^{x}\sin(x)+e^{x}\cos(x))+u_1e^{x}(\cos(x)-\sin(x))+u_2e^{x}(\sin(x)+\cos(x))$. Set $u_1'(e^{x}\cos(x)-e^{x}\sin(x))+u_2'(e^{x}\sin(x)+e^{x}\cos(x)) = 0$. Then $y_p''=u_1'(e^{x}(\cos(x)-\sin(x))-e^{x}(\cos(x)+\sin(x)))+u_2'(e^{x}(\sin(x)+\cos(x))+e^{x}(\cos(x)-\sin(x)))+ \text{(terms involving }u_1\text{ and }u_2\text{)}$. Substitute $y_p$, $y_p'$ and $y_p''$ into the non - homogeneous equation $5y''-10y'+10y = e^{x}\sec(x)$. After simplification, we get the system of equations for $u_1'$ and $u_2'$: $\begin{cases}u_1'(e^{x}\cos(x)-e^{x}\sin(x))+u_2'(e^{x}\sin(x)+e^{x}\cos(x)) = 0\u_1'(- 2e^{x}\sin(x))+u_2'(2e^{x}\cos(x))=\frac{e^{x}\sec(x)}{5}\end{cases}$ Solve this system for $u_1'$ and $u_2'$: From the first equation $u_2'=\frac{(e^{x}\sin(x)-e^{x}\cos(x))u_1'}{e^{x}(\sin(x)+\cos(x))}$. Substitute into the second equation: $u_1'(-2e^{x}\sin(x))+\frac{(e^{x}\sin(x)-e^{x}\cos(x))u_1'}{e^{x}(\sin(x)+\cos(x))}\times(2e^{x}\cos(x))=\frac{e^{x}\sec(x)}{5}$ $u_1'=\frac{-\sin(x)}{5}$ and $u_2'=\frac{\cos(x)}{5}$. Integrate to find $u_1$ and $u_2$: $u_1=\frac{\cos(x)}{5}+K_1$ and $u_2=\frac{\sin(x)}{5}+K_2$. The particular solution $y_p(x)=\frac{1}{5}e^{x}\cos(x)\cos(x)+\frac{1}{5}e^{x}\sin(x)\sin(x)=\frac{1}{5}e^{x}$.

Step3: Find the general solution

The general solution $y(x)=y_c(x)+y_p(x)=e^{x}(C_1\cos(x)+C_2\sin(x))+\frac{1}{5}e^{x}$.

Answer:

$y(x)=e^{x}(C_1\cos(x)+C_2\sin(x))+\frac{1}{5}e^{x}$