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 $36y'' - y=0$. The characteristic equation is $36r^{2}-1 = 0$. Factoring gives $(6r - 1)(6r+1)=0$. So the roots are $r_1=\frac{1}{6}$ and $r_2 =-\frac{1}{6}$. The complementary function $y_c = C_1e^{\frac{x}{6}}+C_2e^{-\frac{x}{6}}$.

Step2: Use variation of parameters

We assume a particular solution of the form $y_p=u_1(x)e^{\frac{x}{6}}+u_2(x)e^{-\frac{x}{6}}$. We have the following system of equations for $u_1'$ and $u_2'$: $u_1'e^{\frac{x}{6}}+u_2'e^{-\frac{x}{6}} = 0$ and $\frac{1}{6}u_1'e^{\frac{x}{6}}-\frac{1}{6}u_2'e^{-\frac{x}{6}}=\frac{1}{36}xe^{\frac{x}{6}}$. From the first equation $u_2'=-u_1'e^{\frac{x}{3}}$. Substituting into the second - equation: $\frac{1}{6}u_1'e^{\frac{x}{6}}+\frac{1}{6}u_1'e^{\frac{x}{6}}=\frac{1}{36}xe^{\frac{x}{6}}$. $\frac{1}{3}u_1'e^{\frac{x}{6}}=\frac{1}{36}xe^{\frac{x}{6}}$, so $u_1'=\frac{1}{12}x$ and $u_1=\frac{1}{24}x^{2}$. Since $u_2'=-u_1'e^{\frac{x}{3}}=-\frac{1}{12}xe^{\frac{x}{3}}$, integrating by parts $u_2=-\frac{1}{4}(x - 3)e^{\frac{x}{3}}$. The particular solution $y_p=\frac{1}{24}x^{2}e^{\frac{x}{6}}-\frac{1}{4}(x - 3)e^{\frac{x}{3}}e^{-\frac{x}{6}}=\frac{1}{24}x^{2}e^{\frac{x}{6}}-\frac{1}{4}(x - 3)e^{\frac{x}{6}}=e^{\frac{x}{6}}(\frac{1}{24}x^{2}-\frac{1}{4}x+\frac{3}{4})$. The general solution is $y = y_c + y_p=C_1e^{\frac{x}{6}}+C_2e^{-\frac{x}{6}}+e^{\frac{x}{6}}(\frac{1}{24}x^{2}-\frac{1}{4}x+\frac{3}{4})$.

Step3: Apply initial conditions

Given $y(0) = 1$, we have $y(0)=C_1 + C_2+\frac{3}{4}=1$, so $C_1 + C_2=\frac{1}{4}$. Differentiating $y$: $y'=\frac{1}{6}C_1e^{\frac{x}{6}}-\frac{1}{6}C_2e^{-\frac{x}{6}}+\frac{1}{6}e^{\frac{x}{6}}(\frac{1}{24}x^{2}-\frac{1}{4}x+\frac{3}{4})+e^{\frac{x}{6}}(\frac{1}{12}x-\frac{1}{4})$. Given $y'(0) = 0$, we have $\frac{1}{6}C_1-\frac{1}{6}C_2+\frac{1}{8}-\frac{1}{4}=0$, so $\frac{1}{6}C_1-\frac{1}{6}C_2=\frac{1}{8}$, or $C_1 - C_2=\frac{3}{4}$. Solving the system $\begin{cases}C_1 + C_2=\frac{1}{4}\C_1 - C_2=\frac{3}{4}\end{cases}$ gives $C_1=\frac{1}{2}$ and $C_2=-\frac{1}{4}$.

Answer:

$y=\frac{1}{2}e^{\frac{x}{6}}-\frac{1}{4}e^{-\frac{x}{6}}+e^{\frac{x}{6}}(\frac{1}{24}x^{2}-\frac{1}{4}x+\frac{3}{4})$