Answer:
[tex]y(x)=3e^{\frac{x}{6}}+\frac{12}{7}xe^{\frac{x}{6}}[/tex] (See attached graph)
Step-by-step explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation [tex]am^2+bm+c=0[/tex] where the values of [tex]m[/tex] are the roots:
[tex]36y''-12y'+y=0\\\\36m^2-12m+1=0\\\\(6m-1)^2=0\\\\6m-1=0\\\\6m=1\\\\m=\frac{1}{6}[/tex]
Since the values of [tex]m[/tex] are equal real roots, then the general solution is [tex]y(x)=C_1e^{m_1x}+C_2xe^{m_1x}[/tex].
Thus, the general solution for our given differential equation is [tex]y(x)=C_1e^{\frac{x}{6}}+C_2xe^{\frac{x}{6}}[/tex].
To account for both initial conditions, take the derivative of [tex]y(x)[/tex], thus, [tex]y'(x)=\frac{C_1}{6}e^{\frac{x}{6}}+\frac{C_2}{6}e^{\frac{x}{6}}+C_2e^{\frac{x}{6}}[/tex]
Now, we can create our system of equations given our initial conditions:
[tex]y(x)=C_1e^{\frac{x}{6}}+\frac{C_2}{6}xe^{\frac{x}{6}}\\ \\y(0)=C_1e^{\frac{0}{6}}+\frac{C_2}{6}(0)e^{\frac{0}{6}}=3\\ \\C_1=3[/tex]
[tex]y'(x)=\frac{C_1}{6}e^{\frac{x}{6}}+\frac{C_2}{6}e^{\frac{x}{6}}+C_2e^{\frac{x}{6}}\\\\y'(0)=\frac{C_1}{6}e^{\frac{0}{6}}+\frac{C_2}{6}e^{\frac{0}{6}}+C_2e^{\frac{0}{6}}=\frac{5}{2}\\ \\\frac{C_1}{6}+\frac{C_2}{6}+C_2=\frac{5}{2}[/tex]
We then solve the system of equations, which becomes easy since we already know that [tex]C_1=3[/tex]:
[tex]\frac{C_1}{6}+\frac{C_2}{6}+C_2=\frac{5}{2}\\\\\frac{3}{6}+\frac{C_2}{6}+C_2=\frac{5}{2}\\\\3+C_2+6C_2=15\\\\7C_2=12\\\\C_2=\frac{12}{7}[/tex]
Thus, our final solution is:
[tex]y(x)=C_1e^{\frac{x}{6}}+C_2xe^{\frac{x}{6}}\\\\y(x)=3e^{\frac{x}{6}}+\frac{12}{7}xe^{\frac{x}{6}}[/tex]
