Answer:
[tex]y(x)=2xe^{3x}[/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]y''-6y'+9y=0\\\\m^2-6m+9=0\\\\(m-3)^2=0\\\\m-3=0\\\\m=3[/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^{3x}+C_2xe^{3x}[/tex].
To account for both initial conditions, take the derivative of [tex]y(x)[/tex], thus, [tex]y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}[/tex]
Now, we can create our system of equations given our initial conditions:
[tex]y(x)=C_1e^{3x}+C_2xe^{3x}\\ \\y(0)=C_1e^{3(0)}+\frac{C_2}{6}(0)e^{3(0)}=0\\ \\C_1=0[/tex]
[tex]y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}\\\\y'(0)=3C_1e^{3(0)}+C_2e^{3(0)}+3C_2(0)e^{3(0)}=2\\\\3C_1+C_2=2[/tex]
We then solve the system of equations, which becomes easy since we already know that [tex]C_1=0[/tex]:
[tex]3C_1+C_2=2\\\\3(0)+C_2=2\\\\C_2=2[/tex]
Thus, our final solution is:
[tex]y(x)=C_1e^{3x}+C_2xe^{3x}\\\\y(x)=2xe^{3x}[/tex]
