Question

Section 5.2 Problem 19:

Answer 1

Answer:

$y(x)=2xe^{3x}$ (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 $am^2+bm+c=0$ where the values of $m$ are the roots:

$y''-6y'+9y=0\\\\m^2-6m+9=0\\\\(m-3)^2=0\\\\m-3=0\\\\m=3$

Since the values of $m$ are equal real roots, then the general solution is $y(x)=C_1e^{m_1x}+C_2xe^{m_1x}$.

Thus, the general solution for our given differential equation is $y(x)=C_1e^{3x}+C_2xe^{3x}$.

To account for both initial conditions, take the derivative of $y(x)$, thus, $y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}$

Now, we can create our system of equations given our initial conditions:

$y(x)=C_1e^{3x}+C_2xe^{3x}\\ \\y(0)=C_1e^{3(0)}+\frac{C_2}{6}(0)e^{3(0)}=0\\ \\C_1=0$

$y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}\\\\y'(0)=3C_1e^{3(0)}+C_2e^{3(0)}+3C_2(0)e^{3(0)}=2\\\\3C_1+C_2=2$

We then solve the system of equations, which becomes easy since we already know that $C_1=0$:

$3C_1+C_2=2\\\\3(0)+C_2=2\\\\C_2=2$

Thus, our final solution is:

$y(x)=C_1e^{3x}+C_2xe^{3x}\\\\y(x)=2xe^{3x}$