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ipn [44]
2 years ago
14

Section 5.2 Problem 19:

Mathematics
1 answer:
rjkz [21]2 years ago
3 0

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}

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