Question

find the unique solution of the second-order initial value problem.
y - 9y + 20y = 0, y(0) = 2, y(0) = 11

y(x) = \square

Explanation:

Step1: Write the characteristic equation

For the differential equation \(y'' - 9y' + 20y = 0\), the characteristic equation is \(r^{2}-9r + 20=0\).
Factor the quadratic equation: \((r - 4)(r - 5)=0\).
Solve for \(r\): \(r_{1}=4\) and \(r_{2}=5\).

Step2: Write the general solution

Since the roots \(r_{1}\) and \(r_{2}\) are real and distinct, the general solution of the differential equation is \(y(x)=C_{1}e^{4x}+C_{2}e^{5x}\).

Step3: Apply the initial condition \(y(0) = 2\)

Substitute \(x = 0\) and \(y(0)=2\) into \(y(x)\):
\(y(0)=C_{1}e^{0}+C_{2}e^{0}=C_{1}+C_{2}\).
So, \(C_{1}+C_{2}=2\).

Step4: Differentiate the general solution

Differentiate \(y(x)=C_{1}e^{4x}+C_{2}e^{5x}\) with respect to \(x\):
\(y'(x)=4C_{1}e^{4x}+5C_{2}e^{5x}\).

Step5: Apply the initial condition \(y'(0) = 11\)

Substitute \(x = 0\) and \(y'(0)=11\) into \(y'(x)\):
\(y'(0)=4C_{1}e^{0}+5C_{2}e^{0}=4C_{1}+5C_{2}\).
So, \(4C_{1}+5C_{2}=11\).

Step6: Solve the system of equations

We have the system \(

$$\begin{cases}C_{1}+C_{2}=2\\4C_{1}+5C_{2}=11\end{cases}$$

\).
From the first equation \(C_{1}=2 - C_{2}\).
Substitute \(C_{1}=2 - C_{2}\) into the second equation:
\(4(2 - C_{2})+5C_{2}=11\).
Expand: \(8-4C_{2}+5C_{2}=11\).
Simplify: \(C_{2}=3\).
Substitute \(C_{2}=3\) into \(C_{1}=2 - C_{2}\), we get \(C_{1}=- 1\).

Answer:

\(y(x)=-e^{4x}+3e^{5x}\)