Question

find the unique solution of the second-order initial value problem
$y + y - 2y = 0$, $y(0) = -1$, $y(0) = -4$

$y(x) = \square$

Explanation:

Step1: Write the characteristic equation

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

Step2: Write the general solution

The general solution of the differential equation is \(y(x)=C_{1}e^{-2x}+C_{2}e^{x}\).

Step3: Use the initial condition \(y(0)=-1\)

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

Step4: Differentiate the general solution

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

Step5: Use the initial condition \(y'(0)=-4\)

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

Step6: Solve the system of equations

We have the system \(

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

\)
Subtract the first equation from the second equation:
\((-2C_{1}+C_{2})-(C_{1}+C_{2})=-4-(-1)\)
\(-2C_{1}+C_{2}-C_{1}-C_{2}=-4 + 1\)
\(-3C_{1}=-3\), so \(C_{1}=1\).
Substitute \(C_{1}=1\) into \(C_{1}+C_{2}=-1\), we get \(1+C_{2}=-1\), so \(C_{2}=-2\).

Step7: Write the particular solution

Substitute \(C_{1}=1\) and \(C_{2}=-2\) into \(y(x)\):
\(y(x)=e^{-2x}-2e^{x}\).

Answer:

\(y(x)=e^{-2x}-2e^{x}\)