Question

solve the initial value problem.
$y + 16y + 64y = 0$, $y(0) = 3$, $y(0) = -25$
$y(x) = \square$

Explanation:

Step1: Write the characteristic equation

For the differential equation \(y'' + 16y' + 64y = 0\), the characteristic equation is \(r^{2}+16r + 64=0\).
Using the quadratic formula \(r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=16\), \(c = 64\), we have \(r=\frac{-16\pm\sqrt{16^{2}-4\times1\times64}}{2\times1}=\frac{-16\pm\sqrt{256 - 256}}{2}=\frac{-16\pm0}{2}=- 8\) (repeated root).

Step2: Write the general solution

Since the characteristic equation has a repeated root \(r=-8\), the general solution of the differential equation is \(y(x)=(C_{1}+C_{2}x)e^{-8x}\).

Step3: Use the initial conditions

• Apply \(y(0) = 3\):

Substitute \(x = 0\) into \(y(x)=(C_{1}+C_{2}x)e^{-8x}\), we get \(y(0)=C_{1}e^{0}\), so \(C_{1}=3\).

• Differentiate \(y(x)\) and apply \(y'(0)=-25\):

Differentiate \(y(x)=(3 + C_{2}x)e^{-8x}\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = 3 + C_{2}x\), \(u^\prime=C_{2}\) and \(v=e^{-8x}\), \(v^\prime=-8e^{-8x}\).
\(y^\prime(x)=C_{2}e^{-8x}-8(3 + C_{2}x)e^{-8x}=(C_{2}-24-8C_{2}x)e^{-8x}\).
Substitute \(x = 0\) into \(y^\prime(x)\), we have \(y^\prime(0)=(C_{2}-24)e^{0}\).
Since \(y^\prime(0)=-25\), then \(C_{2}-24=-25\), so \(C_{2}=-1\).

Answer:

\(y(x)=(3 - x)e^{-8x}\)