Question

solve the initial value problem.
y + 12y + 36y = 0, y(0) = 4, y(0) = -28
y(x) = \square

Explanation:

Step1: Write the characteristic equation

For the differential equation \(y'' + 12y' + 36y = 0\), the characteristic equation is \(r^{2}+12r + 36=0\).
Using the quadratic formula \(r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=12\), \(c = 36\), we have \(r=\frac{-12\pm\sqrt{12^{2}-4\times1\times36}}{2\times1}=\frac{-12\pm\sqrt{144 - 144}}{2}=\frac{-12\pm0}{2}=- 6\) (repeated root).

Step2: Write the general solution

Since the characteristic equation has a repeated root \(r=-6\), the general solution of the differential equation is \(y(x)=(C_{1}+C_{2}x)e^{-6x}\).

Step3: Use the initial condition \(y(0) = 4\)

Substitute \(x = 0\) and \(y(0)=4\) into \(y(x)=(C_{1}+C_{2}x)e^{-6x}\).
We get \(y(0)=(C_{1}+C_{2}\times0)e^{0}=C_{1}\), so \(C_{1}=4\).

Step4: Differentiate \(y(x)\)

Differentiate \(y(x)=(4 + C_{2}x)e^{-6x}\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = 4 + C_{2}x\), \(v=e^{-6x}\), \(u^\prime=C_{2}\), \(v^\prime=-6e^{-6x}\).
\(y^\prime(x)=C_{2}e^{-6x}+(4 + C_{2}x)(-6)e^{-6x}=(C_{2}-24-6C_{2}x)e^{-6x}\).

Step5: Use the initial condition \(y^\prime(0)=-28\)

Substitute \(x = 0\) and \(y^\prime(0)=-28\) into \(y^\prime(x)=(C_{2}-24-6C_{2}x)e^{-6x}\).
We have \(y^\prime(0)=(C_{2}-24)e^{0}=C_{2}-24\).
Set \(C_{2}-24=-28\), then \(C_{2}=- 4\).

Answer:

\(y(x)=(4 - 4x)e^{-6x}\)