Question

Consider the differential equation with the initial conditions given below. If \(y(x)\) is the solution of the equation, the value of the slope, \(\frac{dy}{dx}\), at \(x=\ln(2)\) is ________ (rounded off to three decimal places).
\[ \frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0 \quad \text{with} \quad y|_{x=0}=0 \text{ and } \left.\frac{dy}{dx}\right|_{x=0}=1 \]

Hint:

Solve the repeated-root ODE to get \(y=xe^{-x}\), then differentiate and evaluate at \(x=\ln 2\), using \(e^{-\ln 2}=1/2\).

Answer 1

Correct Answer: 0.153

Step 1: Write the characteristic equation.
The differential equation \(\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}+y=0\) is linear, second order, with constant coefficients. Its characteristic equation is
\[ r^2+2r+1=0 \]

Step 2: Solve the characteristic equation.
This factors as
\[ (r+1)^2=0 \]
giving a repeated root \(r=-1\). A repeated root means the general solution takes the form
\[ y(x)=(C_1+C_2x)e^{-x} \]
where \(C_1\) and \(C_2\) are constants fixed by the initial conditions.

Step 3: Apply the first initial condition.
At \(x=0\), \(y(0)=0\). Substituting \(x=0\) into the general solution:
\[ y(0)=(C_1+0)e^{0}=C_1=0 \]
So \(C_1=0\), and the solution simplifies to \(y(x)=C_2 x e^{-x}\).

Step 4: Apply the second initial condition.
Differentiate \(y(x)=C_2 x e^{-x}\) using the product rule:
\[ \frac{dy}{dx}=C_2 e^{-x}-C_2 x e^{-x}=C_2(1-x)e^{-x} \]
At \(x=0\):
\[ y'(0)=C_2(1-0)e^{0}=C_2=1 \]
So \(C_2=1\), and the particular solution is
\[ y(x)=x e^{-x} \]

Step 5: Write the general slope and evaluate it at \(x=\ln(2)\).
From Step 4, with \(C_2=1\),
\[ \frac{dy}{dx}=(1-x)e^{-x} \]
At \(x=\ln(2)\), note that \(e^{-\ln(2)}=\dfrac{1}{e^{\ln 2}}=\dfrac{1}{2}\). So
\[ \left.\frac{dy}{dx}\right|_{x=\ln2} = (1-\ln 2)\cdot\frac{1}{2} \]
Using \(\ln 2 \approx 0.693147\):
\[ (1-0.693147)\times \frac{1}{2} = 0.306853\times\frac{1}{2} = 0.153427 \]

Final Answer:
Rounded to three decimal places, the slope at \(x=\ln(2)\) is 0.153. \[ \boxed{\frac{dy}{dx}\Big|_{x=\ln2}\approx 0.153} \]