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).
\[ \frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0 \quad \text{with} \quad y|_{x=0}=0 \text{ and } \left.\frac{dy}{dx}\right|_{x=0}=1 \]

Question

If for the solution curve $ y = f(x) $ of the differential equation \[ \frac{dy}{dx} + (\tan x) y = 2 + \sec^2 x, \quad y(\frac{\pi}{3}) = \sqrt{3}, \] $\text{then}$ $ y(\frac{\pi}{4}) $ is equal to:

Answer 1

Instead of guessing the form of the solution from the repeated root, this equation can be solved directly with the Laplace transform, which turns the differential equation into an algebraic one.

Taking the Laplace transform of $y''+2y'+y=0$ with $y(0)=0$ and $y'(0)=1$:

\[ \left[s^2Y(s)-sy(0)-y'(0)\right]+2\left[sY(s)-y(0)\right]+Y(s)=0 \]

Substituting $y(0)=0$ and $y'(0)=1$:

\[ s^2Y(s)-1+2sY(s)+Y(s)=0 \] \[ Y(s)(s^2+2s+1)=1 \] \[ Y(s)=\frac{1}{(s+1)^2} \]

The standard inverse Laplace pair $\mathcal{L}^{-1}\left[\dfrac{1}{(s+a)^2}\right]=te^{-at}$ applies directly here with $a=1$, giving

\[ y(x)=xe^{-x} \]

which matches the solution found from the characteristic equation, confirming the repeated root $r=-1$ was handled correctly.

Now differentiate to get the slope. Using the product rule on $y=xe^{-x}$:

\[ \frac{dy}{dx}=e^{-x}-xe^{-x}=(1-x)e^{-x} \]

At $x=\ln 2$, $e^{-\ln 2}=1/2$, so

\[ \frac{dy}{dx}\bigg|_{x=\ln 2}=(1-\ln 2)\left(\frac{1}{2}\right)=(1-0.693147)(0.5)=0.153427 \]

Rounded to three decimal places, the slope is 0.153. This agrees exactly with the answer obtained from the direct characteristic-equation method, cross-checking the result through a completely different solution technique.

Answer 2