Question

In the differential equation $\dfrac{dy}{dx} + \alpha x y = 0$, $\alpha$ is a positive constant. If $y=1.0$ at $x=0.0$, and $y=0.8$ at $x=1.0$, the value of $\alpha$ is  (rounded off to three decimal places). 

Hint:

For ODEs of the form $\frac{dy}{dx} + f(x)y = 0$, the solution is $y = C e^{-\int f(x) dx}$. Use boundary/initial conditions to fix constants.

Answer 1

Step 1: Write the differential equation.
\[ \frac{dy}{dx} + \alpha x y = 0 \] This is a first-order linear ODE.

Step 2: Rearrange.
\[ \frac{dy}{dx} = -\alpha x y \]

Step 3: Separate variables.
\[ \frac{dy}{y} = -\alpha x dx \]

Step 4: Integrate.
\[ \int \frac{1}{y} \, dy = -\alpha \int x \, dx \] \[ \ln y = -\frac{\alpha x^2}{2} + C \]

Step 5: Exponential solution.
\[ y = C_1 \, e^{-\frac{\alpha x^2}{2}} \]

Step 6: Apply initial condition at $x=0, y=1$.
\[ 1 = C_1 e^0 \Rightarrow C_1 = 1 \] So, \[ y = e^{-\frac{\alpha x^2}{2}} \]

Step 7: Apply condition at $x=1, y=0.8$.
\[ 0.8 = e^{-\frac{\alpha (1)^2}{2}} \Rightarrow \ln(0.8) = -\frac{\alpha}{2} \] \[ \alpha = -2 \ln(0.8) \]

Step 8: Compute numerical value.
\[ \ln(0.8) \approx -0.22314355 \] \[ \alpha = -2 \times (-0.22314355) = 0.4462871 \approx 0.446 \] \[ \boxed{0.446} \]