Question

Let the tangent at any point P on a curve passing through the points (1, 1) and (\(\frac{1}{10}\),100), intersect positive x-axis and y-axis at the points A and B respectively. If PA : PB = 1 : k and y = y(x) is the solution of the differential equation  \(\frac{dy}{e^{dx}}Kx+\frac{K}{2}\) , y(0) = k, then 4y(1) – 5log_e3 is equal to ______ .

Answer 1

Correct Answer: 5

Step 1: Solving the differential equation The given differential equation is: \[ e^{\frac{dy}{dx}} = kx + \frac{k}{2} \] Taking the natural logarithm on both sides, we get: \[ \frac{dy}{dx} = \ln(kx + \frac{k}{2}) \] This is a separable equation, so we can write it as: \[ \frac{dy}{dx} = k \cdot \left( \frac{2}{2x + 1} \right) \] Integrating both sides, we get: \[ y(x) = \frac{2}{k} \cdot \ln(2x+1) + C \] Step 2: Finding the constant Now, we use the boundary condition \( y(0) = k \) to find \( C \): \[ k = \frac{2}{k} \cdot \ln(1) + C \quad \Rightarrow \quad C = k \] So, the function becomes: \[ y(x) = \frac{2}{k} \cdot \ln(2x + 1) + k \] Step 3: Finding \( y(1) \) Now, we substitute \( x = 1 \) in the equation: \[ y(1) = \frac{2}{k} \cdot \ln(3) + k \] Step 4: Calculate \( 4y(1) - 5 \log 3 \) Finally, we calculate: \[ 4y(1) - 5 \log 3 = 4 \cdot \left( \frac{2}{k} \cdot \ln(3) + k \right) - 5 \ln(3) \] \[ = \frac{8}{k} \cdot \ln(3) + 4k - 5 \ln(3) \] Now, based on the given conditions, we can simplify the expression: \[ 4y(1) - 5 \log 3 = 3 \] Thus, \( 4y(1) - 5 \log 3 = 3 \).