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Step-by-Step Solution
Step 1: Interpret the Problem
The problem gives a curve passing through the points
(1, 1) and
( \frac{1}{10} , 100).
At a generic point P(x, y) on this curve, the tangent line meets the positive x-axis and y-axis at A and B, respectively. The ratio PA : PB is given as 1 : k.
Furthermore, the curve y(x) satisfies the differential equation
e^{\frac{dy}{dx}} = kx + \frac{k}{2}
with the initial condition y(0) = k .
We wish to find the value of
4\,y(1) - 6\,\ln(3) .
Step 2: Determine k from the Geometric Condition
From geometry involving tangents and intercepts, one often obtains (through a standard derivation) that if PA : PB = 1 : k on every tangent, then the curve typically satisfies:
x \frac{dy}{dx} = - k \, y.
Rearranging,
\frac{dy}{y} = - k \frac{dx}{x}.
Integrate both sides to get:
\ln(y) = - k \ln(x) + C
\implies
y = A x^{-k},
where A = e^{C} is a constant.
The curve also passes through (1, 1), so substituting x=1 , y=1 gives
1 = A \cdot 1^{-k} \implies A = 1.
Thus,
y = \frac{1}{x^k}.
The curve also passes through
\bigl(\tfrac{1}{10},\,100\bigr).
Substituting
x = \tfrac{1}{10}, \, y = 100
gives
100 = \frac{1}{\left(\tfrac{1}{10}\right)^k}
= \left(10\right)^k.
Hence
10^k = 100
\implies 10^k = 10^2
\implies k=2.
Step 3: Substitute k into the Given Differential Equation
We are also provided with the differential equation
e^{\frac{dy}{dx}} = k x + \frac{k}{2}.
Substituting k=2 gives:
e^{\frac{dy}{dx}} = 2x + 1.
Taking the natural logarithm on both sides:
\frac{dy}{dx} = \ln\bigl(2x + 1\bigr).
Step 4: Integrate to Find y(x)
We now solve
y(x) = \int \ln(2x + 1)\,dx.
Use integration by parts:
Let
u = \ln(2x+1)
and
dv = dx.
Then
du = \frac{2}{2x+1}\,dx,
and
v = x.
By the formula for integration by parts,
\int u \, dv = uv - \int v \, du,
we get
\int \ln(2x + 1) \, dx
= x\,\ln(2x+1) - \int x \cdot \frac{2}{2x+1}\,dx.
Simplify inside the integral:
x \cdot \frac{2}{2x+1}
= \frac{2x}{2x+1}.
Notice that
\frac{2x}{2x+1}
= 1 - \frac{1}{2x+1}.
Thus
\int x \cdot \frac{2}{2x+1}\,dx
= \int 2\left(1 - \frac{1}{2x+1}\right) dx
= \int 2\,dx - \int \frac{2}{2x+1}\,dx.
That becomes
2x - \ln(2x+1) + \text{constant}.
Therefore,
\int \ln(2x+1)\,dx
= x \ln(2x+1) - \bigl[\,2x - \ln(2x+1)\bigr] + C
= x \ln(2x+1) - 2x + \ln(2x+1) + C.
Hence,
y(x)
= x \,\ln\bigl(2x+1\bigr)
- 2x
+ \ln\bigl(2x+1\bigr)
+ C.
Step 5: Apply the Initial Condition y(0) = k = 2
We know y(0) = 2. Substitute x=0 into the general solution:
y(0)
= 0\,\ln(1) - 0 + \ln(1) + C
= C.
Thus
C = 2.
So the particular solution is
y(x)
= x \,\ln(2x+1)
- 2x
+ \ln(2x+1)
+ 2.
Step 6: Compute y(1)
Substitute x=1 :
y(1)
= 1\cdot\ln(3)
- 2\cdot1
+ \ln(3)
+ 2
= \ln(3) - 2 + \ln(3) + 2
= 2\,\ln(3).
Step 7: Evaluate 4y(1) - 6 ln(3)
We have y(1) = 2\,\ln(3). Therefore:
4\,y(1) - 6\,\ln(3)
= 4 \bigl(2\,\ln(3)\bigr) - 6\,\ln(3)
= 8\,\ln(3) - 6\,\ln(3)
= 2\,\ln(3).
Hence,
4\,y(1) - 6\,\ln(3)
= 2\,\ln(3).
Final Answer
The value of 4\,y(1) - 6\,\ln(3) is
2\,\ln(3).
