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Step-by-Step Solution
Step 1: Write the Differential Equation
We are given the differential equation
\cos\bigl(\tfrac{1}{2}\cos^{-1}(e^{-x})\bigr)\,dx = \sqrt{e^{2x} - 1}\,dy
subject to the conditions that the curve intersects the y-axis at y=-1 and the x-axis at (\alpha,0) . We aim to find the value of e^\alpha .
Step 2: Introduce a Substitution
Let
\theta = \cos^{-1}(e^{-x}) .
Because
\theta = \cos^{-1}(e^{-x}) ,
we know
\cos\theta = e^{-x} .
Hence,
\cos\bigl(\tfrac{\theta}{2}\bigr) = \sqrt{\tfrac{1 + \cos\theta}{2}} = \sqrt{\tfrac{1 + e^{-x}}{2}}.
Thus, the left-hand side of the given differential equation becomes
\cos\bigl(\tfrac{\theta}{2}\bigr)\,dx .
Step 3: Match Terms in the Differential Equation
The original differential equation
\cos\bigl(\tfrac{1}{2}\cos^{-1}(e^{-x})\bigr) dx = \sqrt{e^{2x} - 1}\,dy
can be expressed (using the substitution above) as
\sqrt{\tfrac{1 + e^{-x}}{2}}\,dx = \sqrt{e^{2x} - 1}\,dy.
We can rearrange it to separate variables. First, observe:
\sqrt{\tfrac{1 + e^{-x}}{2}} = \sqrt{\tfrac{e^{x} + 1}{2e^{x}}}.
Then we have:
\sqrt{\tfrac{e^{x} + 1}{2 e^{x}}}\,dx = \sqrt{e^{2x} - 1}\,dy.
Rearranging further,
\tfrac{1}{\sqrt{2}}\,\tfrac{dx}{\sqrt{e^{x}}\,\sqrt{e^{x}-1}} = \int dy.
Step 4: Substitute e^x = t
Let
t = e^x \implies \tfrac{dt}{dx} = e^x = t.
Hence,
dx = \tfrac{dt}{t}.
Substitute into the integral:
\tfrac{1}{\sqrt{2}} \int \tfrac{dx}{\sqrt{e^x}\,\sqrt{e^x - 1}}
= \int dy.
With e^x = t , we get:
\sqrt{e^x} = \sqrt{t}, \quad \sqrt{e^x - 1} = \sqrt{t - 1}.
Thus the left side becomes:
\tfrac{1}{\sqrt{2}} \int \tfrac{\tfrac{dt}{t}}{\sqrt{t}\,\sqrt{t - 1}}
= \tfrac{1}{\sqrt{2}} \int \tfrac{dt}{t\sqrt{t}\,\sqrt{t - 1}}.
Simplify inside the integral:
t\sqrt{t} = t^{3/2}.
Hence,
\tfrac{1}{\sqrt{2}} \int \tfrac{dt}{t^{3/2}\,\sqrt{t - 1}} = \int dy.
This integrand can be manipulated further.
Step 5: Another Substitution t = \tfrac{1}{z}
Let
t = \tfrac{1}{z},
then
dt = -\tfrac{1}{z^2}\,dz.
Also,
t - 1 = \tfrac{1}{z} - 1 = \tfrac{1 - z}{z}.
Substituting into the integral:
\int \tfrac{dt}{t\sqrt{t}\,\sqrt{t-1}}
\;=\;
\int \tfrac{-\frac{1}{z^2}\,dz}{\frac{1}{z}\,\sqrt{\frac{1}{z}}\,\sqrt{\frac{1}{z}-1}}
.
Simplify step by step:
\frac{1}{z} \cdot \sqrt{\frac{1}{z}} = \frac{1}{z\sqrt{z}}.
\sqrt{\frac{1}{z} - 1} = \sqrt{\frac{1-z}{z}} = \frac{\sqrt{1-z}}{\sqrt{z}}.
Putting it all together gives:
\int \tfrac{-\frac{1}{z^2}\,dz}{\frac{1}{z\sqrt{z}} \cdot \frac{\sqrt{1-z}}{\sqrt{z}}}
= \int \tfrac{-\frac{1}{z^2}\,dz}{\frac{\sqrt{1-z}}{z\sqrt{z}\sqrt{z}}}
= \int \tfrac{-\frac{1}{z^2}\,dz}{\frac{\sqrt{1-z}}{z z}}
= \int \tfrac{-\frac{1}{z^2}\,dz}{\frac{\sqrt{1-z}}{z^2}}.
This simplifies to
- \int \tfrac{dz}{\sqrt{1 - z}}.
Step 6: Integrate and Introduce the Constant
The integral
- \int \tfrac{dz}{\sqrt{1 - z}} = - \int (1 - z)^{-1/2}\,dz
is equal to
2\sqrt{1 - z}
(up to a constant). Therefore,
2\sqrt{1 - z} = \sqrt{2}\,y + C
for some constant C .
Step 7: Rewrite in Terms of x and y
Recall z = \tfrac{1}{t} = e^{-x} . Thus,
1 - z = 1 - e^{-x}.
Our solution is
2\sqrt{1 - e^{-x}} = \sqrt{2}\,y + C.
Step 8: Apply the Initial Condition (x=0, y=-1)
The curve passes through the y-axis at x=0 . At this point, y=-1 . Substituting x=0 and y=-1 into the solution gives:
2\sqrt{1 - e^{0}} = 2\sqrt{1 - 1} = 0,
on the left-hand side, and
\sqrt{2}\,(-1) + C = -\sqrt{2} + C,
on the right-hand side. Equating them,
0 = -\sqrt{2} + C \implies C = \sqrt{2}.
Step 9: Find the Final Implicit Solution
Therefore, the particular solution is
2\sqrt{1 - e^{-x}} = \sqrt{2}\,(y + 1).
Step 10: Use the Intersection with the x-axis to Find e^\alpha
The curve intersects the x-axis at (\alpha, 0) , so y=0 :
2\sqrt{1 - e^{-\alpha}} = \sqrt{2}\,(0 + 1) = \sqrt{2}.
Divide both sides by 2:
\sqrt{1 - e^{-\alpha}} = \tfrac{1}{\sqrt{2}}.
Square both sides:
1 - e^{-\alpha} = \tfrac{1}{2}.
Hence,
e^{-\alpha} = \tfrac{1}{2},
which implies
e^{\alpha} = 2.
Conclusion
The required value of e^{\alpha} is \boxed{2} .
