Let a curve y=y(x) be given by the solution of the differential equation cos( 1 2 cos^-1(e^-x))dx=(√e^2x-1)dy. If it intersects y-axis at y=-1, and…

We have,

$$\mathrm{cos}(\frac{1}{2}{\mathrm{cos}}^{-1}({e}^{-x}))dx=(\sqrt{{e}^{2x}-1})dy...(i)$$cos(21​cos−1(e−x))dx=(e2x−1​)dy...(i)

Put ${\mathrm{cos}}^{-1}({e}^{-x})=\theta ,\theta \in [0,\pi ]$cos−1(e−x)=θ,θ∈[0,π]

$$\Rightarrow \mathrm{cos}\theta ={e}^{-x}$$⇒cosθ=e−x

$$\Rightarrow 2{\mathrm{cos}}^{2}\frac{\theta }{2}-1={e}^{-x}$$⇒2cos22θ​−1=e−x

$$\Rightarrow \mathrm{cos}(\frac{\theta }{2})=\sqrt{\frac{{e}^{-x}+1}{2}}=\sqrt{\frac{{e}^{x}+1}{2{e}^{x}}}$$⇒cos(2θ​)=2e−x+1​​=2exex+1​​

Hence, by $(i)$(i), we have

$$\mathrm{cos}(\frac{\theta }{2})dx=(\sqrt{{e}^{2x}-1})dy$$cos(2θ​)dx=(e2x−1​)dy

$$\Rightarrow \sqrt{\frac{{e}^{x}+1}{2{e}^{x}}}dx=\sqrt{{e}^{2x}-1}dy$$⇒2exex+1​​dx=e2x−1​dy

$$\Rightarrow (\sqrt{\frac{{e}^{x}+1}{2{e}^{x}}})dx=(\sqrt{{e}^{x}-1})(\sqrt{{e}^{x}+1})dy$$⇒(2exex+1​​)dx=(ex−1​)(ex+1​)dy

$$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{e}^{x}}\sqrt{{e}^{x}-1}}=\int dy$$⇒2​1​∫ex​ex−1​dx​=∫dy

Put ${e}^{x}=t\Rightarrow dt={e}^{x}dx$ex=t⇒dt=exdx

$$\frac{1}{\sqrt{2}}\int \frac{dt}{{e}^{x}\sqrt{{e}^{x}}\sqrt{{e}^{x}-1}}=\int dy$$2​1​∫exex​ex−1​dt​=∫dy

$$\int \frac{dt}{t\sqrt{{t}^{2}-t}}=\sqrt{2}\int dy$$∫tt2−t​dt​=2​∫dy

Put $t=\frac{1}{z}\Rightarrow \frac{dt}{dz}=-\frac{1}{{z}^{2}}$t=z1​⇒dzdt​=−z21​

$$\int \frac{-\frac{dz}{{z}^{2}}}{\frac{1}{z}\sqrt{\frac{1}{{z}^{2}}-\frac{1}{z}}}=\sqrt{2}\int dy$$∫z1​z21​−z1​​−z2dz​​=2​∫dy

$$\Rightarrow -\int \frac{dz}{\sqrt{1-z}}=\sqrt{2}\int dy$$⇒−∫1−z​dz​=2​∫dy

$$\Rightarrow \frac{-2(1-z{)}^{1/2}}{-1}=\sqrt{2}y+c$$⇒−1−2(1−z)1/2​=2​y+c

$$\Rightarrow 2{(1-\frac{1}{t})}^{1/2}=\sqrt{2}y+c$$⇒2(1−t1​)1/2=2​y+c

$$\Rightarrow 2{(1-{e}^{-x})}^{1/2}=\sqrt{2}y+c$$⇒2(1−e−x)1/2=2​y+c

Now, it meets $y-\text{axis}$y−axis at $(0,-1)$(0,−1), hence

$$0=-\sqrt{2}+c\Rightarrow c=\sqrt{2}$$0=−2​+c⇒c=2​

Hence,

$$2{(1-{e}^{-x})}^{1/2}=\sqrt{2}(y+1)$$2(1−e−x)1/2=2​(y+1)

It passes through $(\alpha ,0)$(α,0)

$$2{(1-{e}^{-\alpha })}^{1/2}=\sqrt{2}$$2(1−e−α)1/2=2​

$$\Rightarrow \sqrt{1-{e}^{-\alpha }}=\frac{1}{\sqrt{2}}$$⇒1−e−α​=2​1​

$$\Rightarrow 1-{e}^{-\alpha }=\frac{1}{2}$$⇒1−e−α=21​

$$\Rightarrow {e}^{-\alpha }=\frac{1}{2}\Rightarrow {e}^{\alpha }=2$$⇒e−α=21​⇒eα=2