Given differential equation
The slope is
d y d x = 2 e 2 x − 6 e − x + 9 2 + 9 e − 2 x . \frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}. d x d y = 2 + 9 e − 2 x 2 e 2 x − 6 e − x + 9 . We simplify it first.
Multiply numerator and denominator by e 2 x e^{2x} e 2 x
d y d x = 2 e 4 x − 6 e x + 9 e 2 x 2 e 2 x + 9 . \frac{dy}{dx}=\frac{2e^{4x}-6e^x+9e^{2x}}{2e^{2x}+9}. d x d y = 2 e 2 x + 9 2 e 4 x − 6 e x + 9 e 2 x . Now observe that if we put
t = e x , t=e^x, t = e x ,
d y d x = 2 t 4 + 9 t 2 − 6 t 2 t 2 + 9 . \frac{dy}{dx}=\frac{2t^4+9t^2-6t}{2t^2+9}. d x d y = 2 t 2 + 9 2 t 4 + 9 t 2 − 6 t . But a better simplification is obtained directly by setting
u = e − x . u=e^{-x}. u = e − x .
d y d x = 2 / u 2 − 6 u + 9 2 + 9 u 2 . \frac{dy}{dx}=\frac{2/u^2-6u+9}{2+9u^2}. d x d y = 2 + 9 u 2 2/ u 2 − 6 u + 9 . Instead, let us rewrite the expression in a form suitable for integration by multiplying numerator and denominator by e 2 x e^{2x} e 2 x
d y d x = e 2 x − 3 ⋅ 2 e x 2 e 2 x + 9 + 9 2 + 9 e − 2 x . \frac{dy}{dx}=e^{2x}-3\cdot \frac{2e^x}{2e^{2x}+9}+\frac{9}{2+9e^{-2x}}. d x d y = e 2 x − 3 ⋅ 2 e 2 x + 9 2 e x + 2 + 9 e − 2 x 9 . This is still not clean enough, so let us directly verify a useful decomposition.
We claim:
2 e 2 x − 6 e − x + 9 2 + 9 e − 2 x = e 2 x − 3 e x 1 + 2 9 e 2 x . \frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}
= e^{2x}-\frac{3e^x}{1+\frac{2}{9}e^{2x}}. 2 + 9 e − 2 x 2 e 2 x − 6 e − x + 9 = e 2 x − 1 + 9 2 e 2 x 3 e x . Rather than forcing this, the cleanest route is to substitute t = e x t=e^x t = e x t t t
Since t = e x t=e^x t = e x d x = d t t . dx=\frac{dt}{t}. d x = t d t .
y = ∫ 2 t 2 − 6 / t + 9 2 + 9 / t 2 d x = ∫ 2 t 2 − 6 / t + 9 2 + 9 / t 2 ⋅ d t t . y=\int \frac{2t^2-6/t+9}{2+9/t^2}\,dx
=\int \frac{2t^2-6/t+9}{2+9/t^2}\cdot \frac{dt}{t}. y = ∫ 2 + 9/ t 2 2 t 2 − 6/ t + 9 d x = ∫ 2 + 9/ t 2 2 t 2 − 6/ t + 9 ⋅ t d t . Multiply numerator and denominator by t 2 t^2 t 2
2 t 2 − 6 / t + 9 2 + 9 / t 2 = 2 t 4 + 9 t 2 − 6 t 2 t 2 + 9 . \frac{2t^2-6/t+9}{2+9/t^2}
=\frac{2t^4+9t^2-6t}{2t^2+9}. 2 + 9/ t 2 2 t 2 − 6/ t + 9 = 2 t 2 + 9 2 t 4 + 9 t 2 − 6 t . So
d y d x d x = 2 t 4 + 9 t 2 − 6 t t ( 2 t 2 + 9 ) d t = 2 t 3 + 9 t − 6 2 t 2 + 9 d t . \frac{dy}{dx}\,dx = \frac{2t^4+9t^2-6t}{t(2t^2+9)}dt
=\frac{2t^3+9t-6}{2t^2+9}dt. d x d y d x = t ( 2 t 2 + 9 ) 2 t 4 + 9 t 2 − 6 t d t = 2 t 2 + 9 2 t 3 + 9 t − 6 d t . Hence
y = ∫ 2 t 3 + 9 t − 6 2 t 2 + 9 d t . y=\int \frac{2t^3+9t-6}{2t^2+9}\,dt. y = ∫ 2 t 2 + 9 2 t 3 + 9 t − 6 d t .
Integrate
Now divide:
2 t 3 + 9 t − 6 2 t 2 + 9 = t − 6 2 t 2 + 9 . \frac{2t^3+9t-6}{2t^2+9}=t-\frac{6}{2t^2+9}. 2 t 2 + 9 2 t 3 + 9 t − 6 = t − 2 t 2 + 9 6 . Because
t ( 2 t 2 + 9 ) = 2 t 3 + 9 t . t(2t^2+9)=2t^3+9t. t ( 2 t 2 + 9 ) = 2 t 3 + 9 t . So,
y = ∫ t d t − 6 ∫ d t 2 t 2 + 9 . y=\int t\,dt-6\int \frac{dt}{2t^2+9}. y = ∫ t d t − 6 ∫ 2 t 2 + 9 d t . First integral:
∫ t d t = t 2 2 . \int t\,dt=\frac{t^2}{2}. ∫ t d t = 2 t 2 . Second integral:
∫ d t 2 t 2 + 9 = 1 2 ∫ d t t 2 + 9 2 . \int \frac{dt}{2t^2+9}
=\frac{1}{2}\int \frac{dt}{t^2+\frac{9}{2}}. ∫ 2 t 2 + 9 d t = 2 1 ∫ t 2 + 2 9 d t . Using
∫ d x x 2 + a 2 = 1 a tan − 1 ( x a ) , \int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right), ∫ x 2 + a 2 d x = a 1 tan − 1 ( a x ) , with
a = 3 2 , a=\frac{3}{\sqrt2}, a = 2 3 ,
∫ d t 2 t 2 + 9 = 1 2 ⋅ 2 3 tan − 1 ( 2 t 3 ) = 2 6 tan − 1 ( 2 t 3 ) . \int \frac{dt}{2t^2+9}
=\frac{1}{2}\cdot \frac{\sqrt2}{3}\tan^{-1}\left(\frac{\sqrt2 t}{3}\right)
=\frac{\sqrt2}{6}\tan^{-1}\left(\frac{\sqrt2 t}{3}\right). ∫ 2 t 2 + 9 d t = 2 1 ⋅ 3 2 tan − 1 ( 3 2 t ) = 6 2 tan − 1 ( 3 2 t ) . Therefore,
y = t 2 2 − 6 ⋅ 2 6 tan − 1 ( 2 t 3 ) + C y=\frac{t^2}{2}-6\cdot \frac{\sqrt2}{6}\tan^{-1}\left(\frac{\sqrt2 t}{3}\right)+C y = 2 t 2 − 6 ⋅ 6 2 tan − 1 ( 3 2 t ) + C that is,
y = t 2 2 − 2 tan − 1 ( 2 t 3 ) + C . y=\frac{t^2}{2}-\sqrt2\tan^{-1}\left(\frac{\sqrt2 t}{3}\right)+C. y = 2 t 2 − 2 tan − 1 ( 3 2 t ) + C . Now t = e x t=e^x t = e x
y = e 2 x 2 − 2 tan − 1 ( 2 e x 3 ) + C . y=\frac{e^{2x}}{2}-\sqrt2\tan^{-1}\left(\frac{\sqrt2 e^x}{3}\right)+C. y = 2 e 2 x − 2 tan − 1 ( 3 2 e x ) + C .
Use the point ( 0 , 1 2 + π 2 2 ) \left(0,\frac12+\frac{\pi}{2\sqrt2}\right) ( 0 , 2 1 + 2 2 π )
At x = 0 x=0 x = 0 e x = 1 e^x=1 e x = 1
1 2 + π 2 2 = 1 2 − 2 tan − 1 ( 2 3 ) + C . \frac12+\frac{\pi}{2\sqrt2}
=\frac12-\sqrt2\tan^{-1}\left(\frac{\sqrt2}{3}\right)+C. 2 1 + 2 2 π = 2 1 − 2 tan − 1 ( 3 2 ) + C . So
C = π 2 2 + 2 tan − 1 ( 2 3 ) . C=\frac{\pi}{2\sqrt2}+\sqrt2\tan^{-1}\left(\frac{\sqrt2}{3}\right). C = 2 2 π + 2 tan − 1 ( 3 2 ) . Thus
y = e 2 x 2 − 2 tan − 1 ( 2 e x 3 ) + π 2 2 + 2 tan − 1 ( 2 3 ) . y=\frac{e^{2x}}{2}-\sqrt2\tan^{-1}\left(\frac{\sqrt2 e^x}{3}\right)+\frac{\pi}{2\sqrt2}+\sqrt2\tan^{-1}\left(\frac{\sqrt2}{3}\right). y = 2 e 2 x − 2 tan − 1 ( 3 2 e x ) + 2 2 π + 2 tan − 1 ( 3 2 ) .
Use the second point ( α , 1 2 e 2 α ) \left(\alpha,\frac12 e^{2\alpha}\right) ( α , 2 1 e 2 α )
At x = α x=\alpha x = α
1 2 e 2 α = 1 2 e 2 α − 2 tan − 1 ( 2 e α 3 ) + π 2 2 + 2 tan − 1 ( 2 3 ) . \frac12 e^{2\alpha}
=\frac12 e^{2\alpha}-\sqrt2\tan^{-1}\left(\frac{\sqrt2 e^{\alpha}}{3}\right)+\frac{\pi}{2\sqrt2}+\sqrt2\tan^{-1}\left(\frac{\sqrt2}{3}\right). 2 1 e 2 α = 2 1 e 2 α − 2 tan − 1 ( 3 2 e α ) + 2 2 π + 2 tan − 1 ( 3 2 ) . Cancel 1 2 e 2 α \frac12 e^{2\alpha} 2 1 e 2 α
0 = − 2 tan − 1 ( 2 e α 3 ) + π 2 2 + 2 tan − 1 ( 2 3 ) . 0=-\sqrt2\tan^{-1}\left(\frac{\sqrt2 e^{\alpha}}{3}\right)+\frac{\pi}{2\sqrt2}+\sqrt2\tan^{-1}\left(\frac{\sqrt2}{3}\right). 0 = − 2 tan − 1 ( 3 2 e α ) + 2 2 π + 2 tan − 1 ( 3 2 ) . Multiply by 1 2 \frac{1}{\sqrt2} 2 1
tan − 1 ( 2 e α 3 ) = π 4 + tan − 1 ( 2 3 ) . \tan^{-1}\left(\frac{\sqrt2 e^{\alpha}}{3}\right)=\frac{\pi}{4}+\tan^{-1}\left(\frac{\sqrt2}{3}\right). tan − 1 ( 3 2 e α ) = 4 π + tan − 1 ( 3 2 ) .
Take tangent on both sides
Let
θ = tan − 1 ( 2 3 ) . \theta=\tan^{-1}\left(\frac{\sqrt2}{3}\right). θ = tan − 1 ( 3 2 ) . Then
tan ( π 4 + θ ) = 1 + tan θ 1 − tan θ = 1 + 2 3 1 − 2 3 = 3 + 2 3 − 2 . \tan\left(\frac{\pi}{4}+\theta\right)=\frac{1+\tan\theta}{1-\tan\theta}
=\frac{1+\frac{\sqrt2}{3}}{1-\frac{\sqrt2}{3}}
=\frac{3+\sqrt2}{3-\sqrt2}. tan ( 4 π + θ ) = 1 − tan θ 1 + tan θ = 1 − 3 2 1 + 3 2 = 3 − 2 3 + 2 . Hence
2 e α 3 = 3 + 2 3 − 2 . \frac{\sqrt2 e^{\alpha}}{3}=\frac{3+\sqrt2}{3-\sqrt2}. 3 2 e α = 3 − 2 3 + 2 . Therefore
e α = 3 2 ( 3 + 2 3 − 2 ) . e^{\alpha}=\frac{3}{\sqrt2}\left(\frac{3+\sqrt2}{3-\sqrt2}\right). e α = 2 3 ( 3 − 2 3 + 2 ) .
Match with options
This is exactly Option B .
Final Answer:
3 2 ( 3 + 2 3 − 2 ) \boxed{\frac{3}{\sqrt2}\left(\frac{3+\sqrt2}{3-\sqrt2}\right)} 2 3 ( 3 − 2 3 + 2 )