Related Formula
∫ (1)/(1+y²) dy = ⁻¹y + C$$\int \frac{1}{1+y^2} dy = \tan^{-1}y + C$$
∫ f(x) f'(x) dx = ([f(x)]²)/(2) + C$$\int f(x) f'(x) dx = \frac{[f(x)]^2}{2} + C$$
(A - B) = ( A - B)/(1 + A B)$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Core Logic
Rearrange the given differential equation to separate variables x$x$ and y$y$:
(1+y²)(1+ ₑ x)dx + x dy = 0$$(1+y^2)(1+\log_e x)dx + x dy = 0$$
(1+ ₑ x)/(x) dx + (1)/(1+y²) dy = 0$$\frac{1+\log_e x}{x} dx + \frac{1}{1+y^2} dy = 0$$
Integrate both sides:
∫ (1+ ₑ x)/(x) dx + ∫ (1)/(1+y²) dy = C$$\int \frac{1+\log_e x}{x} dx + \int \frac{1}{1+y^2} dy = C$$
Step 1: Execute Integration
For the first integral, let u = 1+ ₑ x$u = 1+\log_e x$, so du = (1)/(x) dx$du = \frac{1}{x} dx$.
∫ u du = (u²)/(2) = ((1+ ₑ x)²)/(2)$$\int u \, du = \frac{u^2}{2} = \frac{(1+\log_e x)^2}{2}$$
For the second integral:
∫ (1)/(1+y²) dy = ⁻¹y$$\int \frac{1}{1+y^2} dy = \tan^{-1}y$$
Putting them together:
((1+ ₑ x)²)/(2) + ⁻¹y = C$$\frac{(1+\log_e x)^2}{2} + \tan^{-1}y = C$$
Step 2: Apply Boundary Conditions
The curve passes through (1, 1)$(1, 1)$. Substitute x=1, y=1$x=1, y=1$:
((1+ ₑ 1)²)/(2) + ⁻¹(1) = C$$\frac{(1+\log_e 1)^2}{2} + \tan^{-1}(1) = C$$
Since ₑ 1 = 0$\log_e 1 = 0$ and ⁻¹(1) = (π)/(4)$\tan^{-1}(1) = \frac{\pi}{4}$:
(1²)/(2) + (π)/(4) = C ⇒ C = (1)/(2) + (π)/(4)$$\frac{1^2}{2} + \frac{\pi}{4} = C \Rightarrow C = \frac{1}{2} + \frac{\pi}{4}$$
Wait, the solution rearranges differently: ∫ ((1)/(x) + (ln x)/(x)) dx = ln x + ((ln x)²)/(2)$\int (\frac{1}{x} + \frac{\ln x}{x}) dx = \ln x + \frac{(\ln x)^2}{2}$. Let's use this to stay fully synced with the PDF's algebraic step.
ln x + ((ln x)²)/(2) + ⁻¹ y = C$$\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = C$$
At (1,1)$(1,1)$:
ln(1) + ((ln 1)²)/(2) + ⁻¹(1) = C ⇒ 0 + 0 + (π)/(4) = C ⇒ C = (π)/(4)$$\ln(1) + \frac{(\ln 1)^2}{2} + \tan^{-1}(1) = C \Rightarrow 0 + 0 + \frac{\pi}{4} = C \Rightarrow C = \frac{\pi}{4}$$
The general equation simplifies nicely to:
ln x + ((ln x)²)/(2) + ⁻¹ y = (π)/(4)$$\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$
Step 3: Evaluate at x = e
Substitute x=e$x=e$ into the equation to find y(e)$y(e)$:
ln e + ((ln e)²)/(2) + ⁻¹ y = (π)/(4)$$\ln e + \frac{(\ln e)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$
1 + (1²)/(2) + ⁻¹ y = (π)/(4)$$1 + \frac{1^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$
(3)/(2) + ⁻¹ y = (π)/(4)$$\frac{3}{2} + \tan^{-1} y = \frac{\pi}{4}$$
⁻¹ y = (π)/(4) - (3)/(2)$$\tan^{-1} y = \frac{\pi}{4} - \frac{3}{2}$$
y = ((π)/(4) - (3)/(2))$$y = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right)$$
Using the compound angle formula (A-B)$\tan(A-B)$:
y = ( (π/4) - (3/2))/(1 + (π/4) (3/2)) = (1 - (3/2))/(1 + (3/2))$$y = \frac{\tan(\pi/4) - \tan(3/2)}{1 + \tan(\pi/4)\tan(3/2)} = \frac{1 - \tan(3/2)}{1 + \tan(3/2)}$$
Comparing this format to y(e) = (α- (3/2))/(β+ (3/2))$y(e) = \frac{\alpha-\tan(3/2)}{\beta+\tan(3/2)}$ gives:
α = 1, β = 1$$\alpha = 1, \quad \beta = 1$$
Calculate α + 2β$\alpha + 2\beta$:
1 + 2(1) = 3$1 + 2(1) = 3$
Pattern Recognition
For expressions involving separated x$x$ and y$y$ logarithmic terms, always split ∫ (1+ln x)/(x)dx$\int \frac{1+\ln x}{x}dx$ into ∫ (1)/(x) + ∫ (ln x)/(x)$\int \frac{1}{x} + \int \frac{\ln x}{x}$ to avoid carrying a non-zero shift (1)/(2)$\frac{1}{2}$ into the constant C$C$. This allows direct identity matching for tangent sums.
Chapter Mix
Class 12 Mathematics: Differential Equations
Class 11 Mathematics: Trigonometric Functions
