Related Formula
For a quadratic equation Am² + Bm + C = 0$Am^2 + Bm + C = 0$ having equal roots, Discriminant D = 0 ⇒ B² - 4AC = 0$D = 0 \Rightarrow B^2 - 4AC = 0$.
Core Logic
Given quadratic equation in m$m$: f(x)m² - 2f'(x)m + f''(x) = 0$f(x)m^2 - 2f'(x)m + f''(x) = 0$ has equal roots.
D = 0 ⇒ (-2f'(x))² - 4(f(x))(f''(x)) = 0$$D = 0 \Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$$
4(f'(x))² = 4f(x)f''(x) ⇒ (f'(x))² = f(x)f''(x)$$4(f'(x))^2 = 4f(x)f''(x) \Rightarrow (f'(x))^2 = f(x)f''(x)$$
Step 1: Solve the Differential Equation
Rewrite the DE: (f''(x))/(f'(x)) = (f'(x))/(f(x))$\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$
Integrate both sides:
∫ (f''(x))/(f'(x)) dx = ∫ (f'(x))/(f(x)) dx$$\int \frac{f''(x)}{f'(x)} dx = \int \frac{f'(x)}{f(x)} dx$$
ln|f'(x)| = ln|f(x)| + ln|c| ⇒ f'(x) = c · f(x)$$\ln|f'(x)| = \ln|f(x)| + \ln|c| \Rightarrow f'(x) = c \cdot f(x)$$
Using given f(0) = 1$f(0) = 1$ and f'(0) = 2$f'(0) = 2$:
f'(0) = c · f(0) ⇒ 2 = c(1) ⇒ c = 2$$f'(0) = c \cdot f(0) \Rightarrow 2 = c(1) \Rightarrow c = 2$$
Now we have f'(x) = 2f(x) ⇒ (f'(x))/(f(x)) = 2$f'(x) = 2f(x) \Rightarrow \frac{f'(x)}{f(x)} = 2$.
Integrate again:
ln|f(x)| = 2x + d$$\ln|f(x)| = 2x + d$$
Use f(0) = 1 ⇒ ln(1) = 0 + d ⇒ d = 0$f(0) = 1 \Rightarrow \ln(1) = 0 + d \Rightarrow d = 0$.
So, ln f(x) = 2x ⇒ f(x) = e2x$\ln f(x) = 2x \Rightarrow f(x) = e^{2x}$.
Step 2: Investigate increasing interval
Let g(x) = f(ln x - x) = e2(ln x - x)$g(x) = f(\ln x - x) = e^{2(\ln x - x)}$.
For g(x)$g(x)$ to be increasing, g'(x) ≥ 0$g'(x) \geq 0$.
g'(x) = 2e2(ln x - x) · (d)/(dx)(ln x - x)$$g'(x) = 2e^{2(\ln x - x)} \cdot \frac{d}{dx}(\ln x - x)$$
g'(x) = 2e2(ln x - x) ((1)/(x) - 1)$$g'(x) = 2e^{2(\ln x - x)} \left(\frac{1}{x} - 1\right)$$
Since exponential is always positive, g'(x) ≥ 0 ⇒ (1)/(x) - 1 ≥ 0$g'(x) \geq 0 \Rightarrow \frac{1}{x} - 1 \geq 0$.
(1 - x)/(x) ≥ 0$$\frac{1 - x}{x} \geq 0$$
The critical points are x=0, x=1$x=0, x=1$. Based on domain of ln x$\ln x$, x > 0$x > 0$.
Sign scheme yields positive derivative in x in (0, 1]$x \in (0, 1]$.
Therefore, (α, β) = (0, 1) ⇒ α = 0, β = 1$(\alpha, \beta) = (0, 1) \Rightarrow \alpha = 0, \beta = 1$.
Step 3: Final Output
α + β = 0 + 1 = 1$$\alpha + \beta = 0 + 1 = 1$$
Pattern Recognition
The relation (f')² = f · f''$(f')^2 = f \cdot f''$ is a classical indicator of exponential functions (f = Cekx$f = Ce^{kx}$). Solving via double logarithmic integration collapses the differential equation almost instantaneously.
Chapter Mix
Class 12 Maths: Differential Equations
Class 12 Maths: Applications of Derivatives
