Related Formula
For a linear differential equation (dy)/(dx) + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is e∫ P(x) dx$e^{\int P(x) dx}$.
Core Logic
Rewrite the equation as:
(x² - 4)(dy)/(dx) - 2xy = -2x(x² - 4)²$$(x^2 - 4)\frac{dy}{dx} - 2xy = -2x(x^2 - 4)^2$$
Dividing by (x² - 4)²$(x^2 - 4)^2$ to create an exact differential:
((x² - 4)y' - 2xy)/((x² - 4)²) = -2x$$\frac{(x^2 - 4)y' - 2xy}{(x^2 - 4)^2} = -2x$$
Observe that the LHS is the exact derivative of (y)/(x² - 4)$\frac{y}{x^2 - 4}$:
(d)/(dx)((y)/(x² - 4)) = -2x$$\frac{d}{dx}\left(\frac{y}{x^2 - 4}\right) = -2x$$
Integrating both sides:
(y)/(x² - 4) = -x² + C$$\frac{y}{x^2 - 4} = -x^2 + C$$
y = (-x² + C)(x² - 4)$$y = (-x^2 + C)(x^2 - 4)$$
Step 1: Finding C
The curve passes through (3, 15)$(3, 15)$. Substitute x=3, y=15$x=3, y=15$:
15 = (-(3)² + C)(3² - 4)$$15 = (-(3)^2 + C)(3^2 - 4)$$
15 = (-9 + C)(5) 3 = -9 + C C = 12$$15 = (-9 + C)(5) \implies 3 = -9 + C \implies C = 12$$
The function is y = (-x² + 12)(x² - 4)$y = (-x^2 + 12)(x^2 - 4)$.
Step 2: Maxima Calculation
To find the local maximum, differentiate and set y' = 0$y' = 0$:
y' = -2x(x² - 4) + 2x(-x² + 12) = -2x(x² - 4 + x² - 12) = -2x(2x² - 16) = -4x(x² - 8)$$y' = -2x(x^2 - 4) + 2x(-x^2 + 12) = -2x(x^2 - 4 + x^2 - 12) = -2x(2x^2 - 16) = -4x(x^2 - 8)$$
Critical points at x = 0$x = 0$ and x = ± 2√(2)$x = \pm 2\sqrt{2}$. Given domain x > 2$x > 2$, the valid critical point is x = 2√(2)$x = 2\sqrt{2}$.
Check double derivative or nature: It produces a maximum at x = 2√(2)$x = 2\sqrt{2}$.
Substitute x² = 8$x^2 = 8$ into the function for the maximum value:
ymax = (-8 + 12)(8 - 4) = (4)(4) = 16$$y_{\text{max}} = (-8 + 12)(8 - 4) = (4)(4) = 16$$
Pattern Recognition
Any differential equation framed like v u' - u v' = K$v u' - u v' = K$ hints directly at quotient rule exact differentials. Identifying this skips the entire linear integrating factor calculation.
Chapter Mix
Class 12 Maths: Differential Equations
Class 12 Maths: Application of Derivatives
