If the solution curve $y = f(x)$ of the differential equation $(x^2 – 4)y’ – 2xy + 2x(4 – x^2)^2 = 0$, $x > 2$, passes through the point $(3, 15)$, then the local maximum value of $f$ is ________ .
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Solution Steps
1
Rearrange into standard Linear Form
The given equation is $(x^2 – 4)\frac{dy}{dx} – 2xy = -2x(4 – x^2)^2$.
Divide by $(x^2 – 4)$ to get: $\frac{dy}{dx} – \frac{2x}{x^2 – 4}y = -2x(x^2 – 4)$.
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
2
Find the Integrating Factor (I.F.)
$I.F. = e^{\int P(x) dx} = e^{\int -\frac{2x}{x^2 – 4} dx}$.
Let $t = x^2 – 4 \implies dt = 2x dx$. Then $I.F. = e^{-\ln(x^2 – 4)} = e^{\ln(x^2 – 4)^{-1}} = \frac{1}{x^2 – 4}$.
3
Write the General Solution
$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y \cdot \frac{1}{x^2 – 4} = \int -2x(x^2 – 4) \cdot \frac{1}{x^2 – 4} dx + C$.
$\frac{y}{x^2 – 4} = \int -2x dx + C = -x^2 + C$.
4
Determine the Constant $C$
The curve passes through $(3, 15)$. Substitute $x=3$ and $y=15$.
$\frac{15}{3^2 – 4} = -(3^2) + C \implies \frac{15}{5} = -9 + C$.
$3 = -9 + C \implies C = 12$.
5
Express $f(x)$ and find $f'(x)$
$f(x) = (x^2 – 4)(12 – x^2) = 12x^2 – x^4 – 48 + 4x^2 = -x^4 + 16x^2 – 48$.
To find the local maximum, differentiate: $f'(x) = -4x^3 + 32x$.
6
Find Critical Points and Maximize
Set $f'(x) = 0 \implies -4x(x^2 – 8) = 0$. Since $x > 2$, $x = \sqrt{8} = 2\sqrt{2}$.
Check second derivative: $f”(x) = -12x^2 + 32$. At $x = \sqrt{8}$, $f”(\sqrt{8}) = -12(8) + 32 = -96 + 32 = -64 < 0$.
Thus, local maximum occurs at $x^2 = 8$.
7
Calculate final Local Maximum Value
Substitute $x^2 = 8$ into $f(x) = (x^2 – 4)(12 – x^2)$.
$f_{max} = (8 – 4)(12 – 8) = 4 \times 4 = 16$.
Final Answer: 16
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Theory
1. First-Order Linear Differential Equations
A first-order linear differential equation is an equation that can be written in the form $dy/dx + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions of $x$. This form is unique because the dependent variable $y$ and its derivative $dy/dx$ appear only to the first power and are not multiplied together. Solving such equations involves finding an ‘Integrating Factor’ (I.F.) that allows the left-hand side to be expressed as the derivative of a product. This technique is a cornerstone of JEE calculus and is frequently combined with initial value problems or optimization tasks as seen in this problem.
2. The Role of the Integrating Factor (I.F.)
The Integrating Factor is defined as $e^{\int P(x) dx}$. When the entire linear differential equation is multiplied by this factor, the left-hand side becomes the exact derivative of the product $(y \times I.F.)$. This transformation simplifies the differential equation into a direct integration problem: $d/dx(y \cdot I.F.) = Q(x) \cdot I.F.$. It is vital to handle the integration of $P(x)$ carefully, especially when logs and signs are involved, as any error in the I.F. will lead to an incorrect solution curve and subsequent errors in finding local extrema.
3. Maxima and Minima of Functions
To find the local maximum of a function $f(x)$, we employ the First and Second Derivative Tests. First, we find the critical points by solving $f'(x) = 0$. Once critical points are identified, we use the Second Derivative Test: if $f”(c) < 0$ at a critical point $x=c$, then the function has a local maximum at that point. In competitive exams like JEE, after finding the solution curve from a differential equation, students are often asked to find extreme values, requiring a solid grasp of both integral and differential calculus concurrently.
4. Continuity and Domain Constraints
Domain constraints such as $x > 2$ are crucial in differential equations. They ensure that denominators (like $x^2 – 4$) are not zero and that functions like $\ln(x^2 – 4)$ are well-defined within the real number system. When solving for critical points, one must always cross-verify if the solution falls within the given domain. In this problem, $x = \pm \sqrt{8}$ and $x = 0$ were potential critical points, but the constraint $x > 2$ isolated $x = \sqrt{8}$ as the only valid point for investigation, highlighting the importance of domain analysis in complex calculus problems.
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FAQs
1
Why did we divide by $(x^2 – 4)$ at the start?
To transform the equation into the standard linear form $dy/dx + P(x)y = Q(x)$, the coefficient of $dy/dx$ must be 1.
2
How do I handle the negative sign in $P(x) = -2x/(x^2-4)$?
The sign is part of $P(x)$. When calculating $e^{\int P(x) dx}$, the negative sign results in $1/f(x)$ instead of $f(x)$ due to the property $e^{-\ln A} = 1/A$.
3
What if $x=2$?
The problem specifies $x > 2$. If $x=2$, the differential equation becomes singular because the coefficient of $y’$ vanishes.
4
Can I use the substitution $x^2 – 4 = t$ for the whole DE?
It helps in integrating the I.F. and the $Q(x)$ part, but you still need to follow the linear DE method to find the relationship between $y$ and $x$.
5
Is $(3, 15)$ the local maximum?
No, $(3, 15)$ is just a point on the curve used to find the constant $C$. The local maximum occurs at $x = \sqrt{8}$.
6
What is the sign of $f”(x)$ at the critical point?
The second derivative $f”(x) = -12x^2 + 32$. At $x^2=8$, it is $-96+32 = -64$, which is negative, confirming a maximum.
7
How do I integrate $-2x$ easily?
The integral of $x^n$ is $x^{n+1}/(n+1)$. For $-2x^1$, it is $-2(x^2/2) = -x^2$.
8
Why did $Q(x)$ become $-2x(x^2-4)$?
The original term was $2x(4-x^2)^2$. Dividing by $(x^2-4)$ gives $2x(4-x^2)^2 / -(4-x^2) = -2x(4-x^2) = 2x(x^2-4)$.
9
Is this problem common in JEE Main?
Yes, combining differential equations with application of derivatives (maxima/minima) is a very frequent pattern in JEE Main and Advanced.
10
What if I got $C=9$?
You likely made a calculation error in step 4. Double check the substitution $15/5 = -9 + C$.
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