Problem
For x ∈ R, let y(x) be a solution of the differential equation (x² – 5)dy/dx – 2xy = -2x(x² – 5)² such that y(2) = 7. Then, find the maximum value of the function y(x).
Solution
On dividing both sides by (x² – 5), we’ll get dy/dx – 2xy/(x² – 5) = -2x(x² – 5), which is a linear differential equation.
The integration factor is \( e^{\int -\frac{2x}{x^2-5} dx} \) which equals 1/(x² – 5).
So, the solution looks something like y/(x² – 5) = ∫-2xdx or y/(x² – 5) = -x² + c.
Using the condition y(2) = 7, we’ll get the value of c as -3.
Finally, we’ll get y(x) = (5 – x²)(x² + 3).
Now, finding its maximum value is quite straightforward. Two methods to discuss, still.
Method 1
Here, we’ll differentiate y(x), and find the points of its local maxima.
y'(x) comes out to be -4x(x² – 1), which has three roots: 0, 1, and -1.
Now, we’ll find the value of y”(x) at the above values. y”(x) equals -12x² + 4, which is negative at 1 and -1.
So, the maximum value occurs at 1 and -1, which is (5 – 1)(1 + 3) or 16.
Method 2
Here, let’s assume x² as t. So the function becomes (5 – t)(t + 3), which is a quadratic function in t.
Since it’s a downward parabola (the coefficient of t being negative), its maximum value occurs at the midpoint of the roots, i.e. at t = (5 + (-3))/2, or t = 1.
Substituting that, we’ll get the required value as (5 – 1)(1 + 3) or 16 – same as before.
Comments
A rather boring problem – not quite worthy of the JEE Advanced. Finding the maximum value was the only hope to make things a bit more challenging. But that was a letdown too.
Instead, had the solution been something like y(x) = (3 – x²)(x² + 5), there would’ve been atleast 10% more spice in the problem. In this case, the maximum value wouldn’t have occurred at t = -1 (method 2), since that value would be unattainable by x². How would we maximize this, if we were to avoid differentiation?
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