If the curve y = y(x) is the solution of the differential equation 2(x² + x^(5/4)) dy - y(x + x^(1/4

Q: Wait, why did we factor out $x^{5/4}$ from the denominator? It feels like guessing.

It is not guessing, it is pattern recognition! In JEE Advanced, whenever you see fractional powers like $x^{5/4}$ and $x^{1/4}$, look for the greatest common divisor of the exponents. By factoring out $x^{5/4}$, we reveal the term $(x^{3/4} + 1)$ in the denominator, which perfectly matches the structure of the numerator after we multiply by the Integrating Factor. Always look for these 'hidden symmetries'!

Q: Why can't we just solve this using variable separation?

Variable separation only works if you can write the equation as $f(y)dy = g(x)dx$. Here, the $y$ term is tied to an $x$-dependent coefficient, and there is an additive term on the right side. This is a classic linear differential equation. Trying to separate variables here would lead to an algebraic dead end.

Q: How do I know when to use the substitution $t = x^{1/4}$?

Great question! Look at the powers of $x$ in your integral. We had $x^{1/2}$ in the numerator and $x^{3/4}$ in the denominator. The common denominator of these fractional exponents is 4. Setting $t = x^{1/4}$ is the standard 'rationalizing substitution' that clears away all fractional powers, turning the expression into a standard rational function.

Q: Is there a shortcut for the integral $\int \frac{t^2}{t^3 + 1} dt$?

Yes! Notice that the derivative of the denominator $(t^3 + 1)$ is $3t^2$. Since we have $t^2$ in the numerator, this is a perfect candidate for the $\int \frac{f'(t)}{f(t)} dt = \ln|f(t)|$ rule. You don't need partial fractions here; just multiply and divide by 3.

Q: I got the constant $C$ wrong. How do I handle the $\ln 2$ term?

When you plug in $x=1$, the term $\ln(1^{3/4} + 1)$ becomes $\ln(2)$. Since your initial condition also contains a $\ln 2$ term, they will naturally cancel out or combine. If you find yourself with a messy constant, re-check your arithmetic on the $x=1$ substitution—it is designed to simplify beautifully.

Question

If the curve $y = y(x)$ is the solution of the differential equation $2(x^2 + x^{5/4}) dy - y(x + x^{1/4}) dx = 2x^{9/4} dx, x > 0$ which passes through the point $(1, 1 - \frac{4}{3} \log_e 2)$, then the value of $y(16)$ is equal to :

Options:

(A) $4\left(\frac{31}{3} + \frac{8}{3} \log_e 3\right)$

(B) $\left(\frac{31}{3} + \frac{8}{3} \log_e 3\right)$

(C) $4\left(\frac{31}{3} - \frac{8}{3} \log_e 3\right)$

(D) $\left(\frac{31}{3} - \frac{8}{3} \log_e 3\right)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at the equation $2(x^2 + x^{5/4}) dy - y(x + x^{1/4}) dx = 2x^{9/4} dx$, it is natural to feel a sense of intimidation. The fractional powers $x^{5/4}$ and $x^{9/4}$ seem designed to obscure the path.

However, in the world of advanced mathematics, complexity is often just a mask for underlying elegance. Our first mission is to bring this equation into the light of the standard linear form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

By rearranging the terms, we isolate $\frac{dy}{dx}$ and find ourselves staring at:

$$P(x) = -\frac{x + x^{1/4}}{2(x^2 + x^{5/4})}$$

### The Art of Simplification

Here is where the magic happens. If you look closely at the denominator, you can factor out $x^{5/4}$ to reveal $(x^{3/4} + 1)$.

Similarly, the numerator can be simplified by factoring out $x^{1/4}$. Suddenly, the terms collapse like a house of cards:

$$P(x) = -\frac{x^{1/4}(x^{3/4} + 1)}{2x^{5/4}(x^{3/4} + 1)} = -\frac{1}{2x}$$

The $(x^{3/4} + 1)$ terms cancel out entirely, leaving us with the beautifully simple $P(x) = -\frac{1}{2x}$. This is the moment you realize the problem is not a monster; it is a puzzle waiting to be solved.

### The Integrating Factor

With $P(x) = -\frac{1}{2x}$, finding the Integrating Factor (I.F.) becomes a walk in the park. We compute:

$$I.F. = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \ln x} = x^{-1/2}$$

This factor is the key that unlocks the entire equation. When we multiply our original equation by this $I.F.$, the left side becomes the derivative of a product:

$$\frac{d}{dx}(y \cdot x^{-1/2}) = \frac{x^{-1/2} \cdot 2x^{9/4}}{2(x^2 + x^{5/4})} = \frac{x^{7/4}}{x^{5/4}(x^{3/4} + 1)} = \frac{x^{1/2}}{x^{3/4} + 1}$$

On the right side, we are left with the integral $\int \frac{x^{1/2}}{x^{3/4} + 1} dx$. We use the substitution $t = x^{1/4}$, which implies $dt = \frac{1}{4} x^{-3/4} dx$, or $dx = 4t^3 dt$. This transforms our integral into:

$$4 \int \frac{t^2 \cdot t^3}{t^3 + 1} dt = 4 \int \frac{t^5}{t^3 + 1} dt$$

### The Algebraic Masterstroke

Now, we face the integral of a rational function. We use the classic trick of adding and subtracting terms in the numerator:

$$\frac{t^5}{t^3 + 1} = \frac{t^2(t^3 + 1) - t^2}{t^3 + 1} = t^2 - \frac{t^2}{t^3 + 1}$$

This splits our integral into two parts. The first part, $\int t^2 dt$, is trivial. The second part, $\int \frac{t^2}{t^3 + 1} dt$, is a perfect logarithmic derivative.

After integrating and back-substituting $t = x^{1/4}$, we arrive at the general solution:

$$y \cdot x^{-1/2} = \frac{4}{3} x^{3/4} - \frac{4}{3} \ln(x^{3/4} + 1) + C$$

### Final Calculation

We are given the initial condition $(1, 1 - \frac{4}{3} \ln 2)$. Substituting these values allows us to find the constant $C$:

$$1 \cdot (1)^{-1/2} = \frac{4}{3}(1) - \frac{4}{3} \ln(2) + C \Rightarrow C = -\frac{1}{3}$$

Finally, we set our sights on the target: $y(16)$. By plugging in $x = 16$, we note that $16^{3/4} = 8$ and $16^{-1/2} = \frac{1}{4}$:

$$\frac{1}{4} y(16) = \frac{4}{3}(8) - \frac{4}{3} \ln(8 + 1) - \frac{1}{3}$$

$$\frac{1}{4} y(16) = \frac{32}{3} - \frac{4}{3} \ln 9 - \frac{1}{3} = \frac{31}{3} - \frac{8}{3} \ln 3$$

The final resulting answer is:

**$$y(16) = 4 \left( \frac{31}{3} - \frac{8}{3} \ln 3 \right)$$**

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Visualized Solution

$S t an d a r d L in e a r F or m$

$S im pl i f y P (x)$

$S im pl i f y Q (x)$

$I n t e g r a t in g F a c t or (I . F .)$

$G e n er a l S o l u t i o n S e t u p$

$I n t e g r a l S u b s t i t u t i o n$

$A l g e b r ai c M ani p u l a t i o n$

$I n t e g r a t i o n & B a c k - S u b s t i t u t i o n$

$A ppl y in g I ni t ia l C o n d i t i o n s$

$S u b s t i t u t e x = 16$

$S im pl i f y t h e E x p r ess i o n$

$F ina l A n s w er$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Simplification

The Integrating Factor

The Algebraic Masterstroke

Final Calculation

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Visualized Solution

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