Let y = y(x) be the solution curve of the differential equation sin(2x²)logₑ(tan x²)dy + (4xy - 4√2x

Q: Why did we choose $t = x^2$? Couldn't we solve it directly?

Solving it directly would involve a nightmare of chain rules. By setting $t = x^2$, we transform a non-linear looking mess into a standard linear differential equation. Always look for the 'hidden' variable that simplifies the differential operator.

Q: How did the Integrating Factor become $\ln(\tan t)$ so easily?

It's all about the substitution $u = \ln(\tan t)$. When you differentiate $u$, you get $\frac{1}{\tan t} \cdot \sec^2 t = \frac{\cos t}{\sin t} \cdot \frac{1}{\cos^2 t} = \frac{1}{\sin t \cos t} = \frac{2}{\sin 2t}$. This matches the coefficient perfectly!

Q: Why do we reject the negative root or worry about the absolute value?

The problem asks for $|y(\sqrt{\frac{\pi}{3}})|$. In differential equations, the constant $C$ is determined by the initial condition. Once you have the specific solution, the absolute value is just a final step to ensure we provide the magnitude as requested.

Question

Let $y = y(x)$ be the solution curve of the differential equation $\sin(2x^2)\log_e(	an x^2)dy + (4xy - 4\sqrt{2}x\sin(x^2 - \frac{\pi}{4}))dx = 0, 0 < x < \sqrt{\frac{\pi}{2}}$, which passes through the point $(\sqrt{\frac{\pi}{6}}, 1)$. Then $|y(\sqrt{\frac{\pi}{3}})|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:

$$\sin(2x^2)\ln(	an x^2)dy + (4xy - 4\sqrt{2}x\sin(x^2 - \frac{\pi}{4}))dx = 0$$

Your eyes are drawn to the $x^2$ terms. They are everywhere. This is not a coincidence; it is a signal.

In the world of calculus, when you see a repeating argument like $x^2$, you must immediately think of substitution. Let us set $t = x^2$.

Differentiating this gives us $2x dx = dt$. Suddenly, the $x$ terms that were cluttering our equation are absorbed into the differential $dt$.

Our interval $0 < x < \sqrt{\frac{\pi}{2}}$ transforms into $0 < t < \frac{\pi}{2}$. The landscape is now familiar and manageable.

### The Transformation: Entering the Linear Realm

With our substitution in place, the equation sheds its skin. It transforms into:

$$\sin(2t) \ln(	an t) dy + (2y - 2\sqrt{2} \sin(t - \frac{\pi}{4})) dt = 0$$

We are now looking at a structure that screams for the standard linear form:

$$\frac{dy}{dt} + P(t)y = Q(t)$$

By dividing through, we isolate $\frac{dy}{dt}$ and identify our $P(t)$ as:

$$P(t) = \frac{2}{\sin(2t) \ln(	an t)}$$

This looks intimidating, but trust the process. The Integrating Factor, $I.F. = e^{\int P(t) dt}$, is the key that unlocks the door.

### The Magic of the Integrating Factor

This is the moment where the problem rewards your patience. To integrate $P(t)$, we use the substitution $u = \ln(	an t)$.

The derivative $du$ is $\frac{1}{	an t} \cdot \sec^2 t dt$, which simplifies perfectly to $\frac{2}{\sin 2t} dt$. The integral becomes:

$$\int \frac{du}{u} = \ln|u| = \ln|\ln(	an t)|$$

When we exponentiate this to find the $I.F.$, we get $e^{\ln|\ln(	an t)|}$, which is simply $\ln(	an t)$. The complexity vanishes, leaving us with a clean, elegant factor.

### The Final Integration and Conclusion

Now, we multiply the entire equation by our $I.F.$ and integrate. The term $Q(t)$ contains a $\ln(	an t)$ in the denominator, which cancels out our $I.F.$ entirely!

We are left with the integral of:

$$\int \frac{2\sqrt{2} \sin(t - \frac{\pi}{4})}{\sin 2t} dt$$

Expanding the numerator using the compound angle formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$ reveals the beautiful structure of $\sec t - \csc t$.

Integrating this is a standard result: $\ln|\sec t + 	an t| - \ln|\csc t - \cot t|$.

After applying our initial condition to find the constant $C$ and evaluating at $t = \frac{\pi}{3}$, we arrive at $y = -1$. The absolute value is **1**.

You see? The terror was just an illusion. You have mastered the beast.

Enter Numerical Value:

Visualized Solution

The Differential Equation

Substitution: $x^{2} = t$

Transforming the Equation

Standard Linear Form

Calculating the Integrating Factor

The General Solution Setup

Simplifying the Integrand

Integrating the RHS

Applying Initial Condition

Final Calculation for $y (\frac{π}{3})$

Solving for $y$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation

The Magic of the Integrating Factor

The Final Integration and Conclusion

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