Determine the equation of the curve passing through the origin, in the form y = f(x), which satisfie

Q: Wait, why can't we just integrate $\sin(10x + 6y)$ directly? Why do we need this $v$ substitution?

That is a brilliant question! You cannot integrate $\sin(10x + 6y)$ directly with respect to $x$ because $y$ is also a function of $x$. You have a 'nested' dependency where the variable of integration is trapped inside the function. The substitution $v = 10x + 6y$ is the only way to decouple $y$ from the argument of the sine function, allowing us to treat the entire expression as a single, manageable variable.

Q: Why use the half-angle substitution $t = \tan(v/2)$? Is there a faster way?

While there are other methods, the half-angle substitution is the 'universal key' for integrals involving $\sin v$ and $\cos v$. It is robust, systematic, and guaranteed to convert the trigonometric mess into a rational algebraic function. In JEE Advanced, reliability is just as important as speed. It turns a potentially confusing trig integral into a standard quadratic form that you have solved dozens of times before.

Q: Why do we have to solve for $C$ using the origin $(0,0)$? Can't we just leave it as an arbitrary constant?

The problem asks for 'the' equation of the curve passing through the origin. This is a boundary value problem. Without determining $C$, you have an infinite family of curves. By plugging in $(0,0)$, you are selecting the specific member of that family that satisfies the physical constraint given in the problem. It is the difference between a general solution and the specific reality of the curve.

Question

Determine the equation of the curve passing through the origin, in the form $y = f(x)$, which satisfies the differential equation $\frac{dy}{dx} = \sin(10x + 6y)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a differential equation; we are embarking on a journey to uncover a hidden geometric truth.

When you first look at the equation

$$\frac{dy}{dx} = \sin(10x + 6y)$$

it might feel intimidating. It is a non-linear, first-order differential equation where the variables $x$ and $y$ are inextricably tangled inside a trigonometric function.

But here is the secret of the JEE Advanced: complexity is often just a mask for a deeper, simpler structure waiting to be revealed.

### The Art of the Substitution

Imagine you are standing in front of a locked door. You have the key, but you don't know it yet. The key here is the argument of the sine function: $10x + 6y$.

Notice how it is a linear combination? This is a classic signature in differential equations. Whenever you see $\frac{dy}{dx} = f(ax + by + c)$, your intuition should immediately scream: 'Substitution!'

We define a new variable, $v = 10x + 6y$. By doing this, we are essentially changing our perspective. We are no longer looking at the curve in terms of $x$ and $y$ alone, but in terms of a new, unified variable $v$.

To make this transition complete, we must also transform our derivative. Differentiating $v$ with respect to $x$, we get:

$$\frac{dv}{dx} = 10 + 6\frac{dy}{dx}$$

Rearranging this gives us the bridge we need:

$$\frac{dy}{dx} = \frac{1}{6}\left(\frac{dv}{dx} - 10ight)$$

### The Transformation

Now, watch the magic happen. We substitute our new expressions back into the original equation. The equation $\frac{dy}{dx} = \sin(10x + 6y)$ transforms into:

$$\frac{1}{6}\left(\frac{dv}{dx} - 10ight) = \sin v$$

With a little algebraic housekeeping—multiplying by $6$ and adding $10$ to both sides—we arrive at:

$$\frac{dv}{dx} = 10 + 6\sin v$$

Look at that! We have successfully separated the variables. We have moved from a chaotic, tangled mess to the elegant, separable form:

$$\int \frac{dv}{10 + 6\sin v} = \int dx$$

The right side is trivial, but the left side? That is where the real test of your mathematical endurance lies.

### The Calculus Battle

To solve $\int \frac{dv}{10 + 6\sin v}$, we reach for the Weierstrass substitution, the 'surgical tool' of integration. We set $t = 	an\left(\frac{v}{2}ight)$.

This substitution is powerful because it converts trigonometric functions into rational algebraic functions. Using the identity $\sin v = \frac{2t}{1 + t^2}$ and $dv = \frac{2dt}{1 + t^2}$, the integral becomes:

$$\int \frac{\frac{2dt}{1 + t^2}}{10 + 6\left(\frac{2t}{1 + t^2}ight)}$$

When you simplify this, the $(1 + t^2)$ terms cancel out beautifully, leaving you with:

$$\int \frac{2dt}{10t^2 + 12t + 10} = \int \frac{dt}{5t^2 + 6t + 5}$$

This is a standard integral! By completing the square in the denominator, we transform it into:

$$\frac{1}{5} \int \frac{dt}{\left(t + \frac{3}{5}ight)^2 + \left(\frac{4}{5}ight)^2}$$

This is the classic form $\int \frac{du}{u^2 + a^2} = \frac{1}{a}	an^{-1}\left(\frac{u}{a}ight)$.

### The Final Polish

After integrating, we obtain:

$$\frac{1}{4} 	an^{-1}\left(\frac{5t + 3}{4}ight) = x + C$$

Now, we apply the boundary condition. The curve passes through the origin $(0,0)$, which means when $x = 0$, $y = 0$, and consequently $v = 0$ and $t = 0$. Substituting these values, we find:

$$C = \frac{1}{4} 	an^{-1}\left(\frac{3}{4}ight)$$

Finally, we substitute $C$ back and use the inverse trigonometric identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A - B}{1 + AB}ight)$ to combine the terms. After some careful algebraic manipulation to isolate $t$, and then substituting $t = 	an\left(\frac{10x + 6y}{2}ight)$ back to find $y$, we arrive at our destination:

**$$y = \frac{1}{3} \left[	an^{-1} \left(\frac{5 	an 4x}{4 - 3 	an 4x}ight) - 5xight]$$**

This result is more than just a string of symbols. It is the path of a particle, the shape of a curve, and the proof of your persistence. You started with a daunting differential equation and, through the power of substitution and calculus, you tamed it.

Animated Solution for Mathematics - Differential Equations: Determine the equation of the curve passing through the origin, in the form $y = f (x)$ , which satisfies the differential equation $\frac{d y}{d x} = sin (10 x + 6 y)$ .

Visualized Solution

Identifying the Equation $\frac{d y}{d x} = sin (10 x + 6 y)$

Substituting $v = 10 x + 6 y$

Differentiating $v$ with respect to $x$

Transforming the Differential Equation

Separating the Variables

Using Half-Angle Substitution

Simplifying the Integral

Completing the Square

Evaluating the Integral

Applying Boundary Condition $(0, 0)$

Simplifying with Inverse Trig Identities

Solving for $t$ in terms of $x$

Isolating $t$

Final Expression for $y = f (x)$

Key Takeaways

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of the Substitution

The Transformation

The Calculus Battle

The Final Polish

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Visualized Solution

Identifying the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=sin(10x+6y)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=10x+6y

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x