Let y = y(x) be the solution of the differential equation cos x dy/dx + 2y sin x = sin 2x, x ∈ (0, π

Q: Wait, why do we divide by $\cos x$ at the start? Can't we just integrate directly?

Great question! We divide by $\cos x$ because the standard form of a linear differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$, requires the coefficient of $\frac{dy}{dx}$ to be $1$. Without this normalization, the standard formula for the Integrating Factor would not apply. Think of it as preparing the soil before planting the seed.

Q: I got stuck on the integral $\int 2 \sin x \sec^2 x dx$. How do I know to convert it to $\int 2 \tan x \sec x dx$?

That is a classic hurdle! When you see a product of trigonometric functions, look for relationships between them. Since $\sec^2 x = \frac{1}{\cos^2 x}$, you can split it into $\frac{1}{\cos x} \cdot \frac{1}{\cos x}$. Pairing one $\frac{1}{\cos x}$ with $\sin x$ gives you $\tan x$, and the remaining $\frac{1}{\cos x}$ is $\sec x$. It is all about finding the hidden structure!

Q: Why do we ignore the absolute value in $\ln |\sec x|$?

Excellent observation. The problem specifies that $x \in (0, \frac{\pi}{2})$. In this interval, $\cos x$ is positive, which means $\sec x$ is also positive. Since $\sec x > 0$ for all $x$ in this range, the absolute value is redundant. We can safely write $\ln(\sec x)$.

Question

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$, $x \in \left(0, \frac{\pi}{2}\right)$. If $y(\pi/3) = 0$, then $y(\pi/4)$ is equal to :

Options:

(A) $\frac{1}{\sqrt{2}}-1$

(B) $\sqrt{2}-2$

(C) $2-\sqrt{2}$

(D) $2+\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Differential Equations

Differential equations are the heartbeat of physics. They describe how things change, how systems evolve, and how the universe unfolds.

When you first look at the equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$, it might seem like a chaotic jumble of trigonometric functions. But I want you to see past the complexity and recognize the order waiting to be revealed.

### Phase 1: The Art of Normalization

Every linear differential equation has a standard form: $\frac{dy}{dx} + P(x)y = Q(x)$. Our given equation is currently wearing a disguise, as the term $\cos x$ attached to $\frac{dy}{dx}$ is an obstacle.

To clear our path, we divide the entire equation by $\cos x$. This simple, decisive action transforms our equation into:

$$ \frac{dy}{dx} + \frac{2 \sin x}{\cos x} y = \frac{\sin 2x}{\cos x} $$

Using the identity $\sin 2x = 2 \sin x \cos x$, the right side simplifies beautifully. The $\cos x$ terms cancel out, leaving us with $2 \sin x$.

Now, our equation stands in its true, elegant form:

$$ \frac{dy}{dx} + (2 	an x) y = 2 \sin x $$

### Phase 2: The Secret Weapon

Now that we have identified $P(x) = 2 	an x$, we need our secret weapon: the Integrating Factor ($I.F.$). The $I.F.$ is the bridge that allows us to integrate both sides of the equation, defined as $e^{\int P(x) dx}$.

Calculating this, we get $e^{\int 2 	an x dx}$. Since the integral of $	an x$ is $\ln(\sec x)$, our exponent becomes $2 \ln(\sec x)$.

Using the properties of logarithms, this is $\ln(\sec^2 x)$. When we raise $e$ to this power, the $e$ and the $\ln$ cancel out, leaving us with a clean, powerful **$I.F. = \sec^2 x$**.

### Phase 3: The Integration Journey

With our $I.F.$ in hand, the general solution is within reach using the formula $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx$. Substituting our values, we get:

$$ y \sec^2 x = \int (2 \sin x) \sec^2 x dx $$

Do not let this integral intimidate you. By rewriting $\sec^2 x$ as $\frac{1}{\cos^2 x}$, we see the integral as $\int 2 \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} dx$, which is simply $\int 2 	an x \sec x dx$.

This is a standard integral! Since the derivative of $\sec x$ is $\sec x 	an x$, the integral is $2 \sec x + C$. Our general solution is now:

$$ y \sec^2 x = 2 \sec x + C $$

### Phase 4: Finding the Particular Path

We have a family of solutions, but we need the specific one that satisfies the condition $y(\frac{\pi}{3}) = 0$. Substituting $x = \frac{\pi}{3}$ and $y = 0$ into our equation:

$$ 0 \cdot \sec^2\left(\frac{\pi}{3}ight) = 2 \sec\left(\frac{\pi}{3}ight) + C $$

Since $\sec(\frac{\pi}{3}) = 2$, we get $0 = 2(2) + C$, which leads us directly to $C = -4$. Our particular solution is $y \sec^2 x = 2 \sec x - 4$.

### The Final Victory

Finally, we find $y(\frac{\pi}{4})$. Substituting $x = \frac{\pi}{4}$ into our particular solution:

$$ y\left(\frac{\pi}{4}ight) \cdot (\sqrt{2})^2 = 2(\sqrt{2}) - 4 $$

$$ 2 y\left(\frac{\pi}{4}ight) = 2\sqrt{2} - 4 $$

$$ y\left(\frac{\pi}{4}ight) = \sqrt{2} - 2 $$

And there it is. Through systematic steps and a bit of trigonometric grace, we have arrived at the final answer: **$y(\frac{\pi}{4}) = \sqrt{2} - 2$**.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Select Answer:

Visualized Solution

Identifying the Differential Equation

Normalizing the Equation

Simplifying the Right Hand Side

Calculating the Integrating Factor $(I . F .)$

Setting up the General Solution

Solving the Integral

Finding the Constant $C$

The Particular Solution

Calculating $y (\frac{π}{4})$

Final Result

The Sigma Insight: Linear Differential Equations

The Hidden Elegance of Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

If $y = y (x), x \in (0, \frac{π}{2})$ be the solution curve of the differential equation $(sin^{2} 2 x) \frac{d y}{d x} + (8 sin^{2} 2 x + 2 sin 4 x) y = 2 e^{- 4 x} (2 sin 2 x + cos 2 x)$ , with $y (\frac{π}{4}) = e^{- π}$ , then $y (\frac{π}{6})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cos x (lo g_{e} (cos x))^{2} d y + (sin x - 3 y sin x lo g_{e} (cos x)) d x = 0$ $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{l o g _{e} 2}$ , then $y (\frac{π}{6})$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y, y (1) = 0$ . Then $y (3)$ is equal to :