Solution of the differential equation cos x dy = y(sin x - y) dx, 0

Q: Wait, why can't I just separate the variables $x$ and $y$ right at the start? It looks like it might work.

That is a very common temptation! If you try to separate variables, you will find that the $y$ terms are inextricably linked to the $x$ terms inside the parenthesis $(\sin x - y)$. Because of that subtraction, you cannot isolate all $y$ terms on one side and all $x$ terms on the other. This is exactly why we need the Bernoulli method—it is the tool designed specifically to break that coupling.

Q: Why did we choose $t = -\frac{1}{y}$? Why not just $t = \frac{1}{y}$?

Great observation! If you choose $t = \frac{1}{y}$, then $\frac{dt}{dx} = -\frac{1}{y^2} \frac{dy}{dx}$. If you use this, you will end up with a negative sign in front of your $\frac{dt}{dx}$ term in the final linear equation. While it is mathematically correct, choosing $t = -\frac{1}{y}$ makes the coefficient of $\frac{dt}{dx}$ positive, which makes the subsequent integration much cleaner and less prone to sign errors. It is a strategic choice to save you from yourself!

Q: In the final step, we have $-c$. Can I just write $+c$ instead? Does it matter?

In the context of an indefinite integral, $c$ is an arbitrary constant. Whether you write $+c$ or $-c$, it represents 'some constant.' When we multiply by $-y$ and get a term like $-c$, we can simply redefine it as a new constant, say $C$. As long as you acknowledge that the constant is arbitrary, you are mathematically safe. JEE examiners look for the structure of the solution, not the specific sign of the constant.

Question

Solution of the differential equation $\cos x dy = y(\sin x - y) dx, 0 < x < \frac{\pi}{2}$ is

Options:

(A) $y \sec x = 	an x + c$

(B) $y 	an x = \sec x + c$

(C) $	an x = (\sec x + c)y$

(D) $\sec x = (	an x + c)y$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem before us is $\cos x \, dy = y(\sin x - y) \, dx$. At first glance, it looks like a tangled mess of trigonometric functions and variables.

In the world of JEE Advanced, we do not fear the mess; we embrace it. We look for the pattern hidden within the chaos.

### Phase 1: The Standard Form

Our first instinct is to bring order to this equation. We want to see the derivative $\frac{dy}{dx}$ clearly. By dividing both sides by $\cos x \, dx$, we isolate our derivative:

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

Distributing the denominator, we get $\frac{dy}{dx} = y 	an x - y^2 \sec x$. Now, let us rearrange this into a form that feels familiar. Moving the $y$ term to the left, we obtain:

$$ \frac{dy}{dx} - y 	an x = -y^2 \sec x $$

Stop here for a moment. Look at the right-hand side. That $y^2$ term is the intruder. If it were not there, this would be a simple linear differential equation. But it is there, and that makes this a **Bernoulli equation**.

### Phase 2: The Bernoulli Strategy

A Bernoulli equation takes the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$. In our case, $n=2$. This is the classic trap.

Many students try to use separation of variables here and fail. The key to unlocking this is to eliminate the non-linearity. We divide the entire equation by $y^2$, yielding:

$$ \frac{1}{y^2} \frac{dy}{dx} - \frac{1}{y} 	an x = -\sec x $$

This looks strange, but it is the gateway to the solution.

### Phase 3: The Magic of Substitution

Now, we perform the substitution that changes everything. Let $t = -\frac{1}{y}$.

When we differentiate $t$ with respect to $x$, we get $\frac{dt}{dx} = \frac{1}{y^2} \frac{dy}{dx}$. Look at our equation again; that term $\frac{1}{y^2} \frac{dy}{dx}$ is sitting right there, waiting for us!

Substituting $t$ and $\frac{dt}{dx}$ back into our equation, we get:

$$ \frac{dt}{dx} + t 	an x = -\sec x $$

Suddenly, the complexity vanishes. We have transformed a non-linear nightmare into a beautiful, standard linear differential equation.

### Phase 4: The Integrating Factor

We are now in the home stretch. For a linear equation $\frac{dt}{dx} + P(x)t = Q(x)$, the solution relies on the Integrating Factor (I.F.).

Here, $P(x) = 	an x$. The I.F. is defined as:

$$ I.F. = e^{\int P(x) \, dx} = e^{\int 	an x \, dx} $$

We know that the integral of $	an x$ is $\ln|\sec x|$. Thus, our I.F. is $e^{\ln|\sec x|} = \sec x$. This is the magic key that unlocks the equation.

### Phase 5: The Final Integration

We multiply our linear equation by the I.F.:

$$ \sec x \frac{dt}{dx} + t \sec x 	an x = -\sec^2 x $$

The left side is now the derivative of the product $t \sec x$. So, we have:

$$ \frac{d}{dx}(t \sec x) = -\sec^2 x $$

Integrating both sides with respect to $x$, we get $t \sec x = -\int \sec^2 x \, dx$. We know the integral of $\sec^2 x$ is $	an x$. Therefore:

$$ t \sec x = -	an x + c $$

Finally, we substitute $t = -\frac{1}{y}$ back into the equation:

$$ -\frac{1}{y} \sec x = -	an x + c $$

Multiplying by $-y$, we arrive at $\sec x = y(	an x - c)$. By redefining the constant $-c$ as $C$, we reach our elegant final answer:

**$\sec x = y(	an x + C)$**

You have conquered the Bernoulli equation. Remember, the math is not just about the steps; it is about the transformation from confusion to clarity. Keep practicing, and keep believing in your ability to solve the unsolvable.

Animated Solution for Mathematics - Differential Equations: Solution of the differential equation $cos x d y = y (sin x - y) d x, 0 < x < \frac{π}{2}$ is

Select Answer:

Visualized Solution

Initial Differential Equation

Identifying Bernoulli's Equation

Dividing by $y^{2}$

Substitution $t = - \frac{1}{y}$

Linear Differential Equation

Calculating Integrating Factor (I.F.)

Solving the Linear Equation

Evaluating the Integral

Final Solution

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

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:

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

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

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

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 :

If $cos x \frac{d y}{d x} - y sin x = 6 x, (0 < x < \frac{π}{2})$ and $y (\frac{π}{3}) = 0$ , then $y (\frac{π}{6})$ is equal to :-

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solution of the differential equation cosxdy=y(sinx−y)dx,0<x<2π​ is