The Solution of the Differential Equation cos x·dy = y(sin x − y)dx, for 0 < x < π/2, Is

AIEEE 2010. Previous Year Question.

Problem

The solution of the differential equation $\cos x\cdot dy = y(\sin x-y)dx$, for $0<x<\dfrac{\pi}{2}$, is:

Key insight. Once rewritten as $\dfrac{dy}{dx}-y\tan x=-y^2\sec x$, this equation has $y$ appearing with two different powers ($y^1$ and $y^2$) — the signature of a Bernoulli equation. Dividing through by $y^2$ and substituting $t=\dfrac{1}{y}$ converts it into an ordinary linear differential equation in $t$, solvable with the standard integrating factor method.

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Approach

Rewrite the equation in $\dfrac{dy}{dx}$ form and recognise it as a Bernoulli equation in $y$. Divide through by $y^2$ and substitute $t=\dfrac{1}{y}$ to convert it into a linear differential equation in $t$. Find the integrating factor, solve for $t$, and substitute back to express the answer in terms of $y$.

Solution

Step 1 — Rewrite in dy/dx form

$$\cos x\, dy = y(\sin x – y)\,dx \implies \frac{dy}{dx} = \frac{y\sin x – y^2}{\cos x} = y\tan x – y^2\sec x$$

Step 2 — Identify the Bernoulli equation

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

Step 3 — Divide through by y² and substitute t = 1/y

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

With $t=\dfrac{1}{y}$, $\dfrac{dt}{dx} = -\dfrac{1}{y^2}\dfrac{dy}{dx}$:

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

Step 4 — Find the integrating factor

$$\text{I.F.} = e^{\int\tan x\,dx} = e^{\ln|\sec x|} = \sec x$$

Step 5 — Solve the linear equation in t

$$t\cdot\sec x = \int \sec x\cdot\sec x\,dx = \int\sec^2x\,dx = \tan x + c$$

Step 6 — Substitute back t = 1/y

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

$$\sec x = y(\tan x+c)$$

Answer

Common mistakes

• Not recognising the Bernoulli structure and trying to treat the equation as directly separable or linear in $y$. The presence of both $y$ and $y^2$ on the right-hand side is what signals the Bernoulli substitution is needed — attempting standard linear methods on the original equation in $y$ won’t work.
• Forgetting the negative sign when converting $\dfrac{d}{dx}\left(\dfrac{1}{y}\right)$ to $\dfrac{dt}{dx}$. Since $\dfrac{dt}{dx}=-\dfrac{1}{y^2}\dfrac{dy}{dx}$, dividing the original equation by $y^2$ and substituting requires carefully tracking this sign flip to land on the correct linear equation in $t$.

Practise next

• Solve $\sin x\cdot dy = y(\cos x – y)dx$ for $0<x<\dfrac{\pi}{2}$, using the same Bernoulli substitution $t=\dfrac{1}{y}$.