Q.Solve the following differential equation: $\cos \left(\frac{dy}{dx}\right) = a \ (a \in \mathbf{R}); y = 1$ when $x = 0$

The key idea is that the equation is already in the form $\cos(p) = a$, where $p = \frac{dy}{dx}$, so $p$ is constant. This is a special case of Clairaut’s equation (or simply a first-order equation with constant slope).

Since $\cos(p) = a$, we have $p = \cos^{-1}(a)$, a constant. Integrating: …

The equation $\cos(y') = a$ looks unusual — it’s not a polynomial in $y'$, but a transcendental one. The key insight: the derivative $y' = dy/dx$ appears only inside the cosine, and the right-hand side is a constant $a$. This means $y'$ itself must be constant, because the cosine function is not one-to-one over all reals, but for a given $a$, the equation $\cos(t) = a$ has a fixed set of solutions for $t$.

Let’s work through it.

1. Recognize the form.

   The equation is $\cos(y') = a$, where $a \in \mathbb{R}$. Since $a$ is a constant, the value of $y'$ is not free to vary with $x$ or $y$ — it must be one of the angles whose cosine equals $a$. So $y'$ is constant.

   Write $y' = c$, where $c$ satisfies $\cos c = a$.

   Note

   For a real solution $c$ to exist, we need $|a| \le 1$. If $|a| > 1$, the equation has no real solution. The problem likely assumes $a$ is such that a real solution exists.

2. Solve the trivial ODE.

   If $y' = c$ (constant), then integrating gives

where $b$ is the constant of integration. This is the general solution — a family of straight lines.

3. Apply the initial condition. We are given $y = 1$ when $x = 0$. Substitute:

So the particular solution is

where $c$ is any real number such that $\cos c = a$.

4. Express $c$ explicitly. The equation $\cos c = a$ has infinitely many solutions: $c = \pm \arccos(a) + 2n\pi$, $n \in \mathbb{Z}$. But note: $y' = c$ is the slope of the line. All these different $c$ values give different slopes, but they all satisfy the original differential equation because $\cos(c) = a$ for each. So the general solution is actually a family of families: …

Method: Solving an equation given implicitly in $\frac{dy}{dx}$

Use this when the derivative is wrapped inside a function, as in $\cos\!\big(\frac{dy}{dx}\big)=a$ — you first unwrap it to get $\frac{dy}{dx}$ explicitly, then integrate.

Steps

Step 1: Unwrap the derivative

Apply the inverse function to isolate the slope:

a constant, since $a$ is a fixed real number.

Step 2: Integrate directly

With a constant slope, $y=(\cos^{-1}a)\,x + C$ — a straight line whose gradient is $\cos^{-1}a$. …

Common Mistakes

Mistake 1: Not isolating $\dfrac{dy}{dx}$ from inside the cosine first

Why it's wrong: you must apply $\cos^{-1}$ to both sides to get $\frac{dy}{dx}=\cos^{-1}a$ before integrating; integrating $\cos(y')$ term-by-term is meaningless. Correct approach: unwrap the derivative, then integrate.

Mistake 2: Treating $\cos^{-1}a$ as a variable …