Question

The general solution of $\sin^{-1}\left(\frac{dy}{dx}\right) = x + y$ is

• A. $\tan(x+y) - \sec(x+y) = x^2 + c$
• B. $\tan(x+y) + \sec(x+y) = x^2 + c$
• C. $\tan(x+y) + \sec(x+y) = x + c$
• D. $\tan(x+y) - \sec(x+y) = x + c$

Hint:

When standard substitution leads to an integral of the form $\int \frac{1}{1+\sin t} dt$, remember its standard evaluation shorthand: $$\int \frac{1}{1+\sin t} dt = \tan t - \sec t$$ Memorizing common trigonometric integral outcomes helps eliminate multiple algebraic steps during the exam.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a first-order differential equation where the variables $x$ and $y$ are tied together inside an inverse trigonometric expression. We need to find its general solution.

Step 2: Key Formula or Approach:
First, rewrite the equation by isolating the derivative term to get $\frac{dy}{dx} = \sin(x+y)$. Since the terms are in the form $x+y$, we will use the method of substitution by setting $t = x+y$, converting it into a variable separable form.

Step 3: Detailed Explanation:
Given equation: $$\sin^{-1}\left(\frac{dy}{dx}\right) = x + y \Rightarrow \frac{dy}{dx} = \sin(x+y)$$ Let $x + y = t$. Differentiating both sides with respect to $x$: $$1 + \frac{dy}{dx} = \frac{dt}{dx} \Rightarrow \frac{dy}{dx} = \frac{dt}{dx} - 1$$ Substitute this back into the differential equation: $$\frac{dt}{dx} - 1 = \sin t$$ $$\frac{dt}{dx} = 1 + \sin t$$ Separating the variables: $$\frac{1}{1 + \sin t} dt = dx$$ Integrating both sides: $$\int \frac{1}{1 + \sin t} dt = \int dx$$ To integrate the left side, multiply the numerator and denominator by $(1 - \sin t)$: $$\int \frac{1 - \sin t}{(1 + \sin t)(1 - \sin t)} dt = x + c$$ $$\int \frac{1 - \sin t}{1 - \sin^2 t} dt = x + c$$ $$\int \frac{1 - \sin t}{\cos^2 t} dt = x + c$$ $$\int \left(\frac{1}{\cos^2 t} - \frac{\sin t}{\cos^2 t}\right) dt = x + c$$ $$\int (\sec^2 t - \sec t \tan t) dt = x + c$$ Integrating each term gives: $$\tan t - \sec t = x + c$$ Substituting $t = x + y$ back into the equation: $$\tan(x+y) - \sec(x+y) = x + c$$

Step 4: Final Answer:
The general solution is $\tan(x+y) - \sec(x+y) = x + c$, which perfectly matches option (D).