Question:

The solution of the differential equation \( \frac{dy}{dx} = \sin(x+y)\tan(x+y) - 1 \) is

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For equations with $(x+y)$, substitute $v = x+y$ to transform it into a separable equation.
Updated On: Apr 10, 2026
  • $\csc(x+y) + \tan(x+y) = x + c$
  • $x + \csc(x+y) = c$
  • $x + \tan(x+y) = c$
  • $x + \sec(x+y) = c$
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The Correct Option is B

Solution and Explanation

Step 1: Substitution
Put $x+y = z \implies 1 + \frac{dy}{dx} = \frac{dz}{dx}$. Equation becomes $\frac{dz}{dx} - 1 = \sin z \tan z - 1 \implies \frac{dz}{dx} = \sin z \tan z$.
Step 2: Variable Separation

$\frac{dz}{\sin z \tan z} = dx \implies \frac{\cos z}{\sin^{2}z} dz = dx$.
Step 3: Integration

$\int \csc z \cot z dz = \int dx$. $-\csc z = x - c \implies x + \csc(x+y) = c$.
Final Answer: (b)
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