Question

Solve $\cos(x+y)dy=dx$

• A. $y=\tan\frac{x+y}{2}+c$
• B. $y=x\sec(y/x)+c$
• C. $y=-\cos^{-1}(y/x)+c$
• D. $y=\tan(x+y)+c$

Hint:

Use substitution $u=x+y$ for combined-angle differential equations.

Answer 1

The Correct Option is A

Step 1: Read the equation.
We must solve $\cos(x+y)\,dy=dx$. The presence of $x+y$ together suggests substituting a new variable for that sum.

Step 2: Substitute.
Let $u=x+y$. Differentiating with respect to $x$, \[ \frac{du}{dx}=1+\frac{dy}{dx}. \]
Step 3: Rewrite the given equation.
From $\cos(x+y)\,dy=dx$ we get $\dfrac{dy}{dx}=\dfrac{1}{\cos u}=\sec u$.

Step 4: Combine.
Substitute this into the $\dfrac{du}{dx}$ relation: \[ \frac{du}{dx}=1+\sec u. \] Now the variables can be separated.

Step 5: Separate and simplify.
\[ \frac{du}{1+\sec u}=dx. \] Write $1+\sec u=\dfrac{\cos u+1}{\cos u}$, so the left side becomes $\dfrac{\cos u}{1+\cos u}\,du$. Using $1+\cos u=2\cos^2\tfrac{u}{2}$ and simplifying leads to a standard integral.

Step 6: Integrate.
Carrying out the integration gives \[ y=\tan\frac{x+y}{2}+c. \]
\[ \boxed{y=\tan\dfrac{x+y}{2}+c} \]