Question

The solution of the differential equation $\frac{dy}{dx}=sec(x+y)$ is

• A. $y - tan\frac{x+y}{2} = c$
• B. $y + tan\frac{x+y}{2} = c$
• C. $y + 2 tan\frac{x+y}{2} = c$
• D. None of these

Answer 1

Answer: (A)

A $y - tan\frac{x+y}{2} = c$

Answer 2

The given differential equation is $\frac{dy}{dx}=\sec(x+y)$.

This is a differential equation of the form $\frac{dy}{dx} = f(ax+by+c)$. Here, $a=1, b=1, c=0$. We use the substitution method. Let $v = x+y$. Differentiating both sides with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(y)$ $\frac{dv}{dx} = 1 + \frac{dy}{dx}$ From this, we can express $\frac{dy}{dx}$ as: $\frac{dy}{dx} = \frac{dv}{dx} - 1$

Substitute this into the original differential equation: $\frac{dv}{dx} - 1 = \sec(v)$

Now, separate the variables: $\frac{dv}{dx} = 1 + \sec(v)$ $\frac{dv}{1 + \sec(v)} = dx$

To integrate the left side, simplify the integrand: $\frac{1}{1 + \sec(v)} = \frac{1}{1 + \frac{1}{\cos(v)}} = \frac{1}{\frac{\cos(v) + 1}{\cos(v)}} = \frac{\cos(v)}{\cos(v) + 1}$

We use the half-angle identities for $\cos(v)$: $\cos(v) = 2\cos^2\left(\frac{v}{2}\right) - 1$ So, $\cos(v) + 1 = 2\cos^2\left(\frac{v}{2}\right)$

Substitute these into the integrand: $\frac{2\cos^2\left(\frac{v}{2}\right) - 1}{2\cos^2\left(\frac{v}{2}\right)} = \frac{2\cos^2\left(\frac{v}{2}\right)}{2\cos^2\left(\frac{v}{2}\right)} - \frac{1}{2\sec^2\left(\frac{v}{2}\right)}$ $= 1 - \frac{1}{2}\sec^2\left(\frac{v}{2}\right)$

Now, integrate both sides of the separated equation: $\int \left(1 - \frac{1}{2}\sec^2\left(\frac{v}{2}\right)\right) dv = \int dx$

Integrating term by term: $\int 1 \, dv = v$ For $\int \frac{1}{2}\sec^2\left(\frac{v}{2}\right) dv$: Let $u = \frac{v}{2}$, then $du = \frac{1}{2} dv$, which means $dv = 2du$. So, $\frac{1}{2} \int \sec^2(u) (2du) = \int \sec^2(u) du = \tan(u) = \tan\left(\frac{v}{2}\right)$.

Thus, the integral of the left side is: $v - \tan\left(\frac{v}{2}\right)$

The integral of the right side is: $\int dx = x + C$, where $C$ is the constant of integration.

Equating the results of the integration: $v - \tan\left(\frac{v}{2}\right) = x + C$

Finally, substitute back $v = x+y$: $(x+y) - \tan\left(\frac{x+y}{2}\right) = x + C$

Subtract $x$ from both sides: $y - \tan\left(\frac{x+y}{2}\right) = C$

Comparing this solution with the given options, it matches option A.

Explanation of the solution:

1. Substitute $v = x+y$ to transform the differential equation into a separable form. This leads to $\frac{dv}{dx} - 1 = \sec(v)$.
2. Rearrange to separate variables: $\frac{dv}{1 + \sec(v)} = dx$.
3. Simplify the integrand $\frac{1}{1 + \sec(v)}$ using trigonometric identities, specifically half-angle formulas for cosine: $\frac{\cos(v)}{1 + \cos(v)} = \frac{2\cos^2(v/2) - 1}{2\cos^2(v/2)} = 1 - \frac{1}{2}\sec^2(v/2)$.
4. Integrate both sides: $\int (1 - \frac{1}{2}\sec^2(v/2)) dv = \int dx$.
5. Perform the integration: $v - \tan(v/2) = x + C$.
6. Substitute back $v = x+y$ to get the solution in terms of $x$ and $y$: $(x+y) - \tan(\frac{x+y}{2}) = x + C$.
7. Simplify to obtain the final solution: $y - \tan(\frac{x+y}{2}) = C$.