The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $\sec(x-y+1) dy = dx$. Let $v = x-y+1$. Then,$\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1

Vedclass · Accepted Answer

$x + \cot \left(\frac{x-y+1}{2}\right) = c$

Vedclass · Answer

(A) Given the differential equation: $\sec(x-y+1) dy = dx$. Let $v = x-y+1$. Then,$\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. The equation can be rewritten as $\frac{dy}{dx} = \cos(x-y+1)$. Substituting $v$,we get $1 - \frac{dv}{dx} = \cos(v)$. Rearranging the terms: $\frac{dv}{dx} = 1 - \cos(v)$. Separating the variables: $\frac{dv}{1 - \cos(v)} = dx$. Using the identity $1 - \cos(v) = 2\sin^2(\frac{v}{2})$,we have $\frac{dv}{2\sin^2(\frac{v}{2})} = dx$. This simplifies to $\frac{1}{2} \csc^2(\frac{v}{2}) dv = dx$. Integrating both sides: $\int \frac{1}{2} \csc^2(\frac{v}{2}) dv = \int dx$. $-\cot(\frac{v}{2}) = x + c$. Substituting $v = x-y+1$ back: $-\cot(\frac{x-y+1}{2}) = x + c$,which can be written as $x + \cot(\frac{x-y+1}{2}) = c'$ (where $c' = -c$). Thus,the correct option is $A$.

The general solution of the differential equation $\sec(x-y+1) dy = dx$ is

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