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Linear Differential Equations of First Order

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Question

Solution of the differential equation $\frac{d y}{d x} = s i n (x + y) + c o s (x + y)$ is .....

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Solution

$\begin{matrix} \frac{d y}{d x} = sin (x + y) + cos (x + y) S u b s t i t u e x + y = v d i f f . w . r . t x 1 + \frac{d y}{d x} = \frac{d v}{d x} \frac{d y}{d x} = \frac{d v}{d x} - 1 T h e g i v e n d i f f e r e n t i a l e q u a t i o n b e c o m e s \frac{d y}{d x} = sin (x + y) + cos (x + y) \frac{d y}{d x} - 1 = sin v + cos v \frac{d v}{d x} = 1 + sin v + cos v d x = \frac{1}{1 + \frac{2 tan (\frac{v}{2})}{{tan}^{2} (\frac{v}{2})} + \frac{1 - {tan}^{2} (\frac{v}{2})}{1 + {tan}^{2} (\frac{v}{2})}} d x d x = \frac{1 + {tan}^{2} (\frac{v}{2})}{2 (1 + tan (\frac{v}{2}))} d v d x = \frac{{sec}^{2} (\frac{v}{2})}{2 (1 + tan \frac{v}{2})} d v \int d x = \int \frac{{sec}^{2} (\frac{v}{2})}{2 (1 + tan \frac{v}{2})} d v x = log ∣ ∣ 2 tan (\frac{v}{2}) ∣ ∣ + C x = l o g ∣ ∣ 2 tan (\frac{x + y}{2}) ∣ ∣ + C \end{matrix}$

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