The general solution of the differential equation d y-d x-sinx-y-2-sinx-y-2 is -

The correct option is A log_e|tany4|= 2sin x2 +c nbsp;dydx+sinx+y2=sinx y2 ⇒dydx = 2cosx2siny2 ⇒∫dy2siny2 = nbsp; ∫cosx2dx nbsp; ⇒ nbsp;∫dy2siny2 = ...

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Byju's Answer

Standard XII

Mathematics

Linear Differential Equations of First Order

The general s...

Question

The general solution of the differential equation $\frac{d y}{d x} + sin \frac{x + y}{2} = sin \frac{x - y}{2}$ is :

{log}_{e} ∣ ∣ ∣ tan \frac{y}{4} ∣ ∣ ∣ = - 2 sin \frac{x}{2} + c

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{log}_{e} ∣ ∣ ∣ tan \frac{y}{2} ∣ ∣ ∣ = - 2 sin \frac{x}{2} + c

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{log}_{e} ∣ ∣ ∣ tan \frac{y}{4} ∣ ∣ ∣ = 2 sin \frac{x}{2} + c

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{log}_{e} ∣ ∣ ∣ tan \frac{y}{2} ∣ ∣ ∣ = - sin \frac{x}{2} + c

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Solution

The correct option is A ${log}_{e} ∣ ∣ ∣ tan \frac{y}{4} ∣ ∣ ∣ = - 2 sin \frac{x}{2} + c$
$\frac{d y}{d x} + sin \frac{x + y}{2} = sin \frac{x - y}{2} \Rightarrow \frac{d y}{d x} = - 2 cos \frac{x}{2} sin \frac{y}{2} \Rightarrow \int \frac{d y}{2 sin \frac{y}{2}} = - \int cos \frac{x}{2} d x \Rightarrow \int \frac{d y}{2 sin \frac{y}{2}} = - 2 sin \frac{x}{2} \Rightarrow \int \frac{d y}{4 sin \frac{y}{4} cos \frac{y}{4}} = - 2 sin \frac{x}{2} \Rightarrow \int \frac{{sec}^{2} \frac{y}{4} d y}{4 tan \frac{y}{4}} = - 2 sin \frac{x}{2}$
$\Rightarrow {log}_{e} ∣ ∣ ∣ tan \frac{y}{4} ∣ ∣ ∣ = - 2 sin \frac{x}{2} + c$

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The general solution of the differential equation dydx+sinx+y2=sinx−y2 is :

The correct option is A loge∣∣∣tany4∣∣∣=−2sinx2+c dydx+sinx+y2=sinx−y2⇒dydx=−2cosx2siny2⇒∫dy2siny2=−∫cosx2dx⇒∫dy2siny2=−2sinx2⇒∫dy4siny4cosy4=−2sinx2⇒∫sec2y4 dy4tany4=−2sinx2 ⇒loge∣∣∣tany4∣∣∣=−2sinx2+c

The general solution of the differential equation $\frac{d y}{d x} + sin \frac{x + y}{2} = sin \frac{x - y}{2}$ is :