The general solution of the differential equation (d y/d x)+ sin ((x+y/2))= sin ((x-y/2)) is

Question

The general solution of the differential equation (d y/d x)+ sin ((x+y/2))= sin ((x-y/2)) is (A)  ln ( cot (y/2))+2 sin x=c (B)  ln ( cot (y/4))+2 sin (x/2)=C (C)  ln ( tan (y/2))+2 sin x=c (D)  ln ( tan (y/4))+2 sin (x/2)=c

Tardigrade Admin · Accepted Answer

(d y/d x)+ sin ((x+y/2))= sin ((x-y/2)) ⇒ (d y/d x)=-2 cos (x/2) sin (y/2) ⇒ ∫ operatornamecosec (y/2) d y=∫-2 cos (x/2) d x ⇒ 2 ln ( tan (y/4)) =-4 sin (x/2)+c prime ⇒ ln ( tan (y/4))+2 sin (x/2)=c

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

$ln (cot \frac{y}{2}) + 2 sin x = c$

$ln (cot \frac{y}{4}) + 2 sin \frac{x}{2} = C$

$ln (tan \frac{y}{2}) + 2 sin x = c$

$ln (tan \frac{y}{4}) + 2 sin \frac{x}{2} = c$

Q. The general solution of the differential equation dxdy​+sin(2x+y​)=sin(2x−y​) is

ln(cot2y​)+2sinx=c

ln(cot4y​)+2sin2x​=C

ln(tan2y​)+2sinx=c

ln(tan4y​)+2sin2x​=c

dxdy​+sin(2x+y​)=sin(2x−y​)⇒dxdy​=−2cos2x​sin2y​ ⇒∫cosec2y​dy=∫−2cos2x​dx⇒2ln(tan4y​) =−4sin2x​+c′⇒ln(tan4y​)+2sin2x​=c

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

$ln (cot \frac{y}{2}) + 2 sin x = c$

$ln (cot \frac{y}{4}) + 2 sin \frac{x}{2} = C$

$ln (tan \frac{y}{2}) + 2 sin x = c$

$ln (tan \frac{y}{4}) + 2 sin \frac{x}{2} = c$