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

Question

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

Tardigrade Admin · Accepted Answer

Correct answer is (a) ln ( tan ( y /4))= C -2 cos ( x /2) Correct answer is (c) ln ( operatornamecosec ( y /2)- cot ( y /2))= C -2 cos ( x /2) Correct answer is (d) ln ( operatornamecosec ( y /2)+ cos ( y /2))=2 cos ( x /2)- C

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

$ln (tan \frac{y}{4}) = C - 2 cos \frac{x}{2}$

$ln (tan \frac{y}{4}) = 2 cos \frac{x}{2} - C$

$ln (cosec \frac{y}{2} - cot \frac{y}{2}) = C - 2 cos \frac{x}{2}$

$ln (cosec \frac{y}{2} + cos \frac{y}{2}) = 2 cos \frac{x}{2} - C$

Correct answer is (a) $ln (tan \frac{y}{4}) = C - 2 cos \frac{x}{2}$ Correct answer is (c) $ln (cosec \frac{y}{2} - cot \frac{y}{2}) = C - 2 cos \frac{x}{2}$ Correct answer is (d) $ln (cosec \frac{y}{2} + cos \frac{y}{2}) = 2 cos \frac{x}{2} - C$

Q. The general solution of the differential equation dxdy​+cos2x+y​=cos2x−y​ is :

ln(tan4y​)=C−2cos2x​

ln(tan4y​)=2cos2x​−C

ln(cosec2y​−cot2y​)=C−2cos2x​

ln(cosec2y​+cos2y​)=2cos2x​−C