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Question Number 141303 by BHOOPENDRA last updated on 17/May/21

Commented by BHOOPENDRA last updated on 17/May/21

iv

iv

Answered by Dwaipayan Shikari last updated on 17/May/21

M= (((((1−cosθ)/2)  (1/4)sin^2 θ)),((  1              ((1+cosθ)/2))) )  M^2 = (((sin^2 (θ/2)    sin^2 (θ/2)cos^2 (θ/2))),((  1             cos^2 (θ/2))) ) (((sin^2 (θ/2)  sin^2 (θ/2)cos^2 (θ/2))),((  1              cos^2 (θ/2))) )  = (((sin^2 (θ/2)   sin^2 (θ/2) cos^2 (θ/2))),((1           cos^2 (θ/2))) )  M^n = (((sin^2 (θ/2)     sin^2 (θ/2)cos^2 (θ/2))),((  1               cos^2 (θ/2))) )

M=(1cosθ214sin2θ11+cosθ2)M2=(sin2θ2sin2θ2cos2θ21cos2θ2)(sin2θ2sin2θ2cos2θ21cos2θ2)=(sin2θ2sin2θ2cos2θ21cos2θ2)Mn=(sin2θ2sin2θ2cos2θ21cos2θ2)

Answered by Dwaipayan Shikari last updated on 17/May/21

xy′−xy=y  ⇒xy′=(x+1)y  ⇒(dy/dx)=((x+1)/x)y⇒∫(dy/y)=∫((x+1)/x)dx⇒log(y)=x+log(x)+C  ⇒log((y/(xC_1 )))=x⇒y=C_1 xe^x

xyxy=yxy=(x+1)ydydx=x+1xydyy=x+1xdxlog(y)=x+log(x)+Clog(yxC1)=xy=C1xex

Commented by BHOOPENDRA last updated on 17/May/21

tq sir but in upper problem i also need  part two

tqsirbutinupperproblemialsoneedparttwo

Answered by Dwaipayan Shikari last updated on 17/May/21

x^2 y′+3xy=1  ⇒(dy/dx)+((3y)/x)=(1/x^2 )  I.F=e^(∫(3/x)dx) =x^3   x^3 y=∫xdx=(x^2 /2)+C⇒2x^3 y=x^2 +C_1

x2y+3xy=1dydx+3yx=1x2I.F=e3xdx=x3x3y=xdx=x22+C2x3y=x2+C1

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