[x(dy)/(dx)-y=log x,," given that "y(1)=0],[[" Ans."y=-log x-1+x],5]

x(dy)/(dx)-y=log x, given that y=0 when x=1

Find y(x) , if it satisfies the following differential equation (dy)/(dx)=(x-y)^2 , and given that y(1)=1 (A) -ln|(1-x+y)/(1+x-y)|=2(x-1) (B) ln|(2-y)/(2-x)=x+y-1 (C) ln|(1-x+y)/(1+x-y)|=2(x-1) (D) 1/2ln|(1-x+y)/(1+x-y)|+ln|x|=0

Find y(x) , if it satisfies the following differential equation (dy)/(dx)=(x-y)^2 , and given that y(1)=1 (A) -ln|(1-x+y)/(1+x-y)|=2(x-1) (B) ln|(2-y)/(2-x)=x+y-1 (C) ln|(1-x+y)/(1+x-y)|=2(x-1) (D) 1/2ln|(1-x+y)/(1+x-y)|+ln|x|=0

Solve the differential equation (dy)/(dx)=(2x(log x+1))/(sin y+y cos y), given that y=0, whenx =1

Solve the differential equation x backslash(dy)/(dx)-y=log|x|, given that y(1)=0

(dy)/(dx)=(x+y)ln(x+y)-1

solve x((dy)/(dx))=y(log y-log x+1)

If x ^( log y) = log x, then prove that (dy)/(dx) = (y)/(x) ((1- log x log y)/( (log x) ^(2)))

if y=log x^(x) prove that (dy)/(dx)=1+log x