IF `y' = y/x(log y - log x+1),` then the solution of the equation is :

Show that y=a cos(log x)+b sin(log x) is a solution of the differential equation x^2(d^2y)/(dx^2)+x(dy)/(dx)+y=0

Comprehension (Q. No. 6 to Q. No. 8) Let (x_1, y_1)&(x_2, y_2) are the solutions of the equation, (log)_(225)(x)+(log)_(64)(y)=4 and (log)_x(225)-(log)_y(64)=1 (log)_(225)x_1dot(log)_(225)x_2= 2 (b) 4 (c) 6 (d) 8

If (x_1, y_1)&(x_2, y_2) are the solutions of the equaltions, (log)_(225)(x)+log_(64)(y)=4a n d(log)_x(225)-(log)_y(64)=1, (log)_(225)x_1dot(log)_(225)x_2=4 b. (log)_(225)x_1+(log)_(225)x_2=6 c. |(log)_(64)y_1-(log)_(64)y_2|=2sqrt(5) d. (log)_(30)(x_1x_2y_1y_2)=12

If (x_1, y_1)&(x_2, y_2) are the solutions of the equaltions, (log)_(225)(x)+log_(64)(y)=4a n d(log)_x(225)-(log)_y(64)=1, (A) (log)_(225)x_1dot(log)_(225)x_2=4 (B). (log)_(225)x_1+(log)_(225)x_2=6 (C). |(log)_(64)y_1-(log)_(64)y_2|=2sqrt(5) (D). (log)_(30)(x_1x_2y_1y_2)=12

The solution of the equation log ((dy)/(dx))=a x+b y is

If the equation 2^x(1-2^x)+4^y=2^y is solved for y in terms of x where x<0, then the sum of the solution is x(log)_2(1-2^x) (b) x+(log)_2(1-2^x) (log)_2(1-2^x) (d) x(log)_2(2^x+1)

If the equation 2^x+4^y=2^y + 4^x is solved for y in terms of x where x<0, then the sum of the solution is (a) x(log)_2(1-2^x) (b) x+(log)_2(1-2^x) (c) (log)_2(1-2^x) (d) x(log)_2(2^x+1)

If log_y x +log_x y=7, then the value of (log_y x)^2+(log_x y)^2, is

If y=x/(log|cx|) (where c is an arbitrary constant) is the general solution of the differential equation (dy)/(dx)=y/x+varphi(x/y), then the function varphi(x/y) is

The general solution of the equation, x((dy)/(dx)) = y ln (y/x) is