अवकल समीकरण `x (dy)/(dx) + 2y =x^(2) (x ne 0)` का हल जिसके लिए `y(1) = 1` है, है

The solution of the differential equation x(dy)/(dx) + 2y = x^2 (X ne 0) with y(1) =1, is :

(dy)/(dx)=(x+y+1)/(2x+2y+3)

x(2y-1)dx+(x^(2)+1)dy=0

(d^(2)y)/(dx^(2))=x^(2)sin x, givenatx =0 if y=0,(dy)/(dx)=1

x(1+y^(2))dx+y(1+x^(2))dy=0 at (0, 0)

( 1 ) /(x) + (1 ) /( y) = 7 , ( 2)/(x) + ( 3) /( y) = 17 (x ne 0, y ne 0 ) .

(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2));y=0 if x=1

If (dy)/(dx)=(2^(x+y)-2^(x))/(2^(y)),y(0)=1 then y(1) is equal to

The value of k for which the graph of (k- 1)x+ y-2= 0 and (2-k)x-3y+ 1) =0 are parallel is k का मान जिसके लिए (k- 1)x+ y-2= 0 और (2-k)x-3y+ 1) =0 के ग्राफ समांतर हो, है-