1.` xdy/dx+2y=x^2 , xne0`

The solution of the differential equation xdy/dx+2y=x^(2),(xne0) with y(1)=1, is (a) y=x^(2)/4+3/(4x^(2)) (b) y=x^(3)/5+1/(5x^(2)) (c) y=3/4x^(2)+3/(4x^(2)) (d) y=4/5x^(3)+1/(5x^(2))

solve xdy/dx-y=2x^2

Solve the initial value problem: xdy/dx-y= 5x^2+3x, y(0)=3

If y =x+(1)/(x +(1)/(x+.....oo)) then prove that xdy/dx =y/(2y -x).

Solve y=xdy/dx+(dy/dx)^2 .

Discuss the continuity of the function : f(x)={{:((1-cosx)/x^(2)", "xne0),(1", "x=0):} at x = 0.

The solution of (y - (xdy)/dx)= 3 (1-x^(2)(dy)/(dx)) is