(x^(3)+x^(2)+x+1)(dy)/(dx)=2x^(2)+x;y=1" यदि "x=0

(x^(3)+x^(2)+x+1)(dy)/(dx)=2x^(2)+x;y=1 when x=0

Solve (x^(3)+x^(2)+x+1)(dy)/(dx) =2x^(2)+x, " given that " y=1 " when " x =0.

x(dy)/(dx)+y=3x^(2)-2,x>0

x(x^(2)+1)(dy)/(dx)=y(1-x^(2))+x^(3)*ln x

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

Solve the following differential equations (i) (1+y^(2))dx = (tan^(-1)y - x)dy (ii) (x+2y^(3))(dy)/(dx) = y (x-(1)/(y))(dy)/(dx) + y^(2) = 0 (iv) (dy)/(dx)(x^(2)y^(3)+xy) = 1

x(x-1)(dy)/(dx)-(x-2)y=x^(3)(2x-1)

If y=log (x + sqrt(x^(2) + 1)) then show that, (x^(2) + 1) (d^(2)y)/(dx^(2)) + x (dy)/(dx)= 0

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

(y-x(dy)/(dx))=3(1-x^(2)(dy)/(dx))