The differential equation of the curve `xy=ae^(x)+be^(-x)+x^(2)` is
`xy_(2)+ay_(1)+bxy+x^(2)-2=0` .then the value

Find the differential equation of xy=ae^(x)+be^(-x)

The differential equation by eliminating the arbitrary constant from the equation xy=ae^(x)+be^(-x)+x^(2) is xy_(2)+ky_(1)-xy=k then k=

If K is constant such that xy + k = e ^(((x-1)^2)/(2)) satisfies the differential equation x . ( dy )/( dx) = (( x^2 - x-1) y+ (x-1) and y (1) =0 then find the value of K .

The solution of differential equation (1-xy + x^(2) y^(2))dx = x^(2) dy is

(1) y = ae^(4x) -be^(-3x) +c (2) xy = ae^(5x) +be^(-5x)

Solve the differential equations x^(2)dy-(x^(2)+xy-2y^(2))dx=0

Solve the following differential equations : (1) x^(2) (dy)/(dx)= x^(2) + xy + y^(2)

The differential equation obtained by eliminating the constants a and b from xy=ae^(x)+be^(-x)+x^(2) is

The solution of differential equation y'(1+x^(2))=2xy is :

The solution of differential equation (2y+xy^(3))dx+(x+x^(2)y^(2))dy=0 is