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A solution of the differential equation
By actual verification we find that the choice (c), i.e. y = 2x4 satisfies the given differential equation.
If x^{2} + y^{2} = 1, then
Given x^{2} + y^{2} = 1. Differen tiat ing w.r.t . x , we get 2x + 2yy ' = 0 or x + yy '= 0. Again differentiating w.r.t. x, we get 1 + y ' y '+ yy " = 0 or 1 + (y')^{2} + yy" = 0
If y(t) is a solution of then y(1) is equal to
The given differential equation is
Integrating both sides
⇒ ln(1 + y) =  ln(2 + sin x)+C
Put x = 0 andy=1 ⇒ ln(2) =  ln 2+ C ⇒ C = ln4
If y = y(x) and it follows the relation x cos y + y cos x = π then y" (0) =
The solution of primitive integral equation (x^{2} + y^{2}) dy = xy dx is y = y(x). If y (1) = 1 and (x_{0}) = e, then x_{0} is equal to
The given D.E. is (x^{2} + y^{2})dy = xy dx s.t. y (1) = 1 and y(x_{0}) = e
The given eqn can be written as
For th e primitive in tegral equation ydx + y^{2}dy = x dy; x ∈ R, y > 0, y = y(x), y(1) = 1, then y(–3) is
The given eq^{n} is
Now to find y(3), puttin g x =3 in above eqn, we get
But given that y > 0 , ∴ y = 3
The differential equation determines a family of circles with
⇒ ( x  c)^{2} + y^{2} = 1
which is a circle of fixed radius 1 and variable centre (c, 0) lying on xaxis.
The function y = f (x) is the solution of the differential equation satisfying f (0) = 0. Then
Given D.E. can be written as
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