The degree of the differential equation satisfying
putting x = sin A and y = sin B in the given relation, we get
cos ,4 + cos B = a(sin A - sin B)
⇒ A B = 2 cot-1 a
⇒ sin-1 x - sin-1 y = 2 cot-1 a Differentiating w.r.t. x, we get
Clearly, it is a differential equation of degree one
The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of
Differential equation of the family of circles touching the line y = 2 at (0, 2) is
The differential equation of all parabolas whose axis are parallel to the y-axis is
The equation of a member of the family of parabolas having axis parallel toy-ax is is
y = Ax2 + Bx + C ......(1)
where A, B, and C are arbitrary constants
Differentiating equation (1) w.r.t. x, we .....(2)
which on again differentiating w.r.t. jc gives ......(3)
Differentiating (3) w.r.t. x, we get
The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is
) be the centre on y-axis then its radius will be k as it passes through origin. Hence its equation is
The differential equation whose general solution is given by, y = where c1, c2, c3, c4, c5 are arbitrary constants,
The equation of the curves through the point (1, 0) and whose slope is
The solution of the equation log(dy/dx) = ax + by is
Solution of differential equation dy – sin x sin ydx = 0 is
The solution of the equation
Putting u = x - y, we get du/dx = 1 - dy/dx. The given equation can be written as 1 - du/dx = cos u
The general solution of the differential equation
The solutions of (x + y + 1) dy = dx is
Putting x + y 1 = u, we have du = dx + dy and the given equations reduces to u(du - dx) = dx
The slope of the tangent at (x, y) to a curve passing through is given by then the equation of the curve is
On integration, we get
This passes through (1,π/4), therefore 1 = log C.
The solution of (x2 + xy)dy = (x2 + y2)dx is
∴ equation reduces to
The solution of (y + x + 5)dy = (y – x + 1) dx is
The intersection of y – x + 1 = 0 and y + x + 5 = 0 is (-2, -3).
Put x = X - 2, y = Y - 3
The given equation reduces to
putting Y = vX, we get
The slope of the tangent at (x, y) to a curve passing through a point (2, 1) is then the equation of the curve is
∴ equation (1) transforms to
The solution of satisfying y(1) = 1 is given by
Rewriting the given equation as
Hence y2 = x(1 +x) - 1 which represents a system of hyperbola.
Solution of the equation where
The given differential equation can be written as which is linear differential equation of first order.
A function y = f(x) satisfies (x + 1) f ' (x) - 2 (x2 + x)f(x) = 5, then f(x) is
The general solution of the equation
The solution of the differential equation
Integrating, we get the solutions
Which of the following is not the differential equation of family of curves whose tangent from an angle of π/4 with the hyperbola xy = c2?
Tangent to a curve intercepts the y-axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is
The equations of the tangent at the point
The coor dinates of the point P are
The slope of the perpendiculer line
which is the rquried differential equation to the curve at y = f(x).
The curve for which the normal at any point (x, y) and the line joining the origin to that point from an isosceles triangle with the x-axis as base is
It is given that the triangle OPC is an isosceles triangle.
Therefore, OM= MG = sub-normal
On integration, we get x2 -y2 = C, which is a rectangular hyperbola
The curve satisfying the equation and passing through the point (4, - 2 ) is
It passes through the point (4, -2).
A normal at P(x, y) on a curve meets the x-axis at Q and N is the foot of the ordinate at then the equation of curve given that it passes through the point (3,1) is
Let the equation of the curve be y = f(x).
It is lincar differential equation.
The equation of the curve passing through (2, 7/2) and having gradient is
This passes through (2, 7/2),
Thus the equation of the curve is
A normal at any point (x, y) to the curve y = f(x) cuts a triangle of unit area with the axis, the differential equation of the curve is
Equation of normal at point p is
Y – y = (dx/dy)(X – x)
The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is
Equations to the family of parabolas is (y - k)2 = 4a (x - h)
(substing y - k from equations (1))
Hence the order is 2 and the degree is 1.