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The order of the differential equation whose general solution is given by
y = (C_{1} + C_{2}) cos (x+C_{3}) – C_{4}e^{x+C5}, where C_{1}, C_{2}, C_{3}, C_{4}, C_{5}, are arbitrary constants, is
The given solution of D.E. is
[Taking c_{1} + c_{2} = A,c_{4} e^{c5} = B] Thus, there are actually three arbitrary constants and hence this differential equation should be of order 3.
The differential equation representing the family of curves where c is a positive parameter, is of
2 yy_{1} =2c ⇒ c = yy_{1}
Eliminating, c, we get,
It involves only Ist order derivative, its order is 1 but its degree is 3 as y_{1}^{3} is there.
A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, then
Tangent to the curve y = f (x) at (x, y) is
∴ cur ve is x^{3}y = 1 , which also passes through
If y (x) satisfies the differential equation y ' – y tan x = 2x secx and y(0) = 0, then
The given differential equation is
A curve passes through the point Let the slope of the curve at each point (x, y) be
Then the equation of the curve is
Let y (x) be a solution of the differential equation (1 + e^{x}) y ' + ye^{x} = 1 . If y(0) = 2, then which of the following statement is (are) true?
Its solution will be intersection point of y = x + 3 and y = e^{–x}
Clearly there is a critical point in (–1, 0).
Consider the family of all circles whose centers lie on the straight line y = x. If this family of circle is represented by the differential equation Py '' + Qy' + 1=0 , where P, Q are functions of x, y and y' then which of the following statements is (are) true?
Let the equation of circle be x^{2} + y^{2} + 2gx + 2gy + c = 0
⇒ 2x + 2yy' + 2g + 2gy' = 0
⇒ x + yy' + g + gy' = 0 ...(i)
Differentiating again,
Substituting value of g in eqn. (i)
be a differentiable function such that for all x ∈ (0, ∞) and f(1) ≠ 1. Then
A solution curve of the differen tial equation passes through thepoint (1, 3). Then the solution curve
As it passes through (1, 3) ⇒ C = –1 – log 3
..(1)
Intersection of (1) and y = x + 2
∴ (1, 3) is the only intersection point.
Intersection of (1) and y = (x + 2)^{2}
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