Chapter 9
Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, - 5) and (- 4, -2). Also, find its area.
The slope of a line is double of the slope of another line. If tangent of the angle between them is 3 , find the slopes of the lines.
| The degree of the differential equation | 3 | 22 |
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2 sin 1 0d y dy dy dx dxdx + + + = is (A) 3 (B) 2 (C) 1 (D) not defined
A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k - y 1 = m (h - x 1).
The order of the differential equation 2/2 22 3 0d y dyx y dxdx − + = is (A) 2 (B) 1 (C) 0 (D) not defined 9.3. General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the type: x2 + 1 = 0 ... (1) sin² x - cos x = 0 ... (2) Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown x in the given equation, L.H.S. becomes equal to the R.H.S.. Now consider the differential equation 2 0d y y dx + = ... (3) In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. The curve y = φ (x) is called the solution curve (integral curve) of the given differential equation. Consider the function given by y = φ (x) = a sin (x + b), ... (4) where a, b ∈ R. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3). Let a and b be given some particular values say a = 2 and 4b π= , then we get a function y = φ 1(x) = 2sin 4x π + ... (5) When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefore φ 1 is also a solution of equation (3). Function φ consists of two arbitrary constants (parameters) a, b and it is called general solution of the given differential equation. Whereas function φ 1 contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. Example 2 Verify that the function y = e- 3x is a solution of the dif ferential equation 2 6 0d y dy ydxdx + − = Solution Given function is y = e- 3x. Differentiating both sides of equation with respect to x , we get 33 xdy edx −= − ... (1) Now, differentiating (1) with respect to x, we have 2/2 d y dx = 9 e - 3 x Substituting the values of 2 ,d y dy dxdx and y in the given differential equation, we get L.H.S. = 9 e - 3x + (-3 e- 3x) - 6. e- 3x = 9 e- 3x - 9 e - 3x = 0 = R.H.S.. Therefore, the given function is a solution of the given differential equation. Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation 2 0d y y dx + = Solution The given function is y = a cos x + b sin x ... (1) Differentiating both sides of equation (1) with respect to x, successively, we get dy dx = - a sinx + b cosx 2/2 d y dx = - a cos x - b sinx MATHEMA TICS306 Substituting the values of 2/2 d y dx and y in the given differential equation, we get L.H.S. = (- a cos x - b sin x) + (a cos x + b sin x) = 0 = R.H.S. Therefore, the given function is a solution of the given differential equation.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, - 4) and B (8, 0).
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
2( )y′ ′′ + (y″ )3 + (y′)4 + y5 = 0 7. y′′′ + 2y″ + y′ = 0 MATHEMA TICS304
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
y′ + y = ex 9. y″ + (y′)2 + 2y = 0 10. y″ + 2y′ + sin y = 0
Without using distance formula, show that points ( - 2, - 1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).