Chapter 9
1 cos / 1 cos dy x dx x −= + 2. 24 ( 2 2)dy y ydx = − − < <
Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts. (i) x + 7y = 0, (ii) 6x + 3y - 5 = 0, (iii) y = 0.
Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x -x1) + B (y - y 1) = 0.
Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Find the coordinates of the foot of perpendicular from the point ( -1, 3) to the line 3x - 4y - 16 = 0.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2). Find the values of m and c.
If p and q are the lengths of perpendiculars from the origin to the lines θ 2 cos θ sin θ cosk y x=− and x sec θ + y cosec θ = k, respectively, prove that p2 + 4 q2 = k2.
In the triangle ABC with vertices A (2, 3), B (4, -1) and C (1, 2), find the equation and length of altitude from the vertex A.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .1 1 1 2 22 b ap + =
Reduce the following equations into intercept form and find their intercepts on the axes. (i) 3x + 2y - 12 = 0, (ii) 4x - 3y = 6, (iii) 3y + 2 = 0.
Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y - 2).
1 ( 1)dy y ydx + = ≠ 4. sec² x tan y dx + sec 2 y tan x dy = 0
Find the points on the x-axis, whose distances from the line 13 4 x y+ = are 4 units.
Find the distance between parallel lines (i) 15x + 8y - 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) - r = 0.
(ex + e- x) dy - ( ex - e- x) dx = 0 6. 2 2(1 ) (1 )dy x ydx = + +
Find equation of the line parallel to the line 3 4 2 0x y− + = and passing through the point (-2, 3).
Find equation of the line perpendicular to the line x - 7y + 5 = 0 and having x intercept 3.
y log y dx - x dy = 0 8. 5 5dyx ydx = − 9. 1sindy xdx −= 10. ex tan y dx + (1 - ex) sec 2 y dy = 0 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition: 11. 3 2( 1) dyx x x dx+ + + = 2x2 + x; y = 1 when x = 0 12. 2( 1) 1 dyx x dx− = ; y = 0 when x = 2 13. cos dy adx = (a ∈ R); y = 1 when x = 0 14. tandy y xdx = ; y = 1 when x = 0 15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x. 16. For the differential equation ( 2) ( 2)dyxy x ydx = + + , find the solution curve passing through the point (1, -1). 17. Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point. 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, -3). Find the equation of the curve given that it passes through (-2, 1). 19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds. MATHEMA TICS312 20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931). 21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648). 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present? 23. The general solution of the differential equation x ydy edx += is (A) ex + e- y = C (B) ex + ey = C (C) e- x + ey = C (D) e- x + e- y = C 9.4.2 Homogeneous differential equations Consider the following functions in x and y F1 (x, y) = y2 + 2xy, F 2 (x, y) = 2x - 3y , F3 (x, y) = cos y x , F 4 (x, y) = sin x + cos y If we replace x and y by λ x and λ y respectively in the above functions, for any nonzero constant λ , we get F1 (λ x, λ y) = λ 2 (y2 + 2xy) = λ 2 F1 (x, y) F2 (λ x, λ y) = λ (2x - 3 y) = λ F2 (x, y) F3 (λ x, λ y) = cos cosy y x x λ = λ = λ 0 F 3 (x, y) F4 (λ x, λ y) = sin λ x + cos λ y ≠ λ n F4 (x, y), for any n ∈ N Here, we observe that the functions F 1, F 2, F 3 can be written in the form F(λ x, λ y) = λ n F (x, y) but F4 can not be written in this form. This leads to the following definition: A function F( x, y) is said to be homogeneous function of degr ee n if F(λ x, λ y) = λ n F(x, y) for any nonzero constant λ . We note that in the above examples, F 1, F 2, F 3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function. We also observe that F1(x, y) = 2 2 2y y yx x hx xx + = or F1(x, y) = 2 2 21 x xy y hy y + = F2(x, y) = 1 1 32 y yx x hx x − = or F2(x, y) = 1 1 42 3x xy y h y y − = F3(x, y) = 0 0 5cos y yx x h x x = F4(x, y) ≠ 6
| n yx | h | x |
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, for any n ∈ N or F4 (x, y) ≠ 7
| n xy | h | y |
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, for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if F(x, y) = orn n y xx g y hx y A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero. To solve a homogeneous differential equation of the type ( )F ,dy x ydx = = yg x ... (1) We make the substitution y = v.x ... (2) Differentiating equation (2) with respect to x, we get dy dx = dvv x dx+ ... (3) Substituting the value of dy dx from equation (3) in equation (1), we get MATHEMA TICS314 dvv x dx+ = g (v) or dvx dx = g (v) - v ... (4) Separating the variables in equation (4), we get ( ) / dv g v v− = dx x ... (5) Integrating both sides of equation (5), we get ( ) / dv g v v−∫ = 1 Cdxx +∫ ... (6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y x . /handptrtsld1Note If the homogeneous differential equation is in the form F( , )dx x ydy = where, F (x, y) is homogenous function of degree zero, then we make substitution x vy = i.e., x = vy and we proceed further to find the general solution as discussed
| above by writing F( , ) .dx xx | y | h |
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dy y = = Example 10 Show that the differential equation (x - y) dy dx = x + 2y is homogeneous and solve it. Solution The given differential equation can be expressed as dy dx = 2x y x y + − ... (1) Let F(x, y) = 2x y x y + − Now F(λ x, λ y) = 0( 2 ) ( , )( ) x y f x yx y λ + = λ ⋅λ − Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.
| Alternatively, | 21 |
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| y dy x ydx x + = − = yg x ... (2) R.H.S. of differential equation (2) is of the form g y x and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution y = vx ... (3) Differentiating equation (3) with respect to, x we get dy dx = dvv x dx+ ... (4) Substituting the value of y and dy dx in equation (1) we get dvv x dx+ = 1 2 v v + − or dvx dx = 1 2 v vv + −− or dvx dx = 2 1 v v v + + − or 2 1/1 v dv v v − + + = dx x − Integrating both sides of equation (5), we get = or = - log | x |
| yy | x |
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yx x + ... (1) It is a differential equation of the form F( , )dy x ydx = . Here F(x, y) = cos cos
| yy | x | x |
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| yx x + Replacing x by λ x and y by λ y, we get F(λ x, λ y) = [ cos ] [F( , )] cos yy x x x yyx x λ + = λ λ Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution y = vx ... (2) Differentiating equation (2) with respect to x, we get dy dx = dvv x dx+ ... (3) Substituting the value of y and dy dx in equation (1), we get dvv x dx+ = cos 1 cos v v v + or dvx dx = cos 1 cos v v vv + − or dvx dx = 1 cosv or cosv dv = dx x Therefore cosv dv∫ = 1 dxx∫ MATHEMA TICS318 or sin v = log | x | + log |
| x | e | y |
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y e − ... (1) Let F(x, y) = 2 x y x y xe y ye − Then F(λ x, λ y) = 0/2 [F( , )] x y x y xe y x y ye λ − =λ λ Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution x = vy ... (2) Differentiating equation (2) with respect to y, we get dx dy = + dvv y dy Substituting the value of and dxx dy in equation (1), we get dvv y dy + = 2 1 v v v e e − or dvy dy = 2 1 v v
| v | e | v |
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| e / − − or dvy dy = 2 ve − or 2ev dv = dy y − or 2 ve dv⋅∫ = dy y − ∫ or 2 ev = - log | y | + C and replacing v by x y , we get x ye + log |
Find angles between the lines . 1 3 and 1 3= + = +y x y x
The line through the points (h, 3) and (4, 1) intersects the line7 9 19 0x y .− − = at right angle. Find the value of h.