Question 7

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

    , for any n ∈ N or F4 (x, y) ≠ 7

     , 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

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.

yx x   +       ... (1) It is a differential equation of the form F( , )dy x ydx = . Here F(x, y) = cos cos

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