MAS113 CALCULUS II SPRING 2008, QUIZ 5 SOLUTIONS. x 2 dx = 3y + y 3 = x 3 + c. It can be easily verified that the differential equation is exact, as

Transcription

1 MAS113 CALCULUS II SPRING 008, QUIZ 5 SOLUTIONS Quiz 5a Solutions (1) Solve the differential equation y = x 1 + y. (1 + y )y = x = (1 + y ) = x = 3y + y 3 = x 3 + c. () Solve the differential equation (xy + 3x y) + (x + y)x y = 0. d ( xy + 3x y ) = xy + 3x = d ( ) (x + y)x. (xy ψ(x,y) = + 3x y ) = 1 x y + x 3 y + c(y). Also, we must have (x + y)x = ψ y (x,y) = x y + x 3 + c (y). From this, we infer that c (y) = 0 and hence c(y) is a constant function. Therefore, the solution to the differential 1 x y + x 3 y = k. Quiz 5b Solutions x = cost, y = t cost, 0 t π. (1) Solve the differential equation xy = 1 y. y = 1 1 y x = 1 = = arcsin y = ln x + c = y = sin (ln x + c). 1 y x () Solve the differential equation (y cos x + xe y ) + (sinx + x e y 1)y = 0. d (y cos x + xey ) = cosx + xe y = d ( sin x + x e y 1 ). ψ(x,y) = (y cos x + xe y ) = y sin x + x e y + c(y). 1

2 Also, we must have sinx+x e y 1 = ψ y (x,y) = sinx+x e y +c (y). From this, we infer that c (y) = 1 and hence c(y) = y + c. Therefore, the solution to the differential y sin x + x e y y = k. x = 5 sint, y = t, π t π. Quiz 5c Solutions (1) Solve the differential equation y + y sin x = 0. Separating the variables we get y = sin x = y y = sin x = 1 1 = cosx + c = y = y () Solve the differential equation y + (xy e y )y = 0. d y = y = d (xy ey ). ψ(x,y) = y = xy + c(y). cosx + c. Also, we must have xy e y = ψ y (x,y) = xy + c (y). From this, we infer that c (y) = e y and hence c(y) = e y + c. Therefore, the solution to the differential xy e y = k. Quiz 5d Solutions x = sec θ, y = tanθ, π/ θ π/. (1) Solve the differential equation y = x e x y + e y. (y + e y ) = x e x = 1 y + e y = 1 x + e x + c. () Solve the differential equation y x e x cos y + xe x (x + ) siny = 0. d (xex (x + ) siny) = xe x (x + ) cosy = d ( x e x cos y ).

3 ψ(x,y) = x e x cos y = x e x sin y + c(x). Also, we must have xe x (x + ) siny = ψ x (x,y) = (x + x)e x sin y + c (x). From this, we infer that c (x) = 0 and hence c(y) = c. Therefore, the solution to the differential ( k x e x sin y = k or y = arcsin x e x ). x = 1 + t, y = 5 t, t 3. Quiz 5e Solutions (1) Solve the differential equation y x = y(1 + x 3 ). x y = 1 + x 3 = 1 y = 1 3 ln 1 + x3 + c = y = ± 3 ln 1 + x3 + c. () Solve the differential equation ye xy + x + xe xy y = 0. d ( ye xy + x ) = e xy + xe xy = d ( ) xe xy. ψ(x,y) = xe xy = 1 exy + c(x). Also, we must have ye xy + x = ψ x (x,y) = ye xy + c (x). From this, we infer that c (x) = x and hence c(y) = x/ + c. Therefore, the solution to the differential equation is e xy + x = k or y = 1 ln k x. x x = sint, y = 3 cost, 0 t π. Quiz 5f Solutions (1) Solve the differential equation y = (1 + x)(1 + y ). Separating the variables we get 1 + y = (1 + x) = arctany = ln 1 + x + c = y = tan (ln 1 + x + c).

4 () Solve the differential equation (3x y + xy ) + (x 3 + x y)y = 0. d ( 3x y + xy ) = 3x + xy = d ( x 3 + x y ). ψ(x,y) = (x 3 + x y) = x 3 y + 1 x y + c(x). Also, we must have 3x y +xy = ψ x (x,y) = 3x y +xy +c (x). From this, we infer that c (x) = 0 and hence c(y) = c. Therefore, the solution to the differential x 3 y + 1 x y = k. x = 1 + t, y = t 4t, 0 t 5. Quiz 5g Solutions (1) Solve the differential equation y = cos x cos (y). sec y = cos x = 1 tan y = 1 x + 1 sin x + c. 4 () Solve the differential equation ty t + 1 t ( ln(t + 1) ) y = 0. ( ) d ty t + 1 t = t t + 1 = d ( + ln(t + 1) ). dt ψ(t,y) = ( + ln(t + 1) ) = y + y ln(t + 1) + c(t). ty Also, we must have t + 1 t = ψ t(t,y) = ty t c (t). From this, we infer that c (x) = t and hence c(t) = t +c. Therefore, the solution to the differential equation is y + y ln(t + 1) t = k. Quiz 5h Solutions x = cos t, y = cost, 0 t 4π.

5 (1) Solve the differential equation y = cos x 3 + y. Seperating the variables, we get (3 + y) = cos x = y + y = x + c. () Solve the differential equation (xy 9x ) + (y + x + 1)y = 0. d ( ) xy 9x = x = d ( y + x + 1 ). dt ψ(x,y) = (y + x + 1 ) = y + x y + y + c(x). Also, we must have xy 9x = ψ x (x,y) = xy + c (x). From this, we infer that c (x) = 9x and hence c(x) = 3x 3 + c. Therefore, the solution to the differential y + x y + y 3x 3 = k. x = + cost, y = 3 + sint, 0 t π.

6

7