Решенные задачи билета №4 (2 семестр) / 4.doc

Билет№4

y = 1+x² ;

y = 5;

1+x² = 5

x²-4 = 0

x₁ = -2; x₂ = 2;

V_y = 2π_a∫^b (xf(x)dx) ; a>=0;

V = 2π₀∫² (x*(1+x²) dx) = 2π(₀∫² (xdx)+ ₀∫² (x²dx)) =

₂

= 2π(x²/2)+(x³/3)₀ =2π (4/2+8/3) = 2π(2+8/3) = = (28/3) π

Ответ: (28/3)π.

2. y′′-2y′+2y = e^x/sin³x

y′′-2y′+2y = 0

k²-2k+2 = 0

D = 4-4*2 = -4;

k₁ = (2-2i)/2 = 1-i; k₂ = 1+i;

Фундаментальная система решений:

y₁ = e^xcos(x); y₂ = e^xsin(x);

y_o.o. = c₁e^xcos(x) + c₂e^xsin(x);

Общее решение неоднородного уравнения ищем в виде

y_o.н. = c₁(x)e^xcos(x) + c₂(x)e^xsin(x);

c₁(x), c₂(x) находим их системы:

c′₁(x)e^xcos(x)+c′₂(x)e^xsin(x) = 0

c′₁(x)(e^xcos(x)-e^xsin(x))+ c′₂(x)(e^xsin(x)+e^xcos(x)) = e^x/sin³(x);

c′₁(x) = (-sin(x)/cos(x))*c′₂(x);

(-sin(x)/cos(x))*c′₂(x)* (e^xcos(x)-e^xsin(x))+ c′₂(x)(e^xsin(x)+e^xcos(x)) = e^x/sin³(x);

-c′₂(x)* e^xsin(x)+ c′₂(x)*((e^x*sin²(x))/cos(x)) + c′₂(x)* e^xsin(x)- c′₂(x)* e^xcos(x) = (e^x/sin³x);

c′₂(x) e^x ((sin²(x)+cos²(x))/cos(x)) = e^x/sin³(x);

c′₂(x) = cos(x)/sin³(x);

c′₁(x) = (-sin(x)/cos(x))*(cos(x)/sin³(x)) = -1/sin²(x);

c₂(x) = ∫( (cos(x)/sin³(x))dx) = ∫(d(sin(x))/sin³(x)) = -1/(2sin²(x) +c₂;

c₁(x) = ctg(x) +c₁;

y_о.н. = (cos(x)/sin(x)+c₁)* e^xcos(x) + (-1/(2sin²(x))+c₂)*e^xsin(x) =

= (cos²(x)*e^x)/sin(x) +c₁e^xcos(x)-e^x/(2sin(x))+c₂e^xsin(x) =

= (2cos²(x)-1)*e^x)/sinx + c₁e^xcosx +c₂e^xsinx =(cos(2x) *e^x)/sin(x)+c₁e^xcosx+c₂e^xsinx

Ответ: (cos(2x) *e^x)/sin(x)+c₁e^xcosx+c₂e^xsinx