Помочь с решением дифференциальных уравнений (решить самому, если можете):
1) a) 5sin^2xy'=(2/cosy)
б) 2x^2yy'=5 ; y(-1)=-1
2) y'=(y/x) -10(*y/x)^2
3) y'+2*(y/x)=4x
4) y''+y'=x^2+x

Предыдущий
вопрос Следующий
вопрос

1)
a) 5sin^2(xy') = (2/cosy)
sin^2(xy') = 2/(5cosy)
sin^2(xy') = 2/(5cosy)
y' = sin^(-1)(√(2/(5cosy)))/x
y' = sin^(-1)(√(2/(5cosy)))/√(cosy)

б) 2x^2yy' = 5
y' = 5/(2x^2y)
ln|y| = -5/(4x) + C
y = e^(-5/(4x) + C)
y = Ce^(-5/(4x))

2) y' = (y/x) - 10(y/x)^2
Substitute y = vx:
v + xdv/dx = v - 10v^2
xdv/dx = -10v^2
dv/v^2 = -10dx/x
-1/v = -10lnx + C
v = 1/(10lnx + C)
y = x/(10lnx + C)

3) y' + 2(y/x) = 4x
Using the integrating factor method, let μ(x) = e^(∫2/x dx) = e^(2lnx) = x^2
Multiplying both sides by x^2:
x^2y' + 2xy = 4x^3
(x^2y)' = 4x^3
x^2y = x^4 + C
y = x^2 + C/x^2

4) y'' + y' = x^2 + x
Let y = u(x), then y' = u', y'' = u''
u'' + u' = x^2 + x
u'' + u' = x(x + 1)
Using the integrating factor method, let μ(x) = e^(∫1 dx) = e^(x) = x
Multiplying both sides by x:
xu'' + xu' = x^3 + x^2
(xu')' = x^3 + x^2
xu' = (1/4)x^4 + (1/3)x^3 + C
u' = (1/4)x^3 + (1/3)x^2 + C/x
u = (1/16)x^4 + (1/12)x^3 + Clnx + D

Therefore, the solution to the differential equation is y = (1/16)x^4 + (1/12)x^3 + Clnx + D