Question

Решить дифференциальное уравнение:
$y'+4x^{3} y=4y^{2} e^{4x} (1- x^{3} )$

Answer 1

y' + 4x^3 * y = 4y^2 * e^4x * (1 - x^3) | y = 0 - частное решение
y'/y^2 + 4x^3 / y = 4e^4x * (1 - x^3) | z = 1/y, z' = -y'/y^2
z*4x^3 - z' = 4e^4x * (1 - x^3)

1.
z*4x^3 - z' = 0
z'/z = 4x^3
ln|z| = x^4 + ln|C| | C = C(x)
z = C * e^(x^4)
z' = C * 4x^3 * e^(x^4) + C' * e^(x^4)

2.
C * e^(x^4) * 4x^3 - C * 4x^3 * e^(x^4) - C' * e^(x^4) = 4e^4x * (1 - x^3)
C' = -4e^(4x - x^4) * (1 - x^3)
C = -e^(4x - x^4) + C1

z = C1 * e^(x^4) - e^4x
y = 1 / (C1 * e^(x^4) - e^4x)