2 y'' + 7 y' = 0, y(0)=1, y'(0)=-3
Делаем замену y' = f(x), тогда y'' = f'(x).
Получаем уравнение
2 * f' + 7 * f = 0 => 2 * df/dx = -7 * f => df/f = -7/2 dx
int df/f = -7/2 * int dx
ln |f| = -7/2 * x + C1
f = C1 * e^(-7/2 * x)
f = y' => y' = C1 * e^(-7/2 * x) => y = C1 * int e^(-7/2 * x) dx = C1 * e^(-7/2 * x) + C2
y = C1 * e^(-7/2 * x) + C2
y' = -7/2 * C1 * e^(-7/2 * x)
y'(0) = -3 => -7/2 * C1 * e^0 = -3 => C1 = 6/7
y(0) = 1 => C1 * e^0 + C2 = 1 => C2 = 1 - C1 = 1/7
Ответ: y = 6/7 * e^(-7/2 * x) + 1/7.