d+y*d=(x^2)*y*d

$$d y + d = d x^{2} y$$

Перенесём правую часть уравнения в
левую часть уравнения со знаком минус.

Уравнение превратится из
$$d y + d = d x^{2} y$$
в
$$- d x^{2} y + d y + d = 0$$
Это уравнение вида

a*x^2 + b*x + c = 0

Квадратное уравнение можно решить
с помощью дискриминанта.
Корни квадратного уравнения:
$$x_{1} = frac{sqrt{D} – b}{2 a}$$
$$x_{2} = frac{- sqrt{D} – b}{2 a}$$
где D = b^2 – 4*a*c – это дискриминант.
Т.к.
$$a = – d y$$
$$b = 0$$
$$c = d y + d$$
, то

D = b^2 – 4 * a * c =

(0)^2 – 4 * (-d*y) * (d + d*y) = 4*d*y*(d + d*y)

Уравнение имеет два корня.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b – sqrt(D)) / (2*a)

или
$$x_{1} = – frac{1}{d y} sqrt{d y left(d y + dright)}$$
$$x_{2} = frac{1}{d y} sqrt{d y left(d y + dright)}$$

/ / 2 / / 2
_________________________________________________________________________________ | | im(y)*re(y) (1 + re(y))*im(y) im (y) (1 + re(y))*re(y)|| _________________________________________________________________________________ | | im(y)*re(y) (1 + re(y))*im(y) im (y) (1 + re(y))*re(y)||
/ 2 |atan2|————— – —————–, ————— + —————–|| / 2 |atan2|————— – —————–, ————— + —————–||
/ / 2 2 | | 2 2 2 2 2 2 2 2 || / / 2 2 | | 2 2 2 2 2 2 2 2 ||
/ | im (y) (1 + re(y))*re(y)| / im(y)*re(y) (1 + re(y))*im(y) | im (y) + re (y) im (y) + re (y) im (y) + re (y) im (y) + re (y) /| / | im (y) (1 + re(y))*re(y)| / im(y)*re(y) (1 + re(y))*im(y) | im (y) + re (y) im (y) + re (y) im (y) + re (y) im (y) + re (y) /|
x1 = – / |————— + —————–| + |————— – —————–| *cos|——————————————————————————-| – I* / |————— + —————–| + |————— – —————–| *sin|——————————————————————————-|
4 / | 2 2 2 2 | | 2 2 2 2 | 2 / 4 / | 2 2 2 2 | | 2 2 2 2 | 2 /
/ im (y) + re (y) im (y) + re (y) / im (y) + re (y) im (y) + re (y) / / im (y) + re (y) im (y) + re (y) / im (y) + re (y) im (y) + re (y) /

$$x_{1} = – i sqrt[4]{left(frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2} + left(- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2}} sin{left (frac{1}{2} {atan_{2}}{left (- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}},frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} right )} right )} – sqrt[4]{left(frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2} + left(- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2}} cos{left (frac{1}{2} {atan_{2}}{left (- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}},frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} right )} right )}$$

/ / 2 / / 2
_________________________________________________________________________________ | | im(y)*re(y) (1 + re(y))*im(y) im (y) (1 + re(y))*re(y)|| _________________________________________________________________________________ | | im(y)*re(y) (1 + re(y))*im(y) im (y) (1 + re(y))*re(y)||
/ 2 |atan2|————— – —————–, ————— + —————–|| / 2 |atan2|————— – —————–, ————— + —————–||
/ / 2 2 | | 2 2 2 2 2 2 2 2 || / / 2 2 | | 2 2 2 2 2 2 2 2 ||
/ | im (y) (1 + re(y))*re(y)| / im(y)*re(y) (1 + re(y))*im(y) | im (y) + re (y) im (y) + re (y) im (y) + re (y) im (y) + re (y) /| / | im (y) (1 + re(y))*re(y)| / im(y)*re(y) (1 + re(y))*im(y) | im (y) + re (y) im (y) + re (y) im (y) + re (y) im (y) + re (y) /|
x2 = / |————— + —————–| + |————— – —————–| *cos|——————————————————————————-| + I* / |————— + —————–| + |————— – —————–| *sin|——————————————————————————-|
4 / | 2 2 2 2 | | 2 2 2 2 | 2 / 4 / | 2 2 2 2 | | 2 2 2 2 | 2 /
/ im (y) + re (y) im (y) + re (y) / im (y) + re (y) im (y) + re (y) / / im (y) + re (y) im (y) + re (y) / im (y) + re (y) im (y) + re (y) /

$$x_{2} = i sqrt[4]{left(frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2} + left(- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2}} sin{left (frac{1}{2} {atan_{2}}{left (- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}},frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} right )} right )} + sqrt[4]{left(frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2} + left(- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}}right)^{2}} cos{left (frac{1}{2} {atan_{2}}{left (- frac{left(Re{y} + 1right) Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{Re{y} Im{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}},frac{left(Re{y} + 1right) Re{y}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} + frac{left(Im{y}right)^{2}}{left(Re{y}right)^{2} + left(Im{y}right)^{2}} right )} right )}$$

Дано уравнение с параметром:
$$d y + d = d x^{2} y$$
Коэффициент при x равен
$$- d y$$
тогда возможные случаи для y :
$$y < 0$$
$$y = 0$$
Рассмотри все случаи подробнее:
При
$$y < 0$$
уравнение будет
$$d x^{2} = 0$$
его решение
$$x = 0$$
При
$$y = 0$$
уравнение будет
$$d = 0$$
его решение