What is the answer to (x^2D^2+xD)y=0 ?

The answer to (x^2D^2+xD)y=0 is x=0,x=-1/D ,y=0;\quad D\ne 0

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(x^2D^2+xD)y=0

(x^{2} D^{2} + x D) y = 0

x = 0, x = - \frac{1}{D}, y = 0; D \neq = 0

Solution steps

(x^{2} D^{2} + x D) y = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x^{2} D^{2} + x D = 0 or y = 0

Solve

x^{2} D^{2} + x D = 0 : x = 0, x = - \frac{1}{D}; D \neq = 0

y = 0

The solutions are

x = 0, x = - \frac{1}{D}, y = 0; D \neq = 0

BC = 0, 8 D C = 2, 24

(x^{2} - 4) y^{2} = 0

I n (4 - x) + I n (4 x) = 0

2 y m + y = 0

(1 + x) y = 0

What is the answer to (x^2D^2+xD)y=0 ?
The answer to (x^2D^2+xD)y=0 is x=0,x=-1/D ,y=0;\quad D\ne 0