show that y=x^3+3x+1 satisfies y'''+xy''-2y'=0?

Question

please explain...

Answer 1

y=x^3+3x+1,
y'=3x^2+3,
y'' =6x,
y'''=6
Now y'''+xy''-2y'
=6+x*6x-2(3x^2+3)
=6+6x^2-6x^2-6
=0
so y=x^3x+1 satisfies the differencial Eq. y'''+xy''-2y'=0 QED.

Answer 2

y=x^3+3x+1

so

y' = 3x^2 + 3
y'' = 6x
y''' = 6

So y''' + xy'' - 2y' = 6 + x(6x) - 2(3x^2 + 3)
= 6 + 6x^2 - 6x^2 - 6
= 0

Answer 3

y ¹ (x) = 3x² + 3
y " (x) = 6x
y ¹¹¹ (x) = 6
y¹¹¹ + x y " - 2 y ¹ :-
6 + x (6x) - 2 (3x² + 3)
6 + 6x² - 6x² - 6
0 (as required)

Answer 4

You are given function y(x) = x^3+3x+1.

Differentiate it once and produce y'(x).
Differentiate y'(x) and produce y''(x).
Differentiate y''(x) and produce y'''(x).

Substitude expressions for y'(x), y''(x) and y'''(x) into
y'''+xy''-2y',
and show that everthing cancels everyting, and result is zero.