Answer:
(-2,-3)(0,-5)(3,-8)
Step-by-step explanation:
The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.
The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.
This leaves us with the second choice. Recall that
1 + tan²(x) = sec²(x)
and the derivative of tangent,
(tan(x))' = sec²(x)
Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then
y' = sec²(x³/3 + 2) • x²
and substituting y and y' into the ODE gives
sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))
x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)
which is an identity.
So the solution is y = tan(x³/3 + 2).
5x^2 - 6x^2 = - x^2 so its either c or d.
and + 9x - )-3x) = 9x + 3x = + 12x
so the answer s C.
