What is the general solution to the differential equation that generated the slope field?

Choose 1 answer:
$\qquad$
(A) $y=c e^{x}-x-1$
$\qquad$
(B) $y=c e^{\frac{x^{2}}{2}}$
(C) $y=c e^{-\frac{x^{2}}{2}}$
(D) $y^{2}-x^{2}=C$ $\qquad$
(E) $x^{2}+y^{2}=C$ $\qquad$

Question

What is the general solution to the differential equation that generated the slope field?

Choose 1 answer:
$\qquad$
(A) $y=c e^{x}-x-1$
$\qquad$
(B) $y=c e^{\frac{x^{2}}{2}}$
(C) $y=c e^{-\frac{x^{2}}{2}}$
(D) $y^{2}-x^{2}=C$ $\qquad$
(E) $x^{2}+y^{2}=C$ $\qquad$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To determine the general solution to the differential equation that generated the slope field, we need to analyze the given options and match them with the characteristics of the slope field. ‖ ‖ step 2 ⋮ The slope field shows that the slopes are steeper in the top left and bottom right corners, and flatter in the top right and bottom left corners. This suggests a relationship involving $x^2$ and $y^2$. ‖ ‖ step 3 ⋮ Option (D) $y^2 - x^2 = C$ and Option (E) $x^2 + y^2 = C$ both involve $x^2$ and $y^2$. ‖ ‖ step 4 ⋮ The slope field does not suggest a circular symmetry, which would be the case for $x^2 + y^2 = C$. ‖ ‖ step 5 ⋮ Therefore, the correct general solution that matches the slope field is $y^2 - x^2 = C$. ‖ ***D*** ∻Key Concept∻ ⚹Differential Equation Solution⚹ ∻Explanation∻ ⚹The slope field's characteristics suggest a hyperbolic relationship, which matches the form $y^2 - x^2 = C$.⚹◊What is the relationship between the slope field graph and the general solution to the differential equation⍭ Generate me a similar question◊

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