Answer:
D. (3x^2-3x+27)^1/3
Step-by-step explanation:
dy/dx=2x−1/y^2
Cross multiply to get Y's and X's on different sides.
(y^2)dy = (2x-1)dx
Take the integral of each side
∫(y^2)dy = ∫(2x-1)dx
y^3/3 = 2x^2/2 - x + C
Multiply both sides by 3
y^3 = 3x^2 - 3x + C
Plug in (0,3) and solve for C
27 = 0 - 0 + C
C = 27
Plug in C to the equation
y^3 = 3x^2-3x+27
Cube root both sides to get just y
y = √(3x^2-3x+27)
y = (3x^2-3x+27)^1/3
D.
The particular solution of the equation will be "[tex](3x^2-3x+27)^\frac{1}{3}[/tex]".
According to the question,
→ [tex]\frac{dy}{dx}[/tex] = [tex]\frac{2x-1}{y^2}[/tex]
By applying cross-multiplication,
(y²)dy = (2x - 1)dx
By integrating both sides, we get
→ ∫(y²) dy = ∫(2x - 1) dx
[tex]\frac{y^3}{3}[/tex] = [tex]\frac{2x^2}{2}[/tex] - x + C
By multiplying "3" both sides, we get
y³ = 3x² - 3x + c
By substituting "(0,3)" in above equation,
27 = 0 - 0 + C
C = 27
By substituting value of C in the above equation,
y³ = 3x² - 3x + 27
By cubing both sides,
y = √(3x²-3x+27)
= [tex](3x^2-3x+27)^\frac{1}{3}[/tex]
Thus the above answer i.e., "Option D" is correct.
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