Question

solve the differential equation.
\\( \frac { d y } { d x } = 6 x ^ { 2 } \sqrt { y } , y > 0 \\)
choose the correct answer below.
a. \\( \sqrt { y } = 2 x ^ { 3 } + c \\)
b. \\( 2 \sqrt { y } = 6 x ^ { 2 } + c \\)
c. \\( 2 \sqrt { y } = 2 x ^ { 3 } + c \\)
d. \\( - \frac { 1 } { 2 y ^ { 3 / 2 } } = 12 x + c \\)

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{dx}=6x^{2}\sqrt{y}\). We get \(\frac{dy}{\sqrt{y}} = 6x^{2}dx\).

Step2: Integrate both sides

Integrate \(\int y^{-\frac{1}{2}}dy=\int6x^{2}dx\).
Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
For the left - hand side, \(\int y^{-\frac{1}{2}}dy=\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_1=\frac{y^{\frac{1}{2}}}{\frac{1}{2}}+C_1 = 2\sqrt{y}+C_1\).
For the right - hand side, \(\int6x^{2}dx=6\times\frac{x^{2 + 1}}{2+1}+C_2=6\times\frac{x^{3}}{3}+C_2=2x^{3}+C_2\).
Combining the two results (where \(C = C_2 - C_1\)), we get \(2\sqrt{y}=2x^{3}+C\).

Answer:

C. \(2\sqrt{y}=2x^{3}+C\)