Question

solve the differential equation.

\\( \frac { d y } { d x } = 4 x ^ { 4 } \sqrt { y } , y > 0 \\)

choose the correct answer below.

\\( \bigcirc \\) a. \\( \sqrt { y } = \frac { 4 } { 5 } x ^ { 5 } + c \\)

\\( \bigcirc \\) b. \\( - \frac { 1 } { 2 y ^ { 3 / 2 } } = 16 x ^ { 3 } + c \\)

\\( \bigcirc \\) c. \\( 2 \sqrt { y } = 4 x ^ { 4 } + c \\)

\\( \bigcirc \\) d. \\( 2 \sqrt { y } = \frac { 4 } { 5 } x ^ { 5 } + c \\)

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{dx}=4x^{4}\sqrt{y}\).
We get \(\frac{dy}{\sqrt{y}} = 4x^{4}dx\).

Step2: Integrate both sides

Integrate \(\int y^{-\frac{1}{2}}dy=\int4x^{4}dx\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), for the left - hand side: \(\int y^{-\frac{1}{2}}dy=\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_1 = 2y^{\frac{1}{2}}+C_1\).
For the right - hand side: \(\int4x^{4}dx=4\times\frac{x^{4 + 1}}{4+1}+C_2=\frac{4}{5}x^{5}+C_2\).

Answer:

D. \(2\sqrt{y}=\frac{4}{5}x^{5}+C\)