Question

solve the separable differential equation. (use c for any needed constant.)
\\( \frac { d y } { d x } = 5 x y ^ { 2 } \quad ( y \
eq 0 ) \\)

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{dx}=5xy^{2}\). We get \(\frac{dy}{y^{2}} = 5x dx\) (since \(y
eq0\)).

Step2: Integrate both sides

Integrate \(\int y^{- 2}dy=\int5x dx\).
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, \(\int y^{-2}dy=\frac{y^{-2 + 1}}{-2+1}+C_1=-\frac{1}{y}+C_1\).
For the right - hand side, \(\int5x dx=5\times\frac{x^{2}}{2}+C_2=\frac{5}{2}x^{2}+C_2\).
So, \(-\frac{1}{y}=\frac{5}{2}x^{2}+C\) (where \(C = C_2 - C_1\)).

Step3: Solve for \(y\)

First, rewrite the equation as \(\frac{1}{y}=-\frac{5}{2}x^{2}-C\). Then \(y=\frac{1}{- \frac{5}{2}x^{2}-C}\), or \(y =-\frac{2}{5x^{2}+2C}\). Let \(K=-2C\), then \(y=\frac{-2}{5x^{2}+K}\)

Answer:

\(y=\frac{-2}{5x^{2}+C}\)