Question

Find the general solution of the equation: (x^2)(dy/dx)=(y^1/2)(3x+1)

Answer 1

we can rewrite your ODE as follows:
\[\Large \begin{gathered}
{x^2}y' = \left( {3x + 1} \right)\sqrt y \hfill \\
\hfill \\
\frac{{dy}}{{\sqrt y }} = \frac{{3x + 1}}{{{x^2}}}dx \hfill \\
\end{gathered} \]

Answer 2

now we can easily integrate both sides, so we get:
\[\Large 2\sqrt y = 3\ln \left| x \right| - \frac{1}{x} + k\]

Answer 3

finally,dividing ooth sides by 2, we get:
\[\Large \sqrt y = 3\ln \left( {\sqrt x } \right) - \frac{1}{{2x}} + k\]

Answer 4

squaring both sides, we get:
\[\Large y\left( x \right) = {\left( {3\ln \left( {\sqrt x } \right) - \frac{1}{{2x}} + k} \right)^2}\]

Answer 5

where, as usual, k is the arbitrary real constant of integration

Answer 6

Are there any more steps to that?