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

1. Michele_Laino

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

2. Michele_Laino

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

3. Michele_Laino

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

4. Michele_Laino

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

5. Michele_Laino

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

6. anonymous

Are there any more steps to that?