Question

solve the differential equation.

$2\sqrt {xy}\frac {dy}{dx}=1$, $x,y>0$

the solution is
(type an equation.)

Explanation:

Step1: Separate the variables

Given \(2\sqrt{xy}\frac{dy}{dx}=1\), we can rewrite it as \(2\sqrt{y}dy=\frac{1}{\sqrt{x}}dx\) (since \(x,y > 0\)).

Step2: Integrate both sides

Integrate \(\int2y^{\frac{1}{2}}dy=\int x^{-\frac{1}{2}}dx\).
Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)):
For the left - hand side, \(\int2y^{\frac{1}{2}}dy=2\times\frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{4}{3}y^{\frac{3}{2}}+C_1\).
For the right - hand side, \(\int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_2 = 2x^{\frac{1}{2}}+C_2\).
Combining the two results (let \(C = C_2 - C_1\)), we get \(\frac{4}{3}y^{\frac{3}{2}}=2x^{\frac{1}{2}}+C\).
Multiply through by \(\frac{3}{2}\) to simplify: \(2y^{\frac{3}{2}}=3x^{\frac{1}{2}}+C\) (where \(C\) is an arbitrary constant).

Answer:

\(2y^{\frac{3}{2}}=3x^{\frac{1}{2}}+C\)