Use the substitution \(y=ux\), where \(u\) is a function of \(x\), to show that the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac x y + \frac y x
\quad \quad (x > 0, y> 0)
\] that satisfies \(y=2\) when \(x=1\)
is
\[
y= x\, \sqrt{4+2\ln x \, }
( x > \e^{-2}).
\]
Use a substitution to find the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac x y + \frac {2y} x
\quad \quad (x > 0, y > 0)
\] that satisfies \(y=2\) when \(x=1\).
Find the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac {x^2} y + \frac {2y} x
\quad \quad (x> 0, \ y> 0)
\] that satisfies \(y=2\) when \(x=1\).
Solution:
Let \(y = ux\), then \(\frac{\d y}{\d x} = x\frac{\d u}{\d x} = u\) and the differential equation becomes,
\begin{align*}
&& xu' + u &= \frac{1}{u} +u \\
\Rightarrow && u' &= \frac{1}{ux} \\
\Rightarrow && u u' &= \frac1{x} \\
\Rightarrow && \frac12 u^2 &= \ln x + C \\
(x,y) = (1,2): && \frac12 4 &= C \\
\Rightarrow && \frac12 \frac{y^2}{x^2} &= \ln x + 2 \\
\Rightarrow && y^2 &= x^2 \l 2\ln x + 4 \r \\
\Rightarrow && y &= x \sqrt{4 + 2 \ln x} \quad (x > e^{-2})
\end{align*}
Let \(y = ux^2\) then
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{x^2}{y} + \frac{2y}{x} \\
\Rightarrow && u' x^2 + 2x u &= \frac{1}{u} + 2x u \\
\Rightarrow && u' u &= \frac{1}{x^2} \\
\Rightarrow && \frac12 u^2 &= -\frac{1}{x} + C \\
(x,y) = (1,2): && 2 &= C - 1 \\
\Rightarrow && \frac12 \frac{y^2}{x^4} &= 3 - \frac{1}{x} \\
\Rightarrow && y &= x\sqrt{2(3x^2-x)}, \quad (x > \frac13)
\end{align*}