Transformation on the dependent variable (method 2) \(y=v\left ( x\right ) x^{n}\) ode internal name "second_order_change_of_variable_on_y_method_2"

This transformation, if it works, changes the second order ode to an one with missing \(y\), which then can be solved as first order ode by reduction of order. This transformation does not necessarily changes the second order ode to one with constant coefficient like the above general transformation. But to an ode with missing \(y\).

This method assumes\[ y=v\left ( x\right ) x^{n}\] If this transformation changes the ode to one with missing \(y\), then it can be used. Substituting this in (A) results in the following ode where the dependent variable is now \(v\) and not \(y\)\begin {align} x^{n}v^{\prime \prime }+\left ( 2x^{n-1}n+x^{n}p\right ) v^{\prime }+\left ( n\left ( n-1\right ) x^{n-2}+npx^{n-1}+qx^{n}\right ) v & =r\nonumber \\ v^{\prime \prime }+\left ( 2\frac {n}{x}+p\right ) v^{\prime }+\left ( n\left ( n-1\right ) x^{-2}+npx^{-1}+q\right ) v & =\frac {r}{x^{n}} \tag {7} \end {align}

If it happens that \begin {equation} n\left ( n-1\right ) x^{-2}+npx^{-1}+q=0 \tag {7A} \end {equation} For some integer or rational number \(n\), then (7) becomes\begin {equation} v^{\prime \prime }+\left ( 2\frac {n}{x}+p\right ) v^{\prime }=\frac {r}{x^{n}} \tag {7B} \end {equation} Which now can be solved using substitution \(u=v^{\prime }\). \[ u^{\prime }+\left ( 2\frac {n}{x}+p\right ) u=\frac {r}{x^{n}}\] Which is linear first order ode. Once \(u\) is found, then \(v\) is by found integration. Hence \(y\) is now found. To use this method, all what we need is to check if (7A) is true for some number \(n\). Typically one tries \(n=\pm 1\) first and if this does not work, then try to find other values. Example below shows how to apply this method.