4.3.2.0 Transformation on the dependent variable (method 1) y = v(x) z(x)

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Transformation on the dependent variable (method 1) \(y=v\left ( x\right ) z\left ( x\right ) \) ode internal name "second_order_change_of_variable_on_y_method_1"

This is also called Liouville transformation. Book by Einar Hille, ordinary diﬀerential equations in the complex domain. Page 179. This method assumes that\[ y=v\left ( x\right ) z\left ( x\right ) \] Substituting this into (A) results in the following ode where the dependent variable is \(v\) and not \(y\)\begin {equation} v^{\prime \prime }\left ( x\right ) +\left ( p+\frac {2}{z}z^{\prime }\left ( x\right ) \right ) v^{\prime }\left ( x\right ) +\frac {1}{z}\left ( z^{\prime \prime }\left ( x\right ) +pz^{\prime }\left ( x\right ) +qz\left ( x\right ) \right ) v\left ( x\right ) =\frac {r}{z} \tag {6} \end {equation} Assuming that coeﬃcient of \(v^{\prime }\) in (6) zero implies\[ p+\frac {2}{z}z^{\prime }\left ( x\right ) =0 \] Solving gives (where constant of integration is taken as one) \begin {equation} z=e^{-\int \frac {p}{2}dx} \tag {6A} \end {equation} With this choice (6) becomes\[ v^{\prime \prime }+\frac {1}{z}\left ( z^{\prime \prime }+pz^{\prime }+qz\right ) v=\frac {r}{z}\] Substituting \(z\) from (6A) into the above reduces it to (after some algebra) to\begin {equation} v^{\prime \prime }+q_{1}v=r_{1} \tag {6B} \end {equation} Where \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ r_{1} & =\frac {r}{z}\\ & =re^{\frac {1}{2}\int pdx} \end {align*}

\(q_{1}\) is called the Liouville ode invariant. If \(q_{1}\) is constant, then the substitution \(y=\) \(v\left ( x\right ) z\left ( x\right ) \) is used in the original original ode which will result in a constant coeﬃcient ode. In \(y=\) \(v\left ( x\right ) z\left ( x\right ) \) the \(z\left ( x\right ) \) term is known from 6A and \(v\left ( x\right ) \) is the new unknown dependent variable.

The new ode will be in \(v\left ( x\right ) \) but with constant coeﬃcients. Solving it for \(v\left ( x\right ) \) gives \(y\).

Example 1 \begin {equation} y^{\prime \prime }+\frac {2}{x}y^{\prime }+y=\frac {1}{x} \tag {1} \end {equation} In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=\frac {2}{x},q=1,r=\frac {1}{x}\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{-\int \frac {1}{x}dx}\\ & =e^{-\ln x}\\ & =\frac {1}{x} \end {align*} Now we check if \(q_{1}\) is constant or a constant divided by \(x^{2}\). \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =1-\frac {1}{2}\left ( \frac {2}{x}\right ) ^{\prime }-\frac {1}{4}\left ( \frac {2}{x}\right ) ^{2}\\ & =1-\left ( -\frac {1}{x^{2}}\right ) -\frac {1}{4}\frac {4}{x^{2}}\\ & =1-\left ( -\frac {1}{x^{2}}\right ) -\frac {1}{4}\frac {4}{x^{2}}\\ & =1+\frac {1}{x^{2}}-\frac {1}{x^{2}}\\ & =1 \end {align*}

Since \(q_{1}\) is constant, then we can use the change of the variable \(y=v\left ( x\right ) z\left ( x\right ) \) which is\[ y=\frac {v}{x}\] Since \(z=\frac {1}{x}\). Substituting the above into the original ODE (1) gives\begin {align*} \left ( \frac {v}{x}\right ) ^{\prime \prime }+\left ( \frac {2}{x}\left ( \frac {v}{x}\right ) ^{\prime }\right ) +\frac {v}{x} & =\frac {1}{x}\\ \left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) ^{\prime }+\frac {2}{x}\left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) +\frac {v}{x} & =\frac {1}{x}\\ \left ( \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\left ( \frac {v^{\prime }}{x^{2}}-2\frac {v}{x^{3}}\right ) \right ) +\frac {2}{x}\left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) +\frac {v}{x} & =\frac {1}{x}\\ \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\frac {v^{\prime }}{x^{2}}+2\frac {v}{x^{3}}+\frac {2v^{\prime }}{x^{2}}-\frac {2v}{x^{3}}+\frac {v}{x} & =\frac {1}{x}\\ \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\frac {v^{\prime }}{x^{2}}+\frac {2v^{\prime }}{x^{2}}+\frac {v}{x} & =\frac {1}{x}\\ \frac {v^{\prime \prime }}{x}+\frac {v}{x} & =\frac {1}{x}\\ v^{\prime \prime }+v & =1 \end {align*}

This is constant coeﬃcient ODE which is easily solved. If the ode in \(v\left ( x\right ) \) did not come to be constant coeﬃcient then we made a mistake. The solution is \[ v=c_{1}\cos x+c_{2}\sin x+1 \] Hence \begin {align*} y & =\frac {v}{x}\\ & =c_{1}\frac {\cos x}{x}+c_{2}\frac {\sin x}{x}+\frac {1}{x} \end {align*}

Example 2 \begin {align} x^{2}y^{\prime \prime }-x\left ( x+2\right ) y^{\prime }+\left ( x+2\right ) y & =2x^{3}\nonumber \\ y^{\prime \prime }-\frac {x+2}{x}y^{\prime }+\frac {x+2}{x^{2}}y & =2x \tag {1} \end {align} In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=-\frac {x+2}{x},q=\frac {\left ( x+2\right ) }{x^{2}},r=2x\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{\int \frac {x+2}{2x}dx}\\ & =xe^{\frac {x}{2}} \end {align*}

Now we check if Liouville ode invariant \(q_{1}\) is constant or a constant divided by \(x^{2}\). \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\frac {\left ( x+2\right ) }{x^{2}}-\frac {1}{2}\left ( xe^{\frac {x}{2}}\right ) ^{\prime }-\frac {1}{4}\left ( -\frac {x+2}{x}\right ) ^{2}\\ & =-\frac {1}{4} \end {align*}

Since \(q_{1}\) is constant, then we can use the change of the variable \(y=v\left ( x\right ) z\left ( x\right ) \) which is\begin {align*} y & =v\left ( x\right ) z\left ( x\right ) \\ & =v\left ( xe^{\frac {x}{2}}\right ) \end {align*}

Substituting the above into the original ODE (1) gives\begin {align*} y^{\prime \prime }-\frac {x+2}{x}y^{\prime }+\frac {x+2}{x}y & =2x\\ \left ( v\left ( xe^{\frac {x}{2}}\right ) \right ) ^{\prime \prime }-\frac {x+2}{x}\left ( v\left ( xe^{\frac {x}{2}}\right ) \right ) ^{\prime }+\frac {x+2}{x^{2}}v\left ( xe^{\frac {x}{2}}\right ) & =2x \end {align*}

Carrying out the simpliﬁcation gives\[ 4v^{\prime \prime }-v=8e^{-\frac {x}{2}}\] Which is constant coeﬃcient ode. This is easily solved giving the solution\[ v=c_{1}\sinh \left ( \frac {x}{2}\right ) +c_{2}\cosh \left ( \frac {x}{2}\right ) -2xe^{\frac {-x}{2}}\] Hence \begin {align*} y & =v\left ( x\right ) z\left ( x\right ) \\ & =\left ( c_{1}\sinh \left ( \frac {x}{2}\right ) +c_{2}\cosh \left ( \frac {x}{2}\right ) -2xe^{\frac {-x}{2}}\right ) xe^{\frac {x}{2}} \end {align*}

Example 3 \begin {equation} y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y=0 \tag {1} \end {equation} In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=-4x,q=\left ( 4x^{2}-2\right ) ,r=0\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{\int 2xdx}\\ & =e^{x^{2}} \end {align*} Now we check if Liouville ode invariant \(q_{1}\) is constant or a constant divided by \(x^{2}\). \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( 4x^{2}-2\right ) -\frac {1}{2}\left ( -4x\right ) ^{\prime }-\frac {1}{4}\left ( -4x\right ) ^{2}\\ & =\left ( 4x^{2}-2\right ) +2-\frac {1}{4}\left ( 16x^{2}\right ) \\ & =4x^{2}-2+2-4x^{2}\\ & =0 \end {align*}

Substituting the above into the original ODE (1) gives\begin {align*} y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y & =0\\ \left ( ve^{x^{2}}\right ) ^{\prime \prime }-4x\left ( ve^{x^{2}}\right ) ^{\prime }+\left ( 4x^{2}-2\right ) ve^{x^{2}} & =0 \end {align*}

Carrying out the simpliﬁcation gives\[ v^{\prime \prime }=0 \] Which is constant coeﬃcient ode. This is easily solved giving the solution\[ v=c_{1}+c_{2}x \] Hence \begin {align*} y & =v\left ( x\right ) z\left ( x\right ) \\ & =\left ( c_{1}+c_{2}x\right ) e^{x^{2}} \end {align*}

Example 4 \begin {equation} x^{2}y^{\prime \prime }+3xy^{\prime }+y=0 \tag {1} \end {equation} This is of course Euler ode, and we do not need to try this method as solving it as Euler ode is much simpler. But this is just for illustration for the case when the Liouville ode invariant comes out not a constant. In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \begin {equation} y^{\prime \prime }+\frac {3}{x}y^{\prime }+\frac {1}{x^{2}}y=0 \tag {1A} \end {equation} Where now \(p=\frac {3}{x},q=\frac {1}{x^{2}},r=0\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{\frac {-3}{2}\int \frac {1}{x}dx}\\ & =\frac {1}{x^{\frac {3}{2}}} \end {align*} Now we check if Liouville ode invariant \(q_{1}\) is constant. \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( \frac {1}{x^{2}}\right ) -\frac {1}{2}\left ( \frac {3}{x}\right ) ^{\prime }-\frac {1}{4}\left ( \frac {3}{x}\right ) ^{2}\\ & =\left ( \frac {1}{x^{2}}\right ) -\frac {3}{2}\left ( \frac {-1}{x^{2}}\right ) -\frac {1}{4}\left ( \frac {9}{x^{2}}\right ) \\ & =\frac {1}{x^{2}}+\frac {3}{2x^{2}}-\frac {9}{4x^{2}}\\ & =\frac {1}{4x^{2}} \end {align*}

Since \(q_{1}\) is not constant then the ode can not not converted to an ode in \(v\left ( x\right ) \) with constant coeﬃcient.

Example 5 \begin {equation} xy^{\prime \prime }+2y^{\prime }-xy=0 \tag {1} \end {equation} In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \begin {equation} y^{\prime \prime }+\frac {2}{x}y^{\prime }-y=0 \tag {1A} \end {equation} Where now \(p=\frac {2}{x},q=-1,r=0\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{-\int \frac {1}{x}dx}\\ & =\frac {1}{x} \end {align*} Now we check if Liouville ode invariant \(q_{1}\) is constant. \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( -1\right ) -\frac {1}{2}\left ( \frac {2}{x}\right ) ^{\prime }-\frac {1}{4}\left ( \frac {2}{x}\right ) ^{2}\\ & =-1-\left ( -\frac {1}{x^{2}}\right ) -\frac {1}{x^{2}}\\ & =-1+\frac {1}{x^{2}}-\frac {1}{x^{2}}\\ & =-1 \end {align*}

Substituting the above into the original ODE (1A) gives\begin {align*} y^{\prime \prime }+\frac {2}{x}y^{\prime }-y & =0\\ \left ( v\frac {1}{x}\right ) ^{\prime \prime }+\frac {2}{x}\left ( v\frac {1}{x}\right ) ^{\prime }-v\frac {1}{x} & =0 \end {align*}

Carrying out the simpliﬁcation gives\[ v^{\prime \prime }-v=0 \] Which is constant coeﬃcient ode. This is easily solved giving the solution\[ v=c_{1}e^{x}+c_{2}e^{-x}\] Hence \begin {align*} y & =v\left ( x\right ) z\left ( x\right ) \\ & =\left ( c_{1}e^{x}+c_{2}e^{-x}\right ) \frac {1}{x} \end {align*}

Example 6 \begin {equation} y^{\prime \prime }-\frac {1}{\sqrt {x}}y^{\prime }+\left ( \frac {1}{4x}+\frac {1}{4x^{\frac {3}{2}}}-\frac {2}{x^{2}}\right ) y=0 \tag {1} \end {equation} In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=-\frac {1}{\sqrt {x}},q=\left ( \frac {1}{4x}+\frac {1}{4x^{\frac {3}{2}}}-\frac {2}{x^{2}}\right ) ,r=0\). Hence (6A) is\begin {align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{\int \frac {1}{\sqrt {x}}dx}\\ & =e^{2\sqrt {x}} \end {align*} Now we check if Liouville ode invariant \(q_{1}\) is constant. \begin {align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( \frac {1}{4x}+\frac {1}{4x^{\frac {3}{2}}}-\frac {2}{x^{2}}\right ) -\frac {1}{2}\left ( -\frac {1}{\sqrt {x}}\right ) ^{\prime }-\frac {1}{4}\left ( -\frac {1}{\sqrt {x}}\right ) ^{2}\\ & =-\frac {2}{x^{2}} \end {align*}

Not constant. Stop here. This can be solved using Kovacic algorithm.

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