The formula of Liouville - Ostrogradsky

The Ostrogradsky-Liouville formula is a formula linking the Wronsky determinant (Wronskian) for solutions of a differential equation and the coefficients in this equation.

Let there be a differential equation of the form

$y^{(n)}+P_{1}(x)y^{(n-1)}+P_{2}(x)y^{(n-2)}+...+P_{n}(x)y=0,$ ${\ displaystyle y ^ {(n)} + P_ {1} (x) y ^ {(n-1)} + P_ {2} (x) y ^ {(n-2)} + ... + P_ {n} (x) y = 0,}$ $y ^ {{((n)}} + P_ {1} (x) y ^ {{(n-1)}} + P_ {2} (x) y ^ {{(n-2)}} + .. . + P_ {n} (x) y = 0,$

then $W(x)=W(x_{0})e^{-\int _{x_{0}}^{x}P_{1}(\zeta )d\zeta }=Ce^{-\int P_{1}(x)dx},$ ${\ displaystyle W (x) = W (x_ {0}) e ^ {- \ int _ {x_ {0}} ^ {x} P_ {1} (\ zeta) d \ zeta} = Ce ^ {- \ int P_ {1} (x) dx},}$ $W (x) = W (x_ {0}) e ^ {{- \ int _ {{x_ {0}}} ^ {x} P_ {1} (\ zeta) d \ zeta}} = Ce ^ {{ - \ int P_ {1} (x) dx}},$ Where $W(x)$ ${\ displaystyle W (x)}$ $W (x)$ - Vronsky qualifier

For a linear homogeneous system of differential equations

$y'(x)=A(x)y(x),$ ${\ displaystyle y '(x) = A (x) y (x),}$ $y '(x) = A (x) y (x),$ Where $A(x)$ ${\ displaystyle A (x)}$ $A (x)$ - continuous square order matrix $n$ ${\ displaystyle n}$ $n$ , the Liouville-Ostrogradsky formula is valid

$W(x)=W(x_{0})e^{\int _{x_{0}}^{x}\mathop {\rm {tr}} A(\zeta )d\zeta },$ ${\ displaystyle W (x) = W (x_ {0}) e ^ {\ int _ {x_ {0}} ^ {x} \ mathop {\ rm {tr}} A (\ zeta) d \ zeta}, }$ $W (x) = W (x_ {0}) e ^ {{\ int _ {{x_ {0}}} ^ {x} {\ mathop {{\ rm {tr}}}} A (\ zeta) d \ zeta}},$ Where $\mathop {\rm {tr}} A(x)$ ${\ displaystyle \ mathop {\ rm {tr}} A (x)}$ ${\ mathop {{\ rm {tr}}}} A (x)$ - matrix trace $A(x)$ ${\ displaystyle A (x)}$ $A (x)$

Content

Differentiation rule for dimension determinant 2

Derivative of the determinant $\Delta =\Delta (x)={\begin{vmatrix}a_{11}(x)&a_{12}(x)\\a_{21}(x)&a_{22}(x)\end{vmatrix}}=a_{11}a_{22}-a_{12}a_{21}$ ${\ displaystyle \ Delta = \ Delta (x) = {\ begin {vmatrix} a_ {11} (x) & a_ {12} (x) \\ a_ {21} (x) & a_ {22} (x) \ end {vmatrix}} = a_ {11} a_ {22} -a_ {12} a_ {21}}$ by the variable x has the form ${\frac {d\Delta }{dx}}=(a_{11}a_{22}-a_{12}a_{21})'=a_{11}'a_{22}+a_{11}a_{22}'-a_{12}'a_{21}-a_{12}a_{21}'={\begin{vmatrix}a_{11}'&a_{12}'\\a_{21}&a_{22}\end{vmatrix}}+{\begin{vmatrix}a_{11}&a_{12}\\a_{21}'&a_{22}'\end{vmatrix}}$ ${\ displaystyle {\ frac {d \ Delta} {dx}} = (a_ {11} a_ {22} -a_ {12} a_ {21}) '= a_ {11}' a_ {22} + a_ {11 } a_ {22} '- a_ {12}' a_ {21} -a_ {12} a_ {21} '= {\ begin {vmatrix} a_ {11}' & a_ {12} '\\ a_ {21} & a_ {22} \ end {vmatrix}} + {\ begin {vmatrix} a_ {11} & a_ {12} \\ a_ {21} '& a_ {22}' \ end {vmatrix}}}$

Rule of differentiation of a dimension determinant $n$ ${\ displaystyle n}$ $n$

Let be $\Delta =\Delta (x)=\det \left({\begin{matrix}a_{11}(x)&a_{12}(x)&\dots &a_{1n}(x)\\a_{21}(x)&a_{22}(x)&\dots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\dots &a_{nn}(x)\\\end{matrix}}\right)$ ${\ displaystyle \ Delta = \ Delta (x) = \ det \ left ({\ begin {matrix} a_ {11} (x) & a_ {12} (x) & \ dots & a_ {1n} (x) \\ a_ {21} (x) & a_ {22} (x) & \ dots & a_ {2n} (x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ {n1} (x) & a_ {n2} ( x) & \ dots & a_ {nn} (x) \\\ end {matrix}} \ right)}$

Then for the derivative $\Delta '(x)$ ${\ displaystyle \ Delta '(x)}$ right

$\Delta '(x)={\begin{vmatrix}a_{11}'(x)&a_{12}'(x)&\dots &a_{1n}'(x)\\a_{21}(x)&a_{22}(x)&\dots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\dots &a_{nn}(x)\\\end{vmatrix}}+{\begin{vmatrix}a_{11}(x)&a_{12}(x)&\dots &a_{1n}(x)\\a_{21}'(x)&a_{22}'(x)&\dots &a_{2n}'(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\dots &a_{nn}(x)\\\end{vmatrix}}+\dots +{\begin{vmatrix}a_{11}(x)&a_{12}(x)&\dots &a_{1n}(x)\\a_{21}(x)&a_{22}(x)&\dots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}'(x)&a_{n2}'(x)&\dots &a_{nn}'(x)\\\end{vmatrix}}$ ${\ displaystyle \ Delta '(x) = {\ begin {vmatrix} a_ {11}' (x) & a_ {12} '(x) & \ dots & a_ {1n}' (x) \\ a_ {21} ( x) & a_ {22} (x) & \ dots & a_ {2n} (x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ {n1} (x) & a_ {n2} (x) & \ dots & a_ {nn} (x) \\\ end {vmatrix}} + {\ begin {vmatrix} a_ {11} (x) & a_ {12} (x) & \ dots & a_ {1n} (x) \\ a_ {21} '(x) & a_ {22}' (x) & \ dots & a_ {2n} '(x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ {n1} (x) & a_ { n2} (x) & \ dots & a_ {nn} (x) \\\ end {vmatrix}} + \ dots + {\ begin {vmatrix} a_ {11} (x) & a_ {12} (x) & \ dots & a_ {1n} (x) \\ a_ {21} (x) & a_ {22} (x) & \ dots & a_ {2n} (x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ { n1} '(x) & a_ {n2}' (x) & \ dots & a_ {nn} '(x) \\\ end {vmatrix}}}$

(at $i$ ${\ displaystyle i}$ -th term is differentiated $i$ ${\ displaystyle i}$ st row)

Evidence

We use the formula for the complete decomposition of the determinant

$\Delta (x)=\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}a_{1i_{1}}(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}(x)$ ${\ displaystyle \ Delta (x) = \ sum _ {i_ {1}, i_ {2}, \ dots, i_ {n}} (- 1) ^ {P (i_ {1}, i_ {2}, \ dots, i_ {n})} a_ {1i_ {1}} (x) a_ {2i_ {2}} (x) \ cdots a_ {ni_ {n}} (x)}$

The sum is taken from various permutations of numbers $1,2,\dots ,n$ ${\ displaystyle 1,2, \ dots, n}$ , $P(\cdot )$ ${\ displaystyle P (\ cdot)}$ - parity permutation .

Differentiating this expression by $x$ ${\ displaystyle x}$ we get

${\begin{aligned}[l]\Delta '(x)&=\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}{\frac {d\left(a_{1i_{1}}(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}(x)\right)}{dx}}=\\&=\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}\left(a_{1i_{1}}'(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}(x)+\dots +a_{1i_{1}}(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}'(x)\right)=\\&=\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}a_{1i_{1}}'(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}(x)+\\&+\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}a_{1i_{1}}(x)a_{2i_{2}}'(x)\cdots a_{ni_{n}}(x)+\\&+\dots +\\&+\sum _{i_{1},i_{2},\dots ,i_{n}}(-1)^{P(i_{1},i_{2},\dots ,i_{n})}a_{1i_{1}}(x)a_{2i_{2}}(x)\cdots a_{ni_{n}}'(x)\end{aligned}}$ ${\ displaystyle {\ begin {aligned} [l] \ Delta '(x) & = \ sum _ {i_ {1}, i_ {2}, \ dots, i_ {n}} (- 1) ^ {P ( i_ {1}, i_ {2}, \ dots, i_ {n})} {\ frac {d \ left (a_ {1i_ {1}} (x) a_ {2i_ {2}} (x) \ cdots a_ {ni_ {n}} (x) \ right)} {dx}} = \\ & = \ sum _ {i_ {1}, i_ {2}, \ dots, i_ {n}} (- 1) ^ { P (i_ {1}, i_ {2}, \ dots, i_ {n})} \ left (a_ {1i_ {1}} '(x) a_ {2i_ {2}} (x) \ cdots a_ {ni_ {n}} (x) + \ dots + a_ {1i_ {1}} (x) a_ {2i_ {2}} (x) \ cdots a_ {ni_ {n}} '(x) \ right) = \\ & = \ sum _ {i_ {1}, i_ {2}, \ dots, i_ {n}} (- 1) ^ {P (i_ {1}, i_ {2}, \ dots, i_ {n}) } a_ {1i_ {1}} '(x) a_ {2i_ {2}} (x) \ cdots a_ {ni_ {n}} (x) + \\ & + \ sum _ {i_ {1}, i_ { 2}, \ dots, i_ {n}} (- 1) ^ {P (i_ {1}, i_ {2}, \ dots, i_ {n})} a_ {1i_ {1}} (x) a_ { 2i_ {2}} '(x) \ cdots a_ {ni_ {n}} (x) + \\ & + \ dots + \\ & + \ sum _ {i_ {1}, i_ {2}, \ dots, i_ {n}} (- 1) ^ {P (i_ {1}, i_ {2}, \ dots, i_ {n})} a_ {1i_ {1}} (x) a_ {2i_ {2}} ( x) \ cdots a_ {ni_ {n}} '(x) \ end {aligned}}}$

In each sum, the elements are differentiated $i$ ${\ displaystyle i}$ st row and only they. Replacing the sums with determinants, we obtain

$\Delta '(x)={\begin{vmatrix}a_{11}'(x)&a_{12}'(x)&\dots &a_{1n}'(x)\\a_{21}(x)&a_{22}(x)&\dots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\dots &a_{nn}(x)\\\end{vmatrix}}+{\begin{vmatrix}a_{11}(x)&a_{12}(x)&\dots &a_{1n}(x)\\a_{21}'(x)&a_{22}'(x)&\dots &a_{2n}'(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\dots &a_{nn}(x)\\\end{vmatrix}}+\dots +{\begin{vmatrix}a_{11}(x)&a_{12}(x)&\dots &a_{1n}(x)\\a_{21}(x)&a_{22}(x)&\dots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}'(x)&a_{n2}'(x)&\dots &a_{nn}'(x)\\\end{vmatrix}}$ ${\ displaystyle \ Delta '(x) = {\ begin {vmatrix} a_ {11}' (x) & a_ {12} '(x) & \ dots & a_ {1n}' (x) \\ a_ {21} ( x) & a_ {22} (x) & \ dots & a_ {2n} (x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ {n1} (x) & a_ {n2} (x) & \ dots & a_ {nn} (x) \\\ end {vmatrix}} + {\ begin {vmatrix} a_ {11} (x) & a_ {12} (x) & \ dots & a_ {1n} (x) \\ a_ {21} '(x) & a_ {22}' (x) & \ dots & a_ {2n} '(x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ {n1} (x) & a_ { n2} (x) & \ dots & a_ {nn} (x) \\\ end {vmatrix}} + \ dots + {\ begin {vmatrix} a_ {11} (x) & a_ {12} (x) & \ dots & a_ {1n} (x) \\ a_ {21} (x) & a_ {22} (x) & \ dots & a_ {2n} (x) \\\ vdots & \ vdots & \ ddots & \ vdots \\ a_ { n1} '(x) & a_ {n2}' (x) & \ dots & a_ {nn} '(x) \\\ end {vmatrix}}}$

Proof for a second order equation

Let in the equation $y''+p(x)y'+q(x)y=0$ ${\ displaystyle y '' + p (x) y '+ q (x) y = 0}$ the functions $p(x),q(x)$ ${\ displaystyle p (x), q (x)}$ continuous on $[a;b]$ ${\ displaystyle [a; b]}$ , but

$y_{1}=y_{1}(x),y_{2}=y_{2}(x)$ ${\ displaystyle y_ {1} = y_ {1} (x), y_ {2} = y_ {2} (x)}$ - solutions of this equation.

Differentiating Vronsky's determinant, we obtain

${\frac {dW}{dx}}={\frac {d}{dx}}{\begin{vmatrix}y_{1}&y_{2}\\y_{1}'&y_{2}'\end{vmatrix}}={\begin{vmatrix}y_{1}'&y_{2}'\\y_{1}'&y_{2}'\end{vmatrix}}+{\begin{vmatrix}y_{1}&y_{2}\\y_{1}''&y_{2}''\end{vmatrix}}$ ${\ displaystyle {\ frac {dW} {dx}} = {\ frac {d} {dx}} {\ begin {vmatrix} y_ {1} & y_ {2} \\ y_ {1} '& y_ {2}' \ end {vmatrix}} = {\ begin {vmatrix} y_ {1} '& y_ {2}' \\ y_ {1} '& y_ {2}' \ end {vmatrix}} + {\ begin {vmatrix} y_ { 1} & y_ {2} \\ y_ {1} '' & y_ {2} '' \ end {vmatrix}}}$

The first term is 0, since this determinant contains 2 identical lines. Substituting

$y_{1}''=-py_{1}'-qy_{1}$ ${\ displaystyle y_ {1} '' = - py_ {1} '- qy_ {1}}$

$y_{2}''=-py_{2}'-qy_{2}$ ${\ displaystyle y_ {2} '' = - py_ {2} '- qy_ {2}}$

in the second term, we get

${\frac {dW}{dx}}={\begin{vmatrix}y_{1}&y_{2}\\-py_{1}'-qy_{1}&-py_{2}'-qy_{2}\end{vmatrix}}$ ${\ displaystyle {\ frac {dW} {dx}} = {\ begin {vmatrix} y_ {1} & y_ {2} \\ - py_ {1} '- qy_ {1} & - py_ {2}' - qy_ {2} \ end {vmatrix}}}$

Adding the first row multiplied by q to the second, we get

${\frac {dW}{dx}}={\begin{vmatrix}y_{1}&y_{2}\\-py_{1}'&-py_{2}'\end{vmatrix}}=-pW$ ${\ displaystyle {\ frac {dW} {dx}} = {\ begin {vmatrix} y_ {1} & y_ {2} \\ - py_ {1} '& - py_ {2}' \ end {vmatrix}} = -pW}$

solutions are linearly independent, therefore

$W\neq 0\to {\frac {dW}{W}}=-pdx$ ${\ displaystyle W \ neq 0 \ to {\ frac {dW} {W}} = - pdx}$ - differential equation with separable variables.

Integrating, we get

$\ln |W|=-\int p(x)dx+\ln |C|\to \ln \left|{\frac {W}{C}}\right|=-\int p(x)dx\to W=Ce^{-\int p(x)dx}$ ${\ displaystyle \ ln | W | = - \ int p (x) dx + \ ln | C | \ to \ ln \ left | {\ frac {W} {C}} \ right | = - \ int p (x) dx \ to W = Ce ^ {- \ int p (x) dx}}$

Proof for a linear system of ordinary differential equations

Let vector functions ${\mathbf {y} }_{1}(x),{\mathbf {y} }_{2}(x),\dots ,{\mathbf {y} }_{n}(x)$ ${\ displaystyle {\ mathbf {y}} _ {1} (x), {\ mathbf {y}} _ {2} (x), \ dots, {\ mathbf {y}} _ {n} (x) }$ - solutions of the linear ODE system. We introduce the matrix $\Phi$ ${\ displaystyle \ Phi}$ in the following way

$\Phi (x)=\left\|{\begin{matrix}{\mathbf {y} }_{1}(x)&{\mathbf {y} }_{2}(x)&\dots &{\mathbf {y} }_{n}(x)\end{matrix}}\right\|$ ${\ displaystyle \ Phi (x) = \ left \ | {\ begin {matrix} {\ mathbf {y}} _ {1} (x) & {\ mathbf {y}} _ {2} (x) & \ dots & {\ mathbf {y}} _ {n} (x) \ end {matrix}} \ right \ |}$

Then $W(x)\equiv \det \Phi (x)$ ${\ displaystyle W (x) \ equiv \ det \ Phi (x)}$ . Take advantage of the fact that $y_{i}(x)$ ${\ displaystyle y_ {i} (x)}$ - solutions of the ODE system, i.e. ${\mathbf {y} }_{i}'(x)=A(x){\mathbf {y} }_{i}(x)$ ${\ displaystyle {\ mathbf {y}} _ {i} '(x) = A (x) {\ mathbf {y}} _ {i} (x)}$ .

In matrix form, the latter can be represented as $\left\|{\begin{matrix}{\mathbf {y} }_{1}'(x)&{\mathbf {y} }_{2}'(x)&\dots &{\mathbf {y} }_{n}'(x)\end{matrix}}\right\|=\left\|{\begin{matrix}A(x){\mathbf {y} }_{1}(x)&A(x){\mathbf {y} }_{2}(x)&\dots &A(x){\mathbf {y} }_{n}(x)\end{matrix}}\right\|=A(x)\Phi (x)$ ${\ displaystyle \ left \ | {\ begin {matrix} {\ mathbf {y}} _ {1} '(x) & {\ mathbf {y}} _ {2}' (x) & \ dots & {\ mathbf {y}} _ {n} '(x) \ end {matrix}} \ right \ | = \ left \ | {\ begin {matrix} A (x) {\ mathbf {y}} _ {1} ( x) & A (x) {\ mathbf {y}} _ {2} (x) & \ dots & A (x) {\ mathbf {y}} _ {n} (x) \ end {matrix}} \ right \ | = A (x) \ Phi (x)}$

or introducing a derivative of a matrix as a matrix of derivatives of each element

$\Phi '(x)=A(x)\Phi (x)$ ${\ displaystyle \ Phi '(x) = A (x) \ Phi (x)}$

Let be $\varphi _{i}(x)$ ${\ displaystyle \ varphi _ {i} (x)}$ - $i$ ${\ displaystyle i}$ matrix row $\Phi (x)$ ${\ displaystyle \ Phi (x)}$ . Then

$\varphi _{i}'(x)=\sum _{j=1}^{n}a_{ij}(x)\varphi _{j}(x)$ ${\ displaystyle \ varphi _ {i} '(x) = \ sum _ {j = 1} ^ {n} a_ {ij} (x) \ varphi _ {j} (x)}$

The latter means that the derivative of $i$ ${\ displaystyle i}$ matrix row $\Phi (x)$ ${\ displaystyle \ Phi (x)}$ there is a linear combination of all rows of this matrix with coefficients from $i$ ${\ displaystyle i}$ matrix row $A(x)$ ${\ displaystyle A (x)}$ . Consider the determinant of a matrix $\Phi (x)$ ${\ displaystyle \ Phi (x)}$ , wherein $i$ ${\ displaystyle i}$ -th line is differentiated. The determinant does not change if from $i$ ${\ displaystyle i}$ the th row of this matrix subtract the linear combination of all the other rows.

$\left|{\begin{matrix}\varphi _{1}(x)\\\varphi _{2}(x)\\\vdots \\\varphi _{i}'(x)\\\vdots \\\varphi _{n}(x)\\\end{matrix}}\right|=\left|{\begin{matrix}\varphi _{1}(x)\\\varphi _{2}(x)\\\vdots \\\sum _{j=1}^{n}a_{ij}(x)\varphi _{j}(x)\\\vdots \\\varphi _{n}(x)\\\end{matrix}}\right|=\left|{\begin{matrix}\varphi _{1}(x)\\\varphi _{2}(x)\\\vdots \\\sum _{j=1}^{n}a_{ij}(x)\varphi _{j}(x)-\sum _{j\neq i}a_{ij}(x)\varphi _{j}(x)\\\vdots \\\varphi _{n}(x)\\\end{matrix}}\right|=\left|{\begin{matrix}\varphi _{1}(x)\\\varphi _{2}(x)\\\vdots \\a_{ii}(x)\varphi _{i}(x)\\\vdots \\\varphi _{n}(x)\\\end{matrix}}\right|=a_{ii}(x)W(x)$ ${\ displaystyle \ left | {\ begin {matrix} \ varphi _ {1} (x) \\\ varphi _ {2} (x) \\\ vdots \\\ varphi _ {i} '(x) \\ \ vdots \\\ varphi _ {n} (x) \\\ end {matrix}} \ right | = \ left | {\ begin {matrix} \ varphi _ {1} (x) \\\ varphi _ {2 } (x) \\\ vdots \\\ sum _ {j = 1} ^ {n} a_ {ij} (x) \ varphi _ {j} (x) \\\ vdots \\\ varphi _ {n} (x) \\\ end {matrix}} \ right | = \ left | {\ begin {matrix} \ varphi _ {1} (x) \\\ varphi _ {2} (x) \\\ vdots \\ \ sum _ {j = 1} ^ {n} a_ {ij} (x) \ varphi _ {j} (x) - \ sum _ {j \ neq i} a_ {ij} (x) \ varphi _ {j } (x) \\\ vdots \\\ varphi _ {n} (x) \\\ end {matrix}} \ right | = \ left | {\ begin {matrix} \ varphi _ {1} (x) \ \\ varphi _ {2} (x) \\\ vdots \\ a_ {ii} (x) \ varphi _ {i} (x) \\\ vdots \\\ varphi _ {n} (x) \\\ end {matrix}} \ right | = a_ {ii} (x) W (x)}$

Using the differentiation formula for the determinant, we obtain

$W'(x)=a_{11}(x)W(x)+a_{22}(x)W(x)+\dots +a_{nn}(x)W(x)=\operatorname {tr} A(x)W(x)$ ${\ displaystyle W '(x) = a_ {11} (x) W (x) + a_ {22} (x) W (x) + \ dots + a_ {nn} (x) W (x) = \ operatorname {tr} A (x) W (x)}$

The last ordinary differential equation has a solution

$W(x)=W(x_{0})e^{\int _{x_{0}}^{x}\operatorname {tr} A(\zeta )d\zeta }$ ${\ displaystyle W (x) = W (x_ {0}) e ^ {\ int _ {x_ {0}} ^ {x} \ operatorname {tr} A (\ zeta) d \ zeta}}$

Proof for a linear differential equation of arbitrary order

Linear differential equation $n$ ${\ displaystyle n}$ th order

$y^{(n)}(x)+P_{1}(x)y^{(n-1)}(x)+\dots +P_{n-1}(x)y'(x)+P_{n}(x)y(x)=0$ ${\ displaystyle y ^ {(n)} (x) + P_ {1} (x) y ^ {(n-1)} (x) + \ dots + P_ {n-1} (x) y '(x ) + P_ {n} (x) y (x) = 0}$

equivalent to the following system

${\begin{aligned}y_{n-1}'(x)&=-P_{1}(x)y_{n-1}(x)-\dots -P_{n-1}(x)y_{1}(x)-P_{n}(x)y_{0}(x)\\y_{n-2}'(x)&=y_{n-1}\\\vdots \\y_{1}'(x)&=y_{2}\\y_{0}'(x)&=y_{1}\\\end{aligned}}$ ${\ displaystyle {\ begin {aligned} y_ {n-1} '(x) & = - P_ {1} (x) y_ {n-1} (x) - \ dots -P_ {n-1} (x ) y_ {1} (x) -P_ {n} (x) y_ {0} (x) \\ y_ {n-2} '(x) & = y_ {n-1} \\\ vdots \\ y_ {1} '(x) & = y_ {2} \\ y_ {0}' (x) & = y_ {1} \\\ end {aligned}}}$

with matrix $A(x)$ ${\ displaystyle A (x)}$ of the following kind

$A(x)=\left({\begin{matrix}0&1&0&\dots &0\\0&0&1&\dots &0\\0&0&\ddots &\ddots &0\\0&0&\dots &0&1\\-P_{n}(x)&-P_{n-1}(x)&\dots &-P_{2}(x)&-P_{1}(x)\\\end{matrix}}\right)$ ${\ displaystyle A (x) = \ left ({\ begin {matrix} 0 & 1 & 0 & \ dots & 0 \\ 0 & 0 & 1 & \ dots & 0 \\ 0 & 0 & \ ddots & \ ddots & 0 \\ 0 & 0 & \ dots & 0 & 1 \\ - P_ {n} ( x) & - P_ {n-1} (x) & \ dots & -P_ {2} (x) & - P_ {1} (x) \\\ end {matrix}} \ right)}$

Wronskians of the original equation and the system coincide, and the trace of the matrix $A(x)$ ${\ displaystyle A (x)}$ is equal to $-P_{1}(x)$ ${\ displaystyle -P_ {1} (x)}$ . Substituting into the formula for the system we obtain

$W(x)=W(x_{0})e^{-\int _{x_{0}}^{x}P_{1}(\zeta )d\zeta }$ ${\ displaystyle W (x) = W (x_ {0}) e ^ {- \ int _ {x_ {0}} ^ {x} P_ {1} (\ zeta) d \ zeta}}$

Application of the Liouville-Ostrogradsky formula

Let the solution be known $y_{1}(x)$ ${\ displaystyle y_ {1} (x)}$ second-order linear ordinary differential equation, i.e. $n=2$ ${\ displaystyle n = 2}$ . Using the Liouville-Ostrogradsky formula, it is possible to find a solution that is linearly independent of it $y_{2}(x)$ ${\ displaystyle y_ {2} (x)}$ the same system.

We will write the Wronskian:

$Ce^{-\int P_{1}(x)dx}=y_{1}y_{2}'-y_{1}'y_{2}=W.$ ${\ displaystyle Ce ^ {- \ int P_ {1} (x) dx} = y_ {1} y_ {2} '- y_ {1}' y_ {2} = W.}$

${\frac {W}{y_{1}^{2}}}={\frac {y_{1}y_{2}'-y_{1}'y_{2}}{y_{1}^{2}}}=\left({\frac {y_{2}}{y_{1}}}\right)',$ ${\ displaystyle {\ frac {W} {y_ {1} ^ {2}}} = {\ frac {y_ {1} y_ {2} '- y_ {1}' y_ {2}} {y_ {1} ^ {2}}} = \ left ({\ frac {y_ {2}} {y_ {1}}} \ right) ',}$ so

${\frac {y_{2}}{y_{1}}}=\int {\frac {W}{y_{1}^{2}}}dx+B$ ${\ displaystyle {\ frac {y_ {2}} {y_ {1}}} = \ int {\ frac {W} {y_ {1} ^ {2}}} dx + B}$ $\longrightarrow y_{2}=y_{1}\left(\int {\frac {W}{y_{1}^{2}}}dx+B\right)=y_{1}\int {\frac {Ce^{-\int P_{1}(x)dx}}{y_{1}^{2}}}dx+By_{1}$ ${\ displaystyle \ longrightarrow y_ {2} = y_ {1} \ left (\ int {\ frac {W} {y_ {1} ^ {2}}} dx + B \ right) = y_ {1} \ int { \ frac {Ce ^ {- \ int P_ {1} (x) dx}} {y_ {1} ^ {2}}} dx + By_ {1}}$

Since for linear independence $y_{1}(x)$ ${\ displaystyle y_ {1} (x)}$ and $y_{2}(x)$ ${\ displaystyle y_ {2} (x)}$ enough $W\neq 0$ ${\ displaystyle W \ neq 0}$ by accepting $C=1,\,B=0$ ${\ displaystyle C = 1, \, B = 0}$ we get $y_{2}=y_{1}\int {\frac {e^{-\int P_{1}(x)dx}}{y_{1}^{2}}}dx.$ ${\ displaystyle y_ {2} = y_ {1} \ int {\ frac {e ^ {- \ int P_ {1} (x) dx}} {y_ {1} ^ {2}}} dx.}$

Example

Let in the equation $y''-\mathop {\rm {tg}} x\,y'+2y=0$ ${\ displaystyle y '' - \ mathop {\ rm {tg}} x \, y '+ 2y = 0}$ private solution is known $y_{1}=\sin x$ ${\ displaystyle y_ {1} = \ sin x}$ . Using the Liouville-Ostrogradsky formula, we get

$y_{2}=\sin x\int {\frac {dx}{\sin ^{2}xe^{-\int \tan xdx}}}=\sin x\,\ln \left|\tan x+{\frac {1}{\cos x}}\right|\,-1.$ ${\ displaystyle y_ {2} = \ sin x \ int {\ frac {dx} {\ sin ^ {2} xe ^ {- \ int \ tan xdx}}} = \ sin x \, \ ln \ left | \ tan x + {\ frac {1} {\ cos x}} \ right | \, - 1.}$

Then the general solution of the homogeneous equation $y=C_{1}\left(\sin x\,\ln \left|\tan x+{\frac {1}{\cos x}}\right|\,-1\right)+C_{2}\sin x$ ${\ displaystyle y = C_ {1} \ left (\ sin x \, \ ln \ left | \ tan x + {\ frac {1} {\ cos x}} \ right | \, - 1 \ right) + C_ { 2} \ sin x}$

References used

Agafonov S.A., German A.D., Muratova T.V. Differential equations. Textbook for high schools - M. Publishing House of MSTU. Bauman, 1999 .-- 336 p. (Mathematics Series at Technical University; Issue VIII), Chapter 5 paragraph 2.
Romanko V.K. Course of differential equations and calculus of variations. - 2nd ed. - M.: Laboratory of Basic Knowledge, 2001 .-- 344 p.