Differential Equations

Higher Order Equations

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Equations Solvable in Quadratures

  • It is said that the differential equation is solved in quadratures if its general solution is expressed through one or more integrals.

    Next, we consider three types of higher-order equations that are integrated in quadratures.

    Case \(1.\) Equation of Type \(F\left( {x,{y^{\left( n \right)}}} \right) = 0\)

    Assume first that this equation can be transformed into an explicit form for the derivative \({{y^{\left( n \right)}}},\) i.e. expressed as

    \[{y^{\left( n \right)}} = f\left( x \right).\]

    We integrate this equation \(n\) times consecutively in the range from \({x_0}\) to \(x.\) As a result, we obtain the following expressions for the derivatives and the function \(y\left( x \right):\)

    \[{{y^{\left( {n – 1} \right)}}\left( x \right) }={ \int\limits_{{x_0}}^x {f\left( x \right)dx} + {C_1},}\]

    \[{{y^{\left( {n – 2} \right)}}\left( x \right) }={ \int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {f\left( x \right)dx} }+{ {C_1}\left( {x – {x_0}} \right) }+{ {C_2},}\]

    \[\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\]

    \[ {y\left( x \right) }={ \underbrace {\int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {dx} \cdots \int\limits_{{x_0}}^x {f\left( x \right)dx} }_{n\;\text{times}} } + {{C_1}\frac{{{{\left( {x – {x_0}} \right)}^{n – 1}}}}{{\left( {n – 1} \right)!}} + \cdots } + {{C_{n – 1}}\left( {x – {x_0}} \right) }+{ {C_n}.} \]

    The last formula is the general solution of differential equation in quadratures.

    At \(x = {x_0},\) we obtain the particular solution satisfying the initial conditions

    \[
    {y\left( {x = {x_0}} \right) = {C_n},\;\;}\kern-0.3pt
    {y’\left( {x = {x_0}} \right) = {C_{n – 1}},\;\; \ldots,\;\;}\kern-0.3pt
    {{y^{\left( {n – 2} \right)}}\left( {x = {x_0}} \right) = {C_2},\;\;}\kern-0.3pt
    {{y^{\left( {n – 1} \right)}}\left( {x = {x_0}} \right) = {C_1},}
    \]

    where \({C_1},{C_2}, \ldots ,{C_n}\) is a given set of numbers.

    The iterated integral in the expression for \(y\left( x \right)\) can be converted to a single integral. Indeed, in the case \(n = 2,\) we consider the integral

    \[y\left( x \right) = \int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {f\left( \tau \right)d\tau } ,\]

    where \(\tau\) denotes a variable of integration in the inner integral.

    This iterated integral is defined in the triangular region \(D\left( {x,\tau } \right)\) shown in Figure \(1.\)

    The iterated integral is defined in the triangular region D
    Figure 1.

    We can change the order of integration in this integral, using Dirichlet’s formula:

    \[
    {y\left( x \right) }={ \int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {f\left( \tau \right)d\tau } }
    = {\int\limits_{{x_0}}^x {f\left( \tau \right)d\tau } \int\limits_\tau ^x {dx} .}
    \]

    As a result, the double integral reduces to a single integral:

    \[
    {y\left( x \right) }={ \int\limits_{{x_0}}^x {f\left( \tau \right)\left( {x – \tau } \right)d\tau } }
    = {\int\limits_{{x_0}}^x {\frac{{\left( {x – \tau } \right)}}{{1!}}f\left( \tau \right)d\tau } .}
    \]

    Similarly, we can simplify the the triple iterated integral in the case \(n = 3:\)

    \[ {y\left( x \right) }={ \int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {f\left( x \right)dx} } = {\int\limits_{{x_0}}^x {dx} \int\limits_{{x_0}}^x {\frac{{\left( {x – \tau } \right)}}{{1!}}f\left( \tau \right)d\tau } } = {\int\limits_{{x_0}}^x {f\left( \tau \right)d\tau } \int\limits_\tau ^x {\frac{{\left( {x – \tau } \right)}}{{1!}}d\tau } } = {\int\limits_{{x_0}}^x {\left[ {\left. {\left( {\frac{{{{\left( {x – \tau } \right)}^2}}}{{2!}}} \right)} \right|_{x = \tau }^{x = x}} \right]f\left( \tau \right)d\tau } } = {\int\limits_{{x_0}}^x {\frac{{{{\left( {x – \tau } \right)}^2}}}{{2!}}f\left( \tau \right)d\tau } .} \]

    For the iterated integral of arbitrary multiplicity \(n,\) the following expression is valid:

    \[{y\left( x \right) }={ \int\limits_{{x_0}}^x {\frac{{{{\left( {x – \tau } \right)}^{n – 1}}}}{{\left( {n – 1} \right)!}}f\left( \tau \right)d\tau } ,}\]

    which is called the Cauchy formula for iterated integrals.

    The resulting expression is a particular solution of the differential equation \({y^{\left( n \right)}} = f\left( x \right)\) with zero initial conditions:

    \[
    {y\left( {x = {x_0}} \right) = 0,\;\;}\kern-0.3pt
    {y’\left( {x = {x_0}} \right) = 0,\;\; \ldots ,\;\;}\kern-0.3pt
    {{y^{\left( {n – 1} \right)}}\left( {x = {x_0}} \right) = 0.}
    \]

    Accordingly, the general solution of the original equation is described by

    \[ {y\left( x \right) }={ \int\limits_{{x_0}}^x {\frac{{{{\left( {x – \tau } \right)}^{n – 1}}}}{{\left( {n – 1} \right)!}}f\left( \tau \right)d\tau } } + {{C_1}\frac{{{{\left( {x – {x_0}} \right)}^{n – 1}}}}{{\left( {n – 1} \right)!}} + \cdots } + {{C_{n – 1}}\left( {x – {x_0}} \right) }+{ {C_n}.} \]

    Note that the Cauchy formula relates the function \(y\left( x \right)\) and its \(n\)th order derivative \({y^{\left( n \right)}} = f\left( x \right).\) If we assume that \(n\) can be a real number, then we arrive at the concept of fractional order derivative.

    Instead of the factorial \(\left( {n – 1} \right)!\) in the Cauchy’s formula we can write the so-called gamma function \(\Gamma \left( z \right),\) which is continuous and expressed through the improper integral in the form

    \[\Gamma \left( z \right) = \int\limits_0^\infty {{e^{ – t}}{t^{z – 1}}dt} .\]

    A schematic view of the gamma function \(\Gamma \left( z \right)\) for real values of \(z\) is shown in Figure \(2.\)

    A schematic view of the gamma function
    Figure 2.

    For natural values of \(n,\) the following equality holds:

    \[\Gamma \left( n \right) = \left( {n – 1} \right)!\]

    Then the Cauchy formula is represented as follows:

    \[{y\left( {x,z} \right) }={ \int\limits_{{x_0}}^x {\frac{{{{\left( {x – \tau } \right)}^{z – 1}}}}{{\Gamma \left( z \right)}}f\left( \tau \right)d\tau },}\]

    where \(z\) is a real number.

    This formula can be considered as the definition of fractional derivative of order \(z,\) if the original function \(y\left( x \right)\) is known, or as the definition of the integral or fractional order \(z,\) if the corresponding derivative is given.

    We have considered the solution of the explicit differential equation \({y^{\left( n \right)}} = f\left( x \right)\) in quadratures. The implicit equation \(F\left( {x,{y^{\left( n \right)}}} \right) = 0\) can be also integrated if it can be solved with respect to \(x,\) or more generally, presented in parametric form:

    \[{x = \varphi \left( t \right),\;\;}\kern-0.3pt{{y^{\left( n \right)}} = \psi \left( t \right).}\]

    Then, given that

    \[{d{y^{\left( {n – 1} \right)}} = {y^{\left( n \right)}}dx }={ \psi \left( t \right)\varphi’\left( t \right)dt,}\]

    we have

    \[{{y^{\left( {n – 1} \right)}}\left( x \right) }={ \int {\psi \left( t \right)\varphi’\left( t \right)dt} + {C_1}.}\]

    Similarly, we find the other derivatives and the function \(y\left( x \right).\) As a result, we obtain the general solution of the equation in parametric form:

    \[{x = \varphi \left( t \right),\;\;}\kern-0.3pt{y = \Phi \left( {t,{C_1},{C_2}, \ldots ,{C_n}} \right).}\]

    Case \(2.\) Equation of Type \(F\left( {{y^{\left( {n – 1} \right)}},{y^{\left( n \right)}}} \right) = 0\)

    Consider first the case when such an equation can be solved for \({{y^{\left( n \right)}}}:\)

    \[{y^{\left( n \right)}} = f\left( {{y^{\left( {n – 1} \right)}}} \right).\]

    We solve it as follows. We introduce the new variable \(z = {{y^{\left( {n – 1} \right)}}}.\) Then the equation can be written as

    \[z’ = f\left( z \right).\]

    Separating the variables, we find its general solution:

    \[{\int {\frac{{dz}}{{f\left( z \right)}}} = x + {C_1},\;\;} \Rightarrow {z = \varphi \left( {x,{C_1}} \right).}\]

    Returning to the variable \(y,\) we obtain the differential equation of the \(\left( {n – 1} \right)\)th order:

    \[{y^{\left( {n – 1} \right)}} = \varphi \left( {x,{C_1}} \right).\]

    which is solved by the method set out in paragraph \(1\) above.

    The general implicit equation \(F\left( {{y^{\left( {n – 1} \right)}},{y^{\left( n \right)}}} \right) = 0\) can be integrated if it is represented in parametric form as

    \[{{y^{\left( n \right)}} = \varphi \left( t \right),\;\;}\kern-0.3pt{{y^{\left( {n – 1} \right)}} = \psi \left( t \right).}\]

    As \(d{y^{\left( {n – 1} \right)}} = {y^{\left( n \right)}}dx,\) we obtain the following expression for \(x\left( t \right):\)

    \[
    {{dx = \frac{{d{y^{\left( {n – 1} \right)}}}}{{{y^{\left( n \right)}}}} }={ \frac{{\psi’\left( t \right)dt}}{{\varphi \left( t \right)}},\;\;}}\Rightarrow
    {x = \int {\frac{{\psi’\left( t \right)dt}}{{\varphi \left( t \right)}}} + {C_1}.}
    \]

    The expression for \(y\left( t \right)\) is found by successive integration:

    \[
    {{d{y^{\left( {n – 2} \right)}} = {y^{\left( {n – 1} \right)}}dx }={ \frac{{\psi \left( t \right)\psi’\left( t \right)dt}}{{\varphi \left( t \right)}},\;\;}}\Rightarrow
    {{{y^{\left( {n – 2} \right)}} }={ \int {\frac{{\psi \left( t \right)\psi’\left( t \right)dt}}{{\varphi \left( t \right)}}} }+{ {C_2},}}
    \]

    \[\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\]

    \[{dy = y’dx,\;\; }\Rightarrow {y = \int {y’dx} + {C_n}.}\]

    As a result, we obtain the general solution in parametric form.

    Case \(3.\) Equation of Type \(F\left( {{y^{\left( {n – 2} \right)}},{y^{\left( n \right)}}} \right) = 0\)

    Suppose that the equation is solved for \({{y^{\left( n \right)}}}:\)

    \[{y^{\left( n \right)}} = f\left( {{y^{\left( {n – 2} \right)}}} \right).\]

    Introducing the new variable \({{y^{\left( {n – 2} \right)}}} = z,\) we write it as

    \[z^{\prime\prime} = f\left( z \right).\]

    Multiplying both sides by \(z’\) (under the assumption that the equation has no solution \(z’ = 0\)), we obtain:

    \[
    {z’z^{\prime\prime} = f\left( z \right)z’,\;\;}\Rightarrow
    {d\left[ {{{\left( {z’} \right)}^2}} \right] = 2f\left( z \right)dz,\;\;}\Rightarrow
    {{\left( {z’} \right)^2} = 2\int {f\left( z \right)dz} + {C_1},\;\;}\Rightarrow
    {{z’ }={ \sqrt {2\int {f\left( z \right)dz} + {C_1}} ,\;\;}}\Rightarrow
    {{y^{\left( {n – 1} \right)}} }={ \sqrt {2\int {f\left( {{y^{\left( {n – 2} \right)}}} \right)d{y^{\left( {n – 2} \right)}}} + {C_1}} .}
    \]

    It is evident that we have an equation of the form \({y^{\left( {n – 1} \right)}} = f\left( {{y^{\left( {n – 2} \right)}}} \right),\) which was considered in paragraph \(2\) and which can be solved in quadratures.

    If the equation \(z^{\prime\prime} = f\left( z \right)\) has a solution \(z’ = 0,\) then the general solution is given by

    \[
    {{y^{\left( {n – 1} \right)}} = 0,\;\;}\Rightarrow
    {y = {C_1}{x^{n – 2}} }+{ {C_2}{x^{n – 3}} + \cdots }+{ {C_{n – 1}}.}
    \]

    In the case where the differential equation \(F\left( {{y^{\left( {n – 2} \right)}},{y^{\left( n \right)}}} \right) = 0\) admits a parametric representation

    \[{{y^{\left( n \right)}} = \varphi \left( t \right),\;\;}\kern0pt{{y^{\left( {n – 2} \right)}} = \psi \left( t \right),}\]

    its solution is constructed as follows. It follows from the relationships

    \[{d{y^{\left( {n – 1} \right)}} = {y^{\left( n \right)}}dx,\;\;}\kern0pt{d{y^{\left( {n – 2} \right)}} = {y^{\left( {n – 1} \right)}}dx}\]

    that

    \[{{y^{\left( {n – 1} \right)}}d{y^{\left( {n – 1} \right)}} }={ {y^{\left( n \right)}}d{y^{\left( {n – 2} \right)}},}\]

    or in parametric form:

    \[{{y^{\left( {n – 1} \right)}}d{y^{\left( {n – 1} \right)}} }={ \varphi \left( t \right)\psi’\left( t \right)dt.}\]

    Integrating, we find:

    \[{{y^{\left( {n – 1} \right)}} }={ \sqrt {2\int {\varphi \left( t \right)\psi’\left( t \right)dt} + {C_1}} .}\]

    Now we know the parametric expression for the derivatives \({y^{\left( {n – 2} \right)}}\) and \({y^{\left( {n – 1} \right)}},\) that is the problem reduces to type \(2.\)


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the general solution of the differential equation \(y^{\prime\prime\prime} = {x^2} – 1.\)

    Example 2

    Find a particular solution of the equation \({y^{IV}} = \sin x + 1\) with the initial conditions \({x_0} = 0,\) \({y_0} = 1,\) \({y’_0} = {y^{\prime\prime}_0} = {y^{\prime\prime\prime}_0} = 0.\)

    Example 3

    Find the general solution of the equation \({\left( {y^{\prime\prime}} \right)^2} – {\left( {y^{\prime\prime}} \right)^3} = x.\)

    Example 4

    Find a particular solution of the equation \({y^{IV}} – y^{\prime\prime\prime} = 1\) with zero initial conditions: \({x_0} = 0,\) \({y_0} = {y’_0} = {y^{\prime\prime}_0} = {y^{\prime\prime\prime}_0} = 0.\)

    Example 5

    Find the general solution of the differential equation \(y^{\prime\prime\prime} = \sqrt {1 – {{\left( {y^{\prime\prime}} \right)}^2}} .\)

    Example 6

    Construct the general solution of the equation \(y^{\prime\prime\prime}y^{\prime\prime} = 1\) in quadratures.
    Page 1
    Concept
    Page 2
    Problems 1-6