Equations Solvable in Quadratures

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It is said that the differential equation is solved in quadratures if its general solution is expressed in terms of 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

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

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 \(1.\)

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:

\[y\left( {x = {x_0}} \right) = 0,\;\; y'\left( {x = {x_0}} \right) = 0,\;\; \ldots ,\;\; {y^{\left( {n - 1} \right)}}\left( {x = {x_0}} \right) = 0.\]

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 \(2.\)

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:

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

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:

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

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

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

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

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

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

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

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 \(2.\)