# Differential Equations

## Higher Order Equations 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 $$1.$$

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 $$2.$$

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 or tap 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.
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Concept
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Problems 1-6