# 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 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