Differential Equations
Nth Order Equations
Equations Solvable in Quadratures
Theory
Problems 1-6
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
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( 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
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:
Figure 1.
{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:
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
Figure 2.
A schematic view of the gamma function \(\Gamma \left( z \right)\) for real values of \(z\) is shown in Figure \(2\) above. For natural values of \(n,\) the following equality holds:
Then the Cauchy formula is represented as follows:
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:
Then, given that
we have
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:
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 \({{y^{\left( n \right)}}}:\)
We solve it as follows. We introduce the new variable \(z = {{y^{\left( {n – 1} \right)}}}.\) Then the equation can be written as
Separating the variables, we find its general solution:
Returning to the variable \(y,\) we obtain the differential equation of the \(\left( {n – 1} \right)\)th order:
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
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},}}
\]
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 \({{y^{\left( n \right)}}}:\)
Introducing the new variable \({{y^{\left( {n – 2} \right)}}} = z,\) we write it as
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
its solution is constructed as follows. It follows from the relationships
that
or in parametric form:
Integrating, we find:
Now we know the parametric expression for the derivatives \({y^{\left( {n – 2} \right)}}\) and \({y^{\left( {n – 1} \right)}},\) i.e. the problem reduces to type \(2.\)
Solved Problems
Click on problem description to see 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.
Theory
Problems 1-6