Equations Solvable in Quadratures
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.
Example 1.
Find the general solution of the differential equation \[y^{\prime\prime\prime} = {x^2} - 1.\]
Solution.
This equation is of the type \({y^{\left( n \right)}} = f\left( x \right)\) for \(n = 3.\) Its general solution can be written as
Calculate the integral included in this formula:
where \(\alpha,\) \(\beta,\) \(\gamma\) denote the coefficients depending on \({x_0}.\)
Then the general solution of the equation is represented as
Given that \({C_1}\) and \({x_0}\) are arbitrary numbers, the general solution \(y\left( x \right)\) can be rewritten as
Note:
The same answer can be obtained by successive integration of the given differential equation.
Example 2.
Find a particular solution of the equation \[{y^{IV}} = \sin x + 1\] with the initial conditions \({x_0} = 0,\) \({y_0} = 1,\) \({y'_0} = {y^{\prime\prime}_0} = {y^{\prime\prime\prime}_0} = 0.\)
Solution.
We first construct the general solution, successively integrating the given equation:
Substituting the initial values, we determine the coefficients \({C_1} - {C_4}\) from the system of equations:
Hence, the particular solution satisfying the initial conditions has the form:
Example 3.
Find the general solution of the equation \[{\left( {y^{\prime\prime}} \right)^2} - {\left( {y^{\prime\prime}} \right)^3} = x.\]
Solution.
This equation can be solved by the parametric method. We put \(y^{\prime\prime} = t.\) Then
Given that \(d\left( {y'} \right) = y^{\prime\prime}dx,\) we find the derivative \(y'\) expressed in terms of the parameter \(t:\)
Similarly, we perform one more integration:
Thus, the general solution is represented in parametric form as
where \({C_1},\) \({C_2}\) are arbitrary constants.
Example 4.
Find a particular solution of the equation \[{y^{IV}} - y^{\prime\prime\prime} = 1\] with zero initial conditions: \({x_0} = 0,\) \({y_0} = {y'_0} = {y^{\prime\prime}_0} = {y^{\prime\prime\prime}_0} = 0.\)
Solution.
This equation is of type \(2.\) We introduce an intermediate variable \(z = y^{\prime\prime\prime}.\) As a result, we obtain a linear first-order equation:
Its general solution is given by the function
After substitution we obtain the second order differential equation:
that is the equation is converted to type \(1.\) It can be solved by successive integration:
The coefficients \({C_i}\) are determined from the initial conditions:
Thus, a particular solution for the given initial conditions is expressed by the formula
Example 5.
Find the general solution of the differential equation \[y^{\prime\prime\prime} = \sqrt {1 - {{\left( {y^{\prime\prime}} \right)}^2}} .\]
Solution.
This equation is of type \(2.\) We introduce the new variable \(z = y^{\prime\prime}.\) This leads to the first-order equation:
We assume that the function \(z\) is defined on the interval \(\left[ { - 1,1} \right].\) Integrating, we find:
In fact, we have transformed the initial equation to an equation of type \(1.\) The general solution \(y\left( x \right)\) is most easily obtained by double integration of the expression for \(z:\)
where \({C_1},{C_2},{C_3}\) are arbitrary constants.
Example 6.
Construct the general solution of the equation \[y^{\prime\prime\prime}y^{\prime\prime} = 1\] in quadratures.
Solution.
This equation is of type \(3.\) We introduce the new function \(z = y',\) so that the equation is written as
Obviously, \(z = y' \ne 0\) and \(z' = y^{\prime\prime} \ne 0.\) We multiply both sides of this equation by \(2z'\) and integrate once:
It is seen that the equation is converted to type \(1.\) Further, it is convenient to solve it by the parametric method. Let \(z' = t,\) where the variable \(t\) is considered as a parameter. The function \(z\) is expressed in terms of \(t\) as follows:
Thus, the function \(z\) is represented as \(z = \varphi \left( {t,{C_1}} \right).\) The dependence of \(x\) on the parameter \(t\) is also described by a quadrature:
It remains to obtain a parametric expression for the function \(y.\) As
then after integration we have:
So, the general solution of the original equation in parametric form is expressed in terms of quadratures as follows:
where \(\varphi \left( {t,{C_1}} \right) = \pm \frac{1}{{{C_1}}} {e^{{t^2}}}\) and \({C_1},{C_2},{C_3}\) are constants of integration.