Equations Solvable in Quadratures

Solved Problems

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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.

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:

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.

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

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.

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.