Differential Equations

Higher Order Equations

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Equations Solvable in Quadratures

Solved Problems

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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

\[y\left( x \right) = \int\limits_{{x_0}}^x {\frac{{{{\left( {x - \tau } \right)}^2}}}{{2!}}\left( {{\tau ^2} - 1} \right)d\tau } + {C_1}\frac{{{{\left( {x - {x_0}} \right)}^2}}}{{2!}} + {C_2}\left( {x - {x_0}} \right) + {C_3}.\]

Calculate the integral included in this formula:

\[I = \frac{1}{2}\int\limits_{{x_0}}^x {{{\left( {x - \tau } \right)}^2}\left( {{\tau ^2} - 1} \right)d\tau } \frac{1}{2}\int\limits_{{x_0}}^x {\left( {{x^2} - 2x\tau + {\tau ^2}} \right) \left( {{\tau ^2} - 1} \right)d\tau } = \frac{1}{2}\left( {\frac{{{x^5}}}{3} - \frac{{{x^5}}}{2} + \frac{{{x^5}}}{5} - \cancel{x^3} + \cancel{x^3} - \frac{{{x^3}}}{3} + \alpha {x^2} + \beta x + \gamma } \right),\]

where \(\alpha,\) \(\beta,\) \(\gamma\) denote the coefficients depending on \({x_0}.\)

Then the general solution of the equation is represented as

\[y\left( x \right) = \frac{1}{2}\left[ {\left( {\frac{1}{3} - \frac{1}{2} + \frac{1}{5}} \right){x^5} - \frac{{{x^3}}}{3} + \alpha {x^2} + \beta x + \gamma } \right] + {C_1}\frac{{{{\left( {x - {x_0}} \right)}^2}}}{2} + {C_2}\left( {x - {x_0}} \right) + {C_3}.\]

Given that \({C_1}\) and \({x_0}\) are arbitrary numbers, the general solution \(y\left( x \right)\) can be rewritten as

\[y\left( x \right) = \frac{{{x^5}}}{{60}} - \frac{{{x^3}}}{6} + {C_1}{x^2} + {C_2}x + {C_3}.\]

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:

\[y^{\prime\prime\prime} = - \cos x + x + {C_1},\]
\[y^{\prime\prime} = - \sin x + \frac{{{x^2}}}{2} + {C_1}x + {C_2},\]
\[y' = \cos x + \frac{{{x^3}}}{6} + \frac{{{C_1}{x^2}}}{2} + {C_2}x + {C_3},\]
\[y = \sin x + \frac{{{x^4}}}{{24}} + \frac{{{C_1}{x^3}}}{6} + \frac{{{C_2}{x^2}}}{2} + {C_3}x + {C_4}.\]

Substituting the initial values, we determine the coefficients \({C_1} - {C_4}\) from the system of equations:

\[\left\{ \begin{array}{l} 0 = - 1 + {C_1}\\ 0 = {C_2}\\ 0 = 1 + {C_3}\\ 1 = {C_4} \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {C_1} = 0\\ {C_2} = 0\\ {C_3} = - 1\\ {C_4} = 1 \end{array} \right..\]

Hence, the particular solution satisfying the initial conditions has the form:

\[y\left( x \right) = \sin x + \frac{{{x^4}}}{{24}} + \frac{{{x^3}}}{6} - x + 1.\]

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

\[x = {t^2} - {t^3}.\]

Given that \(d\left( {y'} \right) = y^{\prime\prime}dx,\) we find the derivative \(y'\) expressed in terms of the parameter \(t:\)

\[d\left( {y'} \right) = y^{\prime\prime}dx = t\left( {2t - 3{t^2}} \right)dt = \left( {2{t^2} - 3{t^3}} \right)dt,\;\; \Rightarrow y' = \int {\left( {2{t^2} - 3{t^3}} \right)dt} = \frac{{2{t^3}}}{3} - \frac{{3{t^4}}}{4} + {C_1}.\]

Similarly, we perform one more integration:

\[dy = y'dx = \left( {\frac{{2{t^3}}}{3} - \frac{{3{t^4}}}{4} + {C_1}} \right) \left( {2t - 3{t^2}} \right)dt = \left( {\frac{{4{t^3}}}{3} - \frac{{3{t^4}}}{2} + 2{C_1}t - {t^5} + \frac{{9{t^6}}}{4} - 3{C_1}{t^2}} \right)dt,\]
\[\Rightarrow y = \int {\left( {\frac{{4{t^3}}}{3} - \frac{{3{t^4}}}{2} + 2{C_1}t - {t^5} + \frac{{9{t^6}}}{4} - 3{C_1}{t^2}} \right)dt} = \frac{{{t^4}}}{3} - \frac{{3{t^5}}}{{10}} + {C_1}{t^2} - \frac{{{t^6}}}{6} + \frac{{9{t^7}}}{{28}} - {C_1}{t^3} + {C_2} = \frac{{9{t^7}}}{{28}} - \frac{{{t^6}}}{6} - \frac{{3{t^5}}}{{10}} + \frac{{{t^4}}}{3} - {C_1}{t^3} + {C_1}{t^2} + {C_2}.\]

Thus, the general solution is represented in parametric form as

\[\left\{ \begin{array}{l} {x = {t^2} - {t^3}}\\ {y = \frac{{9{t^7}}}{{28}} - \frac{{{t^6}}}{6} - \frac{{3{t^5}}}{{10}} + \frac{{{t^4}}}{3} - {C_1}{t^3} + {C_1}{t^2} + {C_2}} \end{array} \right..\]

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:

\[z' - z = 1.\]

Its general solution is given by the function

\[z = {C_1}{e^x} - 1.\]

After substitution we obtain the second order differential equation:

\[y^{\prime\prime\prime} = {C_1}{e^x} - 1,\]

that is the equation is converted to type \(1.\) It can be solved by successive integration:

\[y^{\prime\prime} = {C_1}{e^x} - x + {C_2},\]
\[y' = {C_1}{e^x} - \frac{{{x^2}}}{2} + {C_2}x + {C_3},\]
\[y = {C_1}{e^x} - \frac{{{x^3}}}{6} + \frac{{{C_2}{x^2}}}{2} + {C_3}x + {C_4}.\]

The coefficients \({C_i}\) are determined from the initial conditions:

\[\left\{ \begin{array}{l} 0 = {C_1} - 1\\ 0 = {C_1} + {C_2}\\ 0 = {C_1} + {C_3}\\ 0 = {C_1} + {C_4} \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {C_1} = 1\\ {C_2} = - 1\\ {C_3} = - 1\\ {C_4} = - 1 \end{array} \right..\]

Thus, a particular solution for the given initial conditions is expressed by the formula

\[y\left( x \right) = {e^x} - \frac{{{x^3}}}{6} - \frac{{{x^2}}}{2} - x - 1.\]

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:

\[z' = \sqrt {1 - {z^2}} .\]

We assume that the function \(z\) is defined on the interval \(\left[ { - 1,1} \right].\) Integrating, we find:

\[\frac{{dz}}{{dx}} = \sqrt {1 - {z^2}} ,\;\; \frac{{dz}}{{\sqrt {1 - {z^2}} }} = dx,\;\; \int {\frac{{dz}}{{\sqrt {1 - {z^2}} }}} = \int {dx} ,\;\; \arcsin z = x + {C_1},\;\; z = \sin \left( {x + {C_1}} \right).\]

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 \(z:\)

\[y^{\prime\prime} = \sin \left( {x + {C_1}} \right),\]
\[y' = - \cos \left( {x + {C_1}} \right) + {C_2},\]
\[y = - \sin \left( {x + {C_1}} \right) + {C_2}x + {C_3},\]

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

\[z^{\prime\prime}z = 1,\;\; \Rightarrow z^{\prime\prime} = \frac{1}{z}.\]

Obviously, \(z = y' \ne 0\) and \(z' = y^{\prime\prime} \ne 0.\) We multiply both sides of this equation by \(2z'\) and integrate once:

\[2z'z^{\prime\prime} = \frac{{2z'}}{z},\;\; \Rightarrow d{\left( {z'} \right)^2} = \frac{{2dz}}{z},\;\; \Rightarrow {\left( {z'} \right)^2} = 2\ln \left| z \right| + \ln {C_1} = \ln \left( {{C_1}{z^2}} \right).\]

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:

\[{t^2} = \ln \left( {{C_1}{z^2}} \right),\;\; \Rightarrow {C_1}{z^2} = {e^{{t^2}}},\;\; \Rightarrow z = \pm \frac{1}{{{C_1}}}{e^{{t^2}}}.\]

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:

\[t = z' = \frac{{dz}}{{dx}},\;\; \Rightarrow dx = \frac{{dz}}{t} = \frac{{\varphi'\left( {t,{C_1}} \right)dt}}{t},\;\; \Rightarrow x = \int {\frac{{\varphi'\left( {t,{C_1}} \right)dt}}{t}} + {C_2}.\]

It remains to obtain a parametric expression for the function \(y.\) As

\[z = \frac{{dy}}{{dx}},\;\; \Rightarrow dy = zdx = \frac{{\varphi \left( {t,{C_1}} \right)\varphi'\left( {t,{C_1}} \right)dt}}{t},\]

then after integration we have:

\[y = \int {\frac{{\varphi \left( {t,{C_1}} \right)\varphi'\left( {t,{C_1}} \right)}}{t}dt} + {C_3}.\]

So, the general solution of the original equation in parametric form is expressed in terms of quadratures as follows:

\[x = \int {\frac{{\varphi'\left( {t,{C_1}} \right)}}{t}dt} + {C_2},\;\;\; y = \int {\frac{{\varphi \left( {t,{C_1}} \right)\varphi'\left( {t,{C_1}} \right)}}{t}dt} + {C_3}.\]

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.

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