# Differential Equations

## 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

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

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.