Cases of Reduction of Order

Solved Problems

Solution.

This equation does not contain the function \(y\) and its first derivative \(y'.\) So we make the change of variable:

We obtain the first-order equation with separable variables:

Integrating, we find the solution:

Returning to the original variable \(y,\) we obtain another differential equation:

Integrating twice, we find the general solution of the original equation:

Solution.

We make the following change:

As a result, we obtain the second-order equation:

Since this equation does not contain the independent variable \(x,\) then we put \(p' = z\left( p \right).\) Hence,

Then the equation can be written as

1. One solution to this equation is given by
   \[z = 0,\;\; \Rightarrow p' = 0,\;\; \Rightarrow p = {C_1},\;\; \Rightarrow y^{\prime\prime} = {C_1},\;\; \Rightarrow y' = {C_1}x + {C_2},\;\; \Rightarrow y = {C_1}{x^2} + {C_2}x + {C_3}.\]
   It is evident that this solution describes a set of parabolas with arbitrary coefficients \({C_1}, {C_2}, {C_3}.\)
2. Now we find the second solution of the differential equation.
   \[5z - 3pz' = 0,\;\; \Rightarrow 3pz' = 5z,\;\; \Rightarrow \frac{{dz}}{z} = \frac{5}{3}\frac{{dp}}{p},\;\; \Rightarrow \int {\frac{{dz}}{z}} = \frac{5}{3}\int {\frac{{dp}}{p}} ,\;\; \Rightarrow \ln \left| z \right| = \frac{5}{3}\ln \left| p \right| + \ln {C_4},\;\; \Rightarrow z = {C_4}{p^{\frac{5}{3}}},\;\; \Rightarrow p' = {C_4}{p^{\frac{5}{3}}}.\]
   The resulting first-order equation is easily integrated:
   \[\int {{p^{ - \frac{5}{3}}}dp} = {C_4}\int {dx} ,\;\; \Rightarrow - \frac{{3{p^{ - \frac{2}{3}}}}}{2} = {C_4}x + {C_5}.\]
   Renaming the constants \({C_4},\) \({C_5},\) the solution can be written as
   \[p = - {\left( {{C_4}x + {C_5}} \right)^{ - \frac{3}{2}}}.\]
   Thus, to determine the second solution, we have the following equation:
   \[y^{\prime\prime} = - {\left( {{C_4}x + {C_5}} \right)^{ - \frac{3}{2}}}.\]
   Integrating twice, we find:
   \[y' = 2{\left( {{C_4}x + {C_5}} \right)^{ - \frac{1}{2}}} + {C_6},\;\; \Rightarrow y = {\left( {{C_4}x + {C_5}} \right)^{\frac{1}{2}}} + {C_6}x + {C_7},\;\; \Rightarrow y = \sqrt {{C_4}x + {C_5}} + {C_6}x + {C_7}.\]

Thus, the general solution of the original equation has two families of functions:

where \({C_1},{C_2}, \ldots ,{C_7}\) are arbitrary numbers.

Solution.

Dividing the equation by \(y'y^{\prime\prime},\) we can express the left side of this equation as a total derivative:

Integrating, we obtain the second-order equation:

In the last equation we make the change:

Hence,

This equation is easily integrated:

We have redefined the constants \({C_1},{C_2}\) in the last expression to simplify the equation.

As a result, we obtain the following expression:

Integrating, we find the function \(y\left( x \right):\)

As the numbers \({C_1}\) and \({C_2}\) are arbitrary, we have omitted here the factor \(\frac{2}{3},\) which appears in the integration of the square root.

At the beginning of the solution, we have lost the solution \(y^{\prime\prime} = 0\) when we divided the equation by \(y^{\prime\prime}.\) It follows that

where \({C_4},{C_5}\) are arbitrary numbers.

So the final answer contains two branches of solutions:

Solution.

It is seen that this equation is homogeneous. Therefore, its order can be reduced by one, using the substitution

Find the derivatives:

After substitution we obtain the second order differential equation:

The order of the new equation can again be reduces by the substitution

Then

As a result, we get the first-order equation for the function \(p\left( z \right):\)

This equation has two solutions.

1. The first solution is determined by the equation
   \[p = 0,\;\; \Rightarrow z' = 0,\;\; \Rightarrow z = {C_1}.\]
   So we can find the function \(y\left( x \right):\)
   \[y\left( x \right) = {C_2}{e^{\int {zdx} }} = {C_2}{e^{\int {{C_1}dx} }} = {C_2}{e^{{C_1}x + {C_3}}} = {C_2}{e^{{C_1}x}}{e^{{C_3}}}.\]
   Here the constant \({C_2}\) can be renamed: \({C_2}{e^{{C_3}}} \to {C_2}.\) Then the first family of solutions has the form:
   \[{y_1}\left( x \right) = {C_2}{e^{{C_1}x}}.\]
2. The second solution is described by a simple equation with separable variables:
   \[p' + 2z = 0,\;\; \Rightarrow \frac{{dp}}{{dz}} = - 2z,\;\; \Rightarrow \int {dp} = - 2\int {zdz} ,\;\; \Rightarrow p = - {z^2} + {C_3} = {C_3} - {z^2}.\]
   Given that \(p = z',\) we find the function \(z\left( x \right):\)
   \[z' = {C_3} - {z^2},\;\; \Rightarrow \frac{{dz}}{{{C_3} - {z^2}}} = dx,\;\; \Rightarrow \int {\frac{{dz}}{{{C_3} - {z^2}}}} = \int {dx} ,\;\; \Rightarrow \frac{1}{{2\sqrt {{C_3}} }}\ln \left| {\frac{{\sqrt {{C_3}} + z}}{{\sqrt {{C_3}} - z}}} \right| = x + {C_4}.\]
   For simplicity, we again replace the constant \(\sqrt {{C_3}} \) to \({C_3}.\) Thus, we obtain:
   \[\frac{1}{{2{C_3}}}\ln \left| {\frac{{{C_3} + z}}{{{C_3} - z}}} \right| = x + {C_4}.\]
   From the last equation we find the explicit expression for \(z\left( x \right):\)
   \[\ln \left| {\frac{{{C_3} + z}}{{{C_3} - z}}} \right| = 2{C_3}x + {C_4},\;\; \Rightarrow \frac{{{C_3} + z}}{{{C_3} - z}} = {e^{2{C_3}x + {C_4}}},\;\; \Rightarrow {C_3} + z = \left( {{C_3} - z} \right){e^{2{C_3}x + {C_4}}},\;\; \Rightarrow {C_3} + z = {C_3}{e^{2{C_3}x + {C_4}}} - z{e^{2{C_3}x + {C_4}}},\;\; \Rightarrow z\left( {{e^{2{C_3}x + {C_4}}} + 1} \right) = {C_3}\left( {{e^{2{C_3}x + {C_4}}} - 1} \right),\;\; \Rightarrow z = {C_3}\frac{{{e^{2{C_3}x + {C_4}}} - 1}}{{{e^{2{C_3}x + {C_4}}} + 1}}.\]
   Now we determine the function \(y\left( x \right)\) by the formula
   \[y\left( x \right) = {C_5}{e^{\int {zdx} }}.\]
   First, we calculate the integral \({\int {zdx} }:\)
   \[I = \int {zdx} = {C_3}\int {\frac{{{e^{2{C_3}x + {C_4}}} - 1}}{{{e^{2{C_3}x + {C_4}}} + 1}}dx} .\]
   We introduce the new variable \(t = {e^{2{C_3}x + {C_4}}}.\) Then
   \[dt = 2{C_3}{e^{2{C_3}x + {C_4}}}dx = 2{C_3}tdx,\;\; \Rightarrow dx = \frac{1}{{2{C_3}}}\frac{{dt}}{t}.\]
   Consequently, the integral is equal to
   \[I = {C_3}\int {\frac{{t - 1}}{{t + 1}}\frac{1}{{2{C_3}}}\frac{{dt}}{t}} = \frac{1}{2}\int {\frac{{t - 1}}{{t\left( {t + 1} \right)}}dt} .\]
   We represent the integrand as a sum of two fractions:
   \[\frac{{t - 1}}{{t\left( {t + 1} \right)}} \equiv \frac{A}{t} + \frac{B}{{t + 1}},\;\; \Rightarrow \frac{{t - 1}}{{t\left( {t + 1} \right)}} \equiv \frac{{A\left( {t + 1} \right) + Bt}}{{t\left( {t + 1} \right)}},\;\; \Rightarrow t - 1 \equiv \left( {A + B} \right)t + A,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {A + B = 1}\\ {A = - 1} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {A = - 1}\\ {B = 2} \end{array}} \right..\]
   Thus, we obtain:
   \[\frac{{t - 1}}{{t\left( {t + 1} \right)}} = \frac{2}{{t + 1}} - \frac{1}{t}.\]
   Calculate the integral:
   \[I = \frac{1}{2}\int {\left( {\frac{2}{{t - 1}} - \frac{1}{t}} \right)dt} = \frac{1}{2} \cdot 2\ln \left| {t + 1} \right| - \frac{1}{2}\ln \left| t \right| = \ln \left| {t + 1} \right| - \frac{1}{2}\ln \left| t \right| = \ln \frac{{\left| {t + 1} \right|}}{{\sqrt {\left| t \right|} }} = \ln \frac{{{e^{2{C_3}x + {C_4}}} + 1}}{{\sqrt {{e^{2{C_3}x + {C_4}}}} }} = \ln \frac{{{e^{2{C_3}x + {C_4}}} + 1}}{{{e^{{C_3}x + {\frac{{{C_4}}}{2}}}}}} = \ln \frac{{{e^{{C_4}}}{e^{2{C_3}x}} + 1}}{{{e^{\frac{{{C_4}}}{2}}}{e^{{C_3}x}}}} = \ln \left( {{e^{\frac{{{C_4}}}{2}}}{e^{{C_3}x}} + \frac{1}{{{e^{\frac{{{C_4}}}{2}}}{e^{{C_3}x}}}}} \right) = \ln \left( {{C_4}{e^{{C_3}x}} + \frac{1}{{{C_4}{e^{{C_3}x}}}}} \right),\]
   where we denoted: \({e^{\frac{{{C_4}}}{2}}} \to {C_4}.\) Then we obtain the following expression for the function \(y\left( x \right):\)
   \[y\left( x \right) = {C_5}{e^{\int {zdx} }} = {C_5}{e^{\ln \left( {{C_4}{e^{{C_3}x}} + \frac{1}{{{C_4}{e^{{C_3}x}}}}} \right)}} = {C_4}{e^{{C_3}x}} + {C_5}{e^{ - {C_3}x}}.\]
   In the last expression we have again redefined the arbitrary constants \({C_4},{C_5}.\) Thus, the second solution of the original equation can be written as
   \[{y_2}\left( x \right) = {C_4}{e^{{C_3}x}} + {C_5}{e^{ - {C_3}x}}.\]

So, the general solution of the differential equation has two branches:

It is seen that the second solution includes the first one. Therefore, the final answer is given by

where \({C_3},\) \({C_4},\) \({C_5}\) are arbitrary numbers.