Differential Equations

2nd Order Equations

Reduction of Order

Page 1
Problem 1
Page 2
Problems 2-7

A second order differential equation is written in general form as

\[F\left( {x,y,y’,y^{\prime\prime}} \right) = 0,\]

where \(F\) is a function of the given arguments.

If the differential equation can be resolved for the second derivative \(y^{\prime\prime},\) it can be represented in the following explicit form:

\[y^{\prime\prime} = f\left( {x,y,y’} \right).\]

In special cases the function \(f\) in the right side may contain only one or two variables. Such
incomplete equations include \(5\) different types:

\[
{y^{\prime\prime} = f\left( x \right),\;\;}\kern-0.3pt
{y^{\prime\prime} = f\left( y \right),\;\;}\kern-0.3pt
{y^{\prime\prime} = f\left( {y’} \right),\;\;}\kern-0.3pt
{y^{\prime\prime} = f\left( {x,y’} \right),\;\;}\kern-0.3pt
{y^{\prime\prime} = f\left( {y,y’} \right).}
\]

With the help of certain substitutions, these equations can be transformed into first order equations.

In the general case of a second order differential equation, its order can be reduced if this equation has a certain symmetry. Below we discuss two types of such equations (cases \(6\) and \(7\)):

  • The function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is a homogeneous function of the arguments \(y,y’,y^{\prime\prime};\)
  • The function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is an exact derivative of the first order function \(\Phi\left( {x,y,y’} \right).\)

Consider these cases of reduction of order in more detail.

Case 1. Equation of type \(y^{\prime\prime} = f\left( x \right)\)

For an equation of type \(y^{\prime\prime} = f\left( x \right),\) its order can be reduced by introducing a new function \(p\left( x \right)\) such that \(y’ = p\left( x \right).\) As a result, we obtain the first order differential equation

\[p’ = f\left( x \right).\]

Solving it, we find the function \(p\left( x \right).\) Then we solve the second equation

\[y’ = p\left( x \right)\]

and obtain the general solution of the original equation.

Case 2. Equation of type \(y^{\prime\prime} = f\left( y \right)\)

The right-hand side of the equation depends only on the variable \(y.\) We introduce a new function \(p\left( y \right),\) setting \(y’ = p\left( y \right).\) Then we can write:

\[{y^{\prime\prime} = \frac{d}{{dx}}\left( {y’} \right) }={ \frac{{dp}}{{dx}} }={ \frac{{dp}}{{dy}}\frac{{dy}}{{dx}} }={ \frac{{dp}}{{dy}}p,}\]

so the equation becomes:

\[\frac{{dp}}{{dy}}p = f\left( y \right).\]

Solving it, we find the function \(p\left( y \right).\) Then we find the solution of the equation \(y’ = p\left( y \right),\) that is, the function \(y\left( x \right).\)

Case 3. Equation of type \(y^{\prime\prime} = f\left( {y’} \right)\)

In this case, to reduce the order we introduce the function \(y’ = p\left( x \right)\) and obtain the equation

\[{y^{\prime\prime} = p’ }={ \frac{{dp}}{{dx}} }={ f\left( p \right),}\]

which is a first order equation with separable variables \(p\) and \(x.\) Integrating, we find the function \(p\left( x \right),\) and then the function \(y\left( x \right).\)

Case 4. Equation of type \(y^{\prime\prime} = f\left( {x,y’} \right)\)

Here we use the substitution \(y’ = p\left( x \right),\) where \( p\left( x \right)\) is a new unknown function. As a result, we obtain the first order equation:

\[p’ = \frac{{dp}}{{dx}} = f\left( {x,p} \right).\]

By integrating, we find the function \( p\left( x \right).\) Next, we solve one more equation of the \(1\)st order

\[y’ = p\left( x \right)\]

and find the general solution \( y\left( x \right).\)

Case 5. Equation of type \(y^{\prime\prime} = f\left( {y,y’} \right)\)

To solve this equation, we introduce a new function \( p\left( y \right),\) setting \( y’ = p\left( y \right),\) similar to case \(2.\) Differentiating this expression with respect to \(x\) leads to the equation

\[
{y^{\prime\prime} = \frac{{d\left( {y’} \right)}}{{dx}} = \frac{{dp}}{{dx}} }
= {\frac{{dp}}{{dy}}\frac{{dy}}{{dx}} }={ \frac{{dp}}{{dy}}p.}
\]

As a result, our original equation is written as an equation of the \(1\)st order

\[p\frac{{dp}}{{dy}} = f\left( {y,p} \right).\]

Solving it, we find the function \( p\left( y \right).\) Then we solve another first order equation

\[y’ = p\left( y \right)\]

and determine the general solution \( y\left( x \right).\)

The above \(5\) cases of reduction of order are not independent. Based on the structure of the equations, it is clear that case \(2\) follows from the case \(5\) and case \(3\) follows from the more general case \(4.\)

Case 6. Function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is homogeneous with respect to the arguments \(y, y’, y^{\prime\prime}\)

If the left side of the differential equation

\[F\left( {x,y,y’,y^{\prime\prime}} \right) = 0\]

satisfies the condition of homogeneity, i.e. the relationship

\[{F\left( {x,ky,ky’,ky^{\prime\prime}} \right) }={ {k^m}F\left( {x,y,y’,y^{\prime\prime}} \right)}\]

is valid for any \(k\), the order of the equation can be reduced by substitution

\[y = {e^{\int {zdx} }}.\]

After the function \( z\left( x \right)\) is found, the original function \( y\left( x \right)\) is determined by the integration formula

\[y\left( x \right) = {C_2}{e^{\int {zdx} }},\]

where \({C_2}\) is the constant of integration.

Case 7. Function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is an exact derivative

If one can find a function \(\Phi\left( {x,y,y’} \right),\) which does not contain the second derivative \(y^{\prime\prime}\) and satisfies the equation

\[{F\left( {x,y,y’,y^{\prime\prime}} \right) }={ \frac{d}{{dx}}\Phi \left( {x,y,y’} \right),}\]

then the solution of the original equation is given by the integral

\[\Phi \left( {x,y,y’} \right) = C.\]

Using this way the second order equation can be reduced to first order equation.

In some cases, the left part of the original equation can be transformed into an exact derivative, using an integrating factor.

Solved Problems

Click on problem description to see solution.

 Example 1

Solve the equation \(y^{\prime\prime} = \sin x + \cos x.\)

 Example 2

Solve the equation \(y^{\prime\prime} = {\large\frac{1}{{4\sqrt y }}\normalsize}.\)

 Example 3

Solve the equation \(y^{\prime\prime} = \sqrt {1 – {{\left( {y’} \right)}^2}} .\)

 Example 4

Solve the equation \(\sqrt x y^{\prime\prime} = {\left( {y’} \right)^2}.\)

 Example 5

Solve the equation \(y^{\prime\prime} = \left( {2y + 3} \right){\left( {y’} \right)^2}.\)

 Example 6

Solve the equation \(yy^{\prime\prime} = \) \({\left( {y’} \right)^2} – {\large\frac{{3{y^2}}}{{\sqrt x }}\normalsize}.\)

 Example 7

Solve the equation \(yy^{\prime\prime} + {\left( {y’} \right)^2} =\) \( 2x + 1.\)

Example 1.

Solve the equation \(y^{\prime\prime} = \sin x + \cos x.\)

Solution.

This example relates to the Case \(1.\) Consider the function \(y’ = p\left( x \right).\) Then \(y^{\prime\prime} = p’.\) Consequently,

\[p’ = \sin x + \cos x.\]

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

\[
{\frac{{dp}}{{dx}} = \sin x + \cos x,\;\;}\Rightarrow
{dp = \left( {\sin x + \cos x} \right)dx,\;\;}\Rightarrow
{{\int {dp} }={ \int {\left( {\sin x + \cos x} \right)dx} ,\;\;}}\Rightarrow
{{p = – \cos x }+{ \sin x + {C_1}.}}
\]

Given that \(y’ = p,\) we integrate one more equation of the \(1\)st order:

\[
{{y’ = – \cos x }+{ \sin x + {C_1},\;\;}}\Rightarrow
{{\int {dy} }={ \int {\left( { – \cos x + \sin x + {C_1}} \right)dx} ,\;\;}}\Rightarrow
{{y = – \sin x }-{ \cos x + {C_1}x + {C_2}.}}
\]

The latter formula gives the general solution of the original differential equation.

Page 1
Problem 1
Page 2
Problems 2-7