Reduction of Order
Solved Problems
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.
Example 2.
Solve the equation \[y^{\prime\prime} = \frac{1}{{4\sqrt y }}.\]
Solution.
This is an equation of type \(2,\) where the right-hand side depends only on the variable \(y.\) We introduce the parameter \(p = y'.\) Then the equation can be written as
\[y^{\prime\prime} = \frac{{dp}}{{dy}}p = \frac{1}{{4\sqrt y }}.\]
We obtain the equation of the \(1\)st order for the function \(p\left( y \right)\) with separable variables. Integrating gives:
\[\frac{{dp}}{{dy}}p = \frac{1}{{4\sqrt y }},\;\; \Rightarrow
2pdp = \frac{{dy}}{{2\sqrt y }},\;\; \Rightarrow
\int {2pdp} = \int {\frac{{dy}}{{2\sqrt y }}} ,\;\; \Rightarrow
{p^2} = \sqrt y + {C_1},\]
where \({C_1}\) is a constant of integration.
Taking the square root of both sides, we find the function \(p\left( y \right):\)
\[p = \pm \sqrt {\sqrt y + {C_1}} .\]
Now recall that \(y' = p\) and solve another equation of the \(1\)st order:
\[y' = \pm \sqrt {\sqrt y + {C_1}} ,\;\; \Rightarrow
\frac{{dy}}{{dx}} = \pm \sqrt {\sqrt y + {C_1}} .\]
Separate the variables and integrate:
\[\frac{{dy}}{{\sqrt {\sqrt y + {C_1}} }} = \pm dx,\;\; \Rightarrow
\int {\frac{{dy}}{{\sqrt {\sqrt y + {C_1}} }}} = \pm \int {dx} .\]
To calculate the integral on the left-hand side, make the replacement:
\[\sqrt y + {C_1} = z,\;\; \Rightarrow
dz = \frac{{dy}}{{2\sqrt y }},\;\; \Rightarrow
dy = 2\sqrt y dz = 2\left( {z - {C_1}} \right)dz.\]
Then the left-hand integral is equal to
\[\int {\frac{{dy}}{{\sqrt {\sqrt y + {C_1}} }}}
= \int {\frac{{2\left( {z - {C_1}} \right)dz}}{{\sqrt z }}}
= 2\int {\left( {\frac{z}{{\sqrt z }} - \frac{{{C_1}}}{{\sqrt z }}} \right)dz}
= 2\int {\left( {{z^{\frac{1}{2}}} - {C_1}{z^{ - \frac{1}{2}}}} \right)dz}
= 2\left( {\frac{{{z^{\frac{3}{2}}}}}{{\frac{3}{2}}} - {C_1}\frac{{{z^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right)
= \frac{4}{3}{z^{\frac{3}{2}}} - 4{C_1}{z^{\frac{1}{2}}}
= \frac{4}{3}\sqrt {{{\left( {\sqrt y + {C_1}} \right)}^3}} - 4{C_1}\sqrt {\sqrt y + {C_1}} .\]
As a result, we obtain the following algebraic equation:
\[\frac{4}{3}\sqrt {{{\left( {\sqrt y + {C_1}} \right)}^3}} - 4{C_1}\sqrt {\sqrt y + {C_1}} = {C_2} \pm x,\]
where \({C_1}, {C_2}\) are constants of integration.
The last expression is the general solution of the differential equation in implicit form.
Example 3.
Solve the equation \[y^{\prime\prime} = \sqrt {1 - {{\left( {y'} \right)}^2}}.\]
Solution.
This equation does not contain the function \(y\) and the independent variable \(x\) (Case \(3\)). Therefore, we set \(y' = p\left( x \right).\) Then this equation takes the form
\[y^{\prime\prime} = p' = \sqrt {1 - {p^2}} .\]
The resulting first-order equation for the function \(p\left( x \right)\) is a separable equation and can be easily integrated:
\[\frac{{dp}}{{dx}} = \sqrt {1 - {p^2}} ,\;\; \Rightarrow
\frac{{dp}}{{\sqrt {1 - {p^2}} }} = dx,\;\; \Rightarrow
\int {\frac{{dp}}{{\sqrt {1 - {p^2}} }}} = \int {dx} ,\;\; \Rightarrow
\arcsin p = x + {C_1},\;\; \Rightarrow
p = \sin \left( {x + {C_1}} \right).\]
Replacing \(p\) by \(y',\) we obtain
\[y' = \sin \left( {x + {C_1}} \right).\]
Integrating again, we find the general solution of the original differential equation:
\[\frac{{dy}}{{dx}} = \sin \left( {x + {C_1}} \right),\;\; \Rightarrow
dy = \sin \left( {x + {C_1}} \right)dx,\;\; \Rightarrow
\int {dy} = \int {\sin \left( {x + {C_1}} \right)dx} ,\;\; \Rightarrow
y = - \cos \left( {x + {C_1}} \right) + {C_2},\;\; \Rightarrow
y = {C_2} - \cos \left( {x + {C_1}} \right).\]
Example 4.
Solve the equation \[\sqrt x y^{\prime\prime} = \left( {y'} \right)^2.\]
Solution.
This equation does not explicitly include the variable \(y\), i.e. it corresponds to the type \(4\) in our classification. We introduce the new variable \(y' = p\left( x \right).\) The original equation is transformed into the first order equation:
\[\sqrt x p' = {p^2},\]
which is solved by separation of variables:
\[\sqrt x \frac{{dp}}{{dx}} = {p^2},\;\; \Rightarrow
\frac{{dp}}{{{p^2}}} = \frac{{dx}}{{\sqrt x }},\;\; \Rightarrow
\int {\frac{{dp}}{{{p^2}}}} = \int {\frac{{dx}}{{\sqrt x }}} ,\;\; \Rightarrow
- \frac{1}{p} = 2\sqrt x + {C_1},\;\; \Rightarrow
p = y' = \frac{{ - 1}}{{2\sqrt x + {C_1}}}.\]
Integrating the resulting equation once more yields the function \(y\left( x \right):\)
\[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{2\sqrt x + {C_1}}},\;\; \Rightarrow
dy = - \frac{{dx}}{{2\sqrt x + {C_1}}},\;\; \Rightarrow
y = - \int {\frac{{dx}}{{2\sqrt x + {C_1}}}} .\]
To compute the last integral we make the substitution: \(x = {t^2},\;dx = 2tdt.\) As a result, we have
\[y = - \int {\frac{{dx}}{{2\sqrt x + {C_1}}}}
= - \int {\frac{{2tdt}}{{2t + {C_1}}}}
= - \int {\frac{{2t + {C_1} - {C_1}}}{{2t + {C_1}}}dt}
= - \int {\left( {1 - \frac{{{C_1}}}{{2t + {C_1}}}} \right)dt}
= - t + {C_1}\int {\frac{{dt}}{{2t + {C_1}}}}
= - t + \frac{{{C_1}}}{2}\int {\frac{{d\left( {2t + {C_1}} \right)}}{{2t + {C_1}}}}
= - t + \frac{{{C_1}}}{2}\ln \left| {2t + {C_1}} \right| + {C_2}.\]
Returning to the variable \(x,\) we finally obtain
\[y = - \sqrt x + \frac{{{C_1}}}{2}\ln \left| {2\sqrt x + {C_1}} \right| + {C_2}.\]
Example 5.
Solve the equation \[y^{\prime\prime} = \left( {2y + 3} \right){\left( {y'} \right)^2}.\]
Solution.
This equation does not explicitly contain the independent variable \(x,\) that is refers to the Case \(5.\) Let \(y' = p\left( y \right).\) Then the equation can be written as
\[p' = \left( {2y + 3} \right){p^2}.\]
Separate variables and integrate:
\[\frac{{dp}}{{{p^2}}} = \left( {2y + 3} \right)dy,\;\; \Rightarrow
\int {\frac{{dp}}{{{p^2}}}} = \int {\left( {2y + 3} \right)dy} ,\;\; \Rightarrow
- \frac{1}{p} = {y^2} + 3y + {C_1},\;\; \Rightarrow
p = y' = \frac{{ - 1}}{{{y^2} + 3y + {C_1}}}.\]
Integrating again, we obtain the final solution in implicit form:
\[\left( {{y^2} + 3y + {C_1}} \right)dy = - dx,\;\; \Rightarrow
\int {\left( {{y^2} + 3y + {C_1}} \right)dy} = - \int {dx} ,\;\; \Rightarrow
{y^3} + \frac{{3{y^2}}}{2} + {C_1}y + {C_2} = - x,\;\; \Rightarrow
2{y^3} + 3{y^2} + {C_1}y + {C_2} + 2x = 0,\]
where \({C_1}, {C_2}\) are constants of integration.
Example 6.
Solve the equation \[yy^{\prime\prime} = \left( {y'} \right)^2 - \frac{{3{y^2}}}{{\sqrt x }}.\]
Solution.
The equation satisfies the condition of homogeneity. Therefore, we make the following change of variable: \(y = {e^{\int {zdx} }}.\) The derivatives will be equal
\[y' = z{e^{\int {zdx} }},\]
\[y^{\prime\prime} = z'{e^{\int {zdx} }} + {z^2}{e^{\int {zdx} }}
= \left( {z' + {z^2}} \right){e^{\int {zdx} }}.\]
Then the differential equation becomes:
\[{e^{\int {zdx} }}\left( {z' + {z^2}} \right){e^{\int {zdx} }}
= {\left( {z{e^{\int {zdx} }}} \right)^2} - \frac{{3{{\left( {{e^{\int {zdx} }}} \right)}^2}}}{{\sqrt x }},\;\; \Rightarrow
\cancel{e^{2\int {zdx} }}\left( {z' + {z^2}} \right) = \cancel{e^{2\int {zdx} }}\left( {{z^2} - \frac{3}{{\sqrt x }}} \right),\;\; \Rightarrow
z' + \cancel{z^2} = \cancel{z^2} - \frac{3}{{\sqrt x }},\;\; \Rightarrow
z' = - \frac{3}{{\sqrt x }}.\]
It is easy to find the function \(z\left( x \right):\)
\[\frac{{dz}}{{dx}} = - \frac{3}{{\sqrt x }},\;\; \Rightarrow
dz = - \frac{{3dx}}{{\sqrt x }},\;\; \Rightarrow
\int {dz} = - 3\int {\frac{{dx}}{{\sqrt x }}} ,\;\; \Rightarrow
z = - 6\sqrt x + {C_1}.\]
The original function \(y\left( x \right)\) is defined by the formula
\[y\left( x \right) = {C_2}{e^{\int {zdx} }}.\]
The calculations give the following answer:
\[y\left( x \right) = {C_2}{e^{\int {zdx} }}
= {C_2}{e^{\int {\left( {{C_1} - 6\sqrt x } \right)dx} }}
= {C_2}{e^{{C_1}x - \frac{{6{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}}}
= {C_2}{e^{{C_1}x - 4\sqrt {{x^3}} }}.\]
Note that in addition to the general solution, the differential equation also contains the singular solution \(y = 0.\)
Example 7.
Solve the equation \[yy^{\prime\prime} + {\left( {y'} \right)^2} = 2x + 1.\]
Solution.
You may notice that the left side of the equation is the derivative of \(yy'.\) Therefore, denoting \(z = yy',\) we obtain the following differential equation:
\[{\left( {yy'} \right)^\prime } = 2x + 1,\;\; \Rightarrow z' = 2x + 1.\]
The last equation is easily solved by separation of variables:
\[\frac{{dz}}{{dx}} = 2x + 1,\;\; \Rightarrow
dz = \left( {2x + 1} \right)dx,\;\; \Rightarrow
\int {dz} = \int {\left( {2x + 1} \right)dx} ,\;\; \Rightarrow
z = {x^2} + x + {C_1}.\]
Now we integrate one more equation for \(y\left( x \right):\)
\[yy' = {x^2} + x + {C_1},\;\; \Rightarrow
\int {ydy} = \int {\left( {{x^2} + x + {C_1}} \right)dx} ,\;\; \Rightarrow
\frac{{{y^2}}}{2} = \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + {C_1}x + {C_2},\;\; \Rightarrow
3{y^2} = 2{x^3} + 3{x^2} + {C_1}x + {C_2},\]
where \({C_1}, {C_2}\) are arbitrary constants.
