Differential Equations

Second Order Equations

2nd Order Diff Equations Logo

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.

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