Differential equations – Ordinary differential equations – Reduction of order

The method of reduction of order is based on the same principle as the method of variation of the parameter that we have seen for first-order linear differential equations. In that case the arbitrary integration constant in the general solution of the homogeneous differential equation is replaced by a function in order to find a particular solution (or the general solution) of the ninhomogeneous differential equation.

The method can also be used to find a second, linear independent, solution of a homogeneous second-order linear differential equation with constant coefficients whose characteristic polynomial has two equal real zeros.

Example: Consider \(y''(t)+4y'(t)+4y(t)=0\) with characteristic equation \(r^2+4r+4=0\) or equivalently \((r+2)^2=0\). Hence: \(y(t)=ce^{-2t}\) is a solution for each \(c\in\mathbb{R}\). Now set:

\[y(t)=u(t)e^{-2t}\quad\Longrightarrow\quad y'(t)=u'(t)e^{-2t}-2u(t)e^{-2t}\quad\Longrightarrow\quad y''(t)=u''(t)e^{-2t}-4u'(t)e^{-2t}+4u(t)e^{-2t}.\]

Substitution then gives:

\[u''(t)e^{-2t}-4u'(t)e^{-2t}+4u(t)e^{-2t}+4u'(t)e^{-2t}-8u(t)e^{-2t}+4u(t)e^{-2t}=0\quad\Longrightarrow\quad u''(t)=0.\]

This implies that \(u(t)=c_1+c_2t\). Finally we have: \(y(t)=u(t)e^{-2t}=c_1e^{-2t}+c_2te^{-2t}\).

If we apply the same principle to an Euler equation, whose characteristic equation has two equal real solutions, then the name method of reduction of order becomes clear.

Example: Consider \(t^2y''(t)+7ty'(t)+9y(t)=0\) with \(t>0\). The characteristic equation is: \(r(r-1)+7r+9=0\) or equivalently \((r+3)^2=0\). Hence: \(y(t)=ct^{-3}\) is a solution for each \(c\in\mathbb{R}\). Now set:

\[y(t)=u(t)t^{-3}\quad\Longrightarrow\quad y'(t)=u'(t)t^{-3}-3u(t)t^{-4}\quad\Longrightarrow\quad y''(t)=u''(t)t^{-3}-6u'(t)t^{-4}+12u(t)t^{-5}.\]

Substitution then gives:

\[t^{-1}u''(t)-6t^{-2}u'(t)+12t^{-3}u(t)+7t^{-2}u'(t)-21t^{-3}u(t)+9t^{-3}u(t)=0\quad\Longrightarrow\quad tu''(t)+u'(t)=0.\]

Now set \(u'(t)=v(t)\), then we have: \(tv'(t)+v(t)=0\). This is a first-order differential equation, which explains the name reduction of order. This differential equation is separable and has general solution \(v(t)=\displaystyle\frac{c_2}{t}\). This implies that \(u(t)=c_1+c_2\ln(t)\). Finally we have: \(y(t)=u(t)t^{-3}=c_1t^{-3}+c_2t^{-3}\ln(t)\).

This principle works generally: if \(u(t)\) is a constant, then it is a solution of the homogeneous differential equation. In that case the derivatives of \(u(t)\) would all be zero. This means that in the general case all terms without derivatives of \(u(t)\) cancel. Then for a second-order differential equation only the terms with \(u'(t)\) and \(u''(t)\) remain (the terms with \(u(t)\) cancel). For the unknown function \(v(t)=u'(t)\) a first-order differential equation arises. The method of reduction of order can generally be used to find a second, linear independent, solution (or the general solution) of a homogeneous second-order linear differential equation for which one solution is known.

Examples:

1) Consider \(t^2y''(t)-t(t+2)y'(t)+(t+2)y(t)=0\) with \(t>0\). Note that \(y(t)=t\) is a solution. Now set:

\[y(t)=tu(t)\quad\Longrightarrow\quad y'(t)=tu'(t)+u(t)\quad\Longrightarrow\quad y''(t)=tu''(t)+2u'(t).\]

Substitution then gives:

\[t^3u''(t)+2t^2u'(t)-t^2(t+2)u'(t)-t(t+2)u(t)+t(t+2)u(t)=0\quad\Longrightarrow\quad t^3(u''(t)+u'(t))=0\quad\Longrightarrow\quad u''(t)-u'(t)=0.\]

Now set \(v(t)=u'(t)\), then we have: \(v'(t)-v(t)=0\). Hence: \(v(t)=c_2e^t\). This implies that \(u(t)=c_1+c_2e^t\). Finally we have: \(y(t)=tu(t)=c_1t+c_2te^t\).

2) Consider \(ty''(t)-(1+2t)y'(t)+(1+t)y(t)=0\) with \(t>0\). Note that \(y(t)=e^t\) is a solution. Now set:

\[y(t)=e^tu(t)\quad\Longrightarrow\quad y'(t)=e^tu'(t)+e^tu(t)\quad\Longrightarrow\quad y''(t)=e^tu''(t)+2e^tu'(t)+e^tu(t).\]

Substitution then gives:

\[te^tu''(t)+2te^tu'(t)+te^tu(t)-(1+2t)e^tu'(t)-(1+2t)e^tu(t)+(1+t)e^tu(t)=0\quad\Longrightarrow\quad tu''(t)-u'(t)=0.\]

Now set \(v(t)=u'(t)\), then we have: \(tv'(t)-v(t)=0\). Hence: \(v(t)=c_2t\). This implies that \(u(t)=c_1+\frac{1}{2}c_2t^2\). Finally we have: \(y(t)=tu(t)=c_1e^t+\frac{1}{2}c_2t^2e^t\).

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Last modified on April 19, 2021