Differential equations – Ordinary differential equations – Higher-order linear differential equations

For higher-order linear linear differential equations the theory is quite similar to that of second-order linear differential equations. An \(n^{\text{th}}\)-order linear differential equation can be written in the (standard) form

\[y^{(n)}(t)+p_1(t)y^{(n-1)}(t)+\cdots+p_{n-1}(t)y'(t)+p_n(t)y(t)=g(t)\]

with \(p_1,p_2,\ldots,p_n\) and \(g\) continuous functions. This differential equation is called homogeneous if \(g(t)\equiv0\) and otherwise nonhomogeneous.

We start with the existence and uniqueness theorem:

Theorem: If \(p_1,p_2,\ldots,p_n\) and \(g\) are continuous on an open interval \(I\) and \(t_0\in I\), then there exists exactly one function \(y(t)\) that satisfies both the differential equation and the initial conditions \(y(t_0)=y_0,\;y'(t_0)=y_0',\;\ldots,\;y^{(n-1)}(t_0)=y_0^{(n-1)}\).

Moreover, for a homogeneous linear differential equation we have the super position principle:

Theorem: If \(y_1,\;y_2,\;\ldots,\;y_n\) are solutions of the homogeneous linear differential equation

\[y^{(n)}(t)+p_1(t)y^{(n-1)}(t)+\cdots+p_{n-1}(t)y'(t)+p_n(t)y(t)=0,\]

then \(y(t)=c_1y_1(t)+c_2y_2(t)+\cdots+c_ny_n(t)\) is also a solution for every choice of \(c_1,c_2,\ldots,c_n\in\mathbb{R}\).

Definition: If \(y_1,\;y_2,\;\ldots,\;y_n\) are solutions of the homogeneous linear differential equation

\[y^{(n)}(t)+p_1(t)y^{(n-1)}(t)+\cdots+p_{n-1}(t)y'(t)+p_n(t)y(t)=0,\]

then

\[W(y_1,y_2,\ldots,y_n)(t):=\begin{vmatrix}y_1(t)&y_2(t)&\ldots&y_n(t)\\y_1'(t)&y_2'(t)&\ldots&y_n'(t)\\\vdots&\vdots&&\vdots\\ y_1^{(n-1)}(t)&y_2^{(n-1)}(t)&\ldots&y_n^{(n-1)}(t)\end{vmatrix}\]

is the Wronskian determinant or the Wronskian of the solutions \(y_1,\;y_2,\;\ldots,\;y_n\). Now we have Abel's theorem:

Theorem: \(W(y_1,y_2,\ldots,y_n)(t)=c\cdot e^{-\int p_1(t)\,dt}\) for some \(c\in\mathbb{R}\).

This means that the determinant is either equal to zero for all \(t\) (if \(c=0\)) or unequal to zero for all \(t\) (if \(c\neq0\)).

For homogeneous linear differential equations with constant coefficients the substitution of \(y(t)=e^{rt}\) leads to an \(n^{\text{th}}\)-degree characteristic polynomial. If we assume that the coefficients are real, then this polynomial has \(n\) zeros, counted with multiplicity, where nonreal zeros only occur in complex conjugate pairs.

Examples:

1) The characteristic equation of \(y^{(3)}(t)-2y''(t)-5y'(t)+6y(t)=0\) is \(r^3-2r^2-5r+6=0\). If this equation has an integere solution, then this should be a divisor of \(6\). So try \(\pm1\), \(\pm2\), \(\pm3\) and \(\pm6\) as a possible solution. Note that \(r=1\) is a solution, so we have: \(r^3-2r^2-5r+6=(r-1)(r^2-r-6)=(r-1)(r+2)(r-3)\). Then the general solution is \(y(t)=c_1e^t+c_2e^{-2t}+c_3e^{3t}\).

2) The characteristic equation of \(y^{(4)}-2y''(t)+y(t)=0\) is \(r^4-2r^2+1=0\) or equivalently \((r^2-1)^2=0\). So the solutions are \(r=\pm1\), which both occur twice. Hence the general solution is: \(y(t)=c_1e^t+c_2te^t+c_3e^{-t}+c_4te^{-t}\).

3) The characteristic equation of \(y^{(4)}-y(t)=0\) is \(r^4-1=0\) or equivalently \((r^2-1)(r^2+1)=0\). So the solutions are \(r=\pm1\) and \(r=\pm i\). Hence the general solution is: \(y(t)=c_1e^t+c_2e^{-t}+c_3\cos(t)+c_4\sin(t)\).

For nonhomogeneous linear differential equations with constant coefficients we can use the method of undetermined coefficients to find a particular solution if the right-hand side \(g(t)\) is an exponential function, a polynomial, a sine, a cosine or a combination of these functions.

Examples:

1) Consider \(y^{(3)}(t)-y''(t)+y'(t)-y(t)=e^{-t}\). The characteristic equation is: \(r^3-r^2+r-1=0\) or equivalently \((r-1)(r^2+1)=0\). So the general solution of the homogeneous differential equation is: \(y_h(t)=c_1e^t+c_2\cos(t)+c_3\sin(t)\). Now set \(y_p(t)=Ae^{-t}\), then we have: \(y_p'(t)=-Ae^{-t}\), \(y_p''(t)=Ae^{-t}\) and \(y_p^{(3)}(t)=-Ae^{-t}\). Substitution then gives: \(-4Ae^{-t}=e^{-t}\) and therefore \(A=-\frac{1}{4}\). Hence a particular solution is: \(y_p(t)=-\frac{1}{4}e^{-t}\). Then the general solution is: \(y(t)=y_p(t)+y_h(t)=-\frac{1}{4}e^{-t}+c_1e^t+c_2\cos(t)+c_3\sin(t)\).

2) Consider \(y^{(3)}(t)-y''(t)+y'(t)-y(t)=e^t\). Hence (see previous example): \(y_h(t)=c_1e^t+c_2\cos(t)+c_3\sin(t)\). Now set \(y_p(t)=Ate^t\), then we have: \(y_p'(t)=A(t+1)e^t\), \(y''(t)=A(t+2)e^t\) and \(y^{(3)}(t)=A(t+3)e^t\). Substitution then gives: \(A(t+3-t-2+t+1-t)e^t=e^t\) or equivalently \(2Ae^t=e^t\). Hence: \(A=\frac{1}{2}\). So a particular solution is: \(y_p(t)=\frac{1}{2}te^t\). Then the general solution is: \(y(t)=y_p(t)+y_h(t)=\frac{1}{2}te^t+c_1e^t+c_2\cos(t)+c_3\sin(t)\).

3) Consider \(y^{(4)}(t)-5y''(t)+4y(t)=8t^3\). The characteristic equation is: \(r^4-5r^2+4=0\) or equivalently \((r^2-1)(r^2-4)=0\). So the general solution of the homogeneous differential equation is: \(y_h(t)=c_1e^t+c_2^{-t}+c_3e^{2t}+c_4e^{-2t}\). Now set \(y_p(t)=At^3+Bt^2+Ct+D\), then we have: \(y_p'(t)=3At^2+2Bt+C\), \(y_p''(t)=6At+2B\), \(y_p^{(3)}(t)=6A\) and \(y^{(4)}(t)=0\). Substitution then gives:

\[-30At-10B+4At^3+4Bt^2+4Ct+4D=8t^3.\]

This implies that \(4A=8\), \(4B=0\), \(4C-30A=0\) and \(4D-10B=0\). Hence: \(A=2\), \(B=0\), \(C=15\) and \(D=0\). So a particular solution is: \(y_p(t)=2t^3+15t\). Then the general solution is: \(y(t)=y_p(t)+y_h(t)=2t^3+15t+c_1e^t+c_2^{-t}+c_3e^{2t}+c_4e^{-2t}\).

We also have the method of variation of parameters: if \(y_1,\;y_2,\;\ldots,\;y_n\) are linear independent solutions of the homogeneous differential equation

\[y^{(n)}(t)+p_1(t)y^{(n-1)}(t)+\cdots+p_{n-1}(t)y'(t)+p_n(t)y(t)=0,\]

then the Wronskian determinant or Wronskian of these solutions is

\[W(y_1,y_2,\ldots,y_n)(t):=\begin{vmatrix}y_1(t)&y_2(t)&\ldots&y_n(t)\\y_1'(t)&y_2'(t)&\ldots&y_n'(t)\\\vdots&\vdots&&\vdots\\ y_1^{(n-1)}(t)&y_2^{(n-1)}(t)&\ldots&y_n^{(n-1)}(t)\end{vmatrix}\neq0\]

and so \(y_h(t)=c_1y_1(t)+c_2y_2(t)+\cdots+c_ny_n(t)\) is the general solution of the homogeneous differential equation.

Now suppose that \(y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)+\cdots+u_n(t)y_n(t)\), then we have:

\[y'(t)=\underbrace{u_1'(t)y_1(t)+u_2'(t)y_2(t)+\cdots+u_n'(t)y_n(t)}_{\text{stel dit gelijk aan nul}}+u_1(t)y_1'(t)+u_2(t)y_2'(t)+\cdots+u_n(t)y_n'(t),\] \[y''(t)=\underbrace{u_1'(t)y_1'(t)+u_2'(t)y_2'(t)+\cdots+u_n'(t)y_n'(t)}_{\text{stel dit gelijk aan nul}}+u_1(t)y_1''(t)+u_2(t)y_2''(t)+\cdots+u_n(t)y_n''(t),\]

and so on until

\[y^{(n-1)}(t)=\underbrace{u_1'(t)y_1^{(n-2)}(t)+u_2'(t)y_2^{(n-2)}(t)+\cdots+u_n'(t)y_n^{(n-2)}(t)}_{\text{stel dit gelijk aan nul}}+u_1(t)y_1^{(n-1)}(t)+u_2(t)y_2^{(n-1)}(t)+\cdots+u_n(t)y_n^{(n-1)}(t).\]

Then we have:

\[y^{(n)}(t)=u_1'(t)y_1^{(n-1)}(t)+u_2'(t)y_2^{(n-1)}(t)+\cdots+u_n'(t)y_n^{(n-1)}(t)+u_1(t)y_1^{(n)}(t)+u_2(t)y_2^{(n)}(t)+\cdots+u_n(t)y_n^{(n)}(t).\]

Substitution into the (standard form of the) nonhomogeneous differential equation then finally gives (the terms with \(u_1(t),\;\ldots,\;u_n(t)\) cancel):

\[u_1'(t)y_1^{(n-1)}(t)+u_2'(t)y_2^{(n-1)}(t)+\cdots+u_n'(t)y_n^{(n-1)}(t)=g(t).\]

Then finally we have:

\[\begin{pmatrix}y_1(t)&y_2(t)&\ldots&y_n(t)\\y_1'(t)&y_2'(t)&\ldots&y_n'(t)\\\vdots&\vdots&&\vdots\\ y_1^{(n-1)}(t)&y_2^{(n-1)}(t)&\ldots&y_n^{(n-1)}(t)\end{pmatrix}\begin{pmatrix}u_1'(t)\\u_2'(t)\\\vdots\\u_n'(t)\end{pmatrix} =\begin{pmatrix}0\\0\\\vdots\\g(t)\end{pmatrix}.\]

Since the Wronskian is unequal to zero, this coefficient matrix is invertable and this system has exactly one solution for \(u_1'(t),\;\ldots,\;u_n'(t)\). Integration then gives \(u_1(t),\;\ldots,\;u_n(t)\) each including an arbitrary integration constant. Then the general solution is: \(y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)+\cdots+u_n(t)y_n(t)\).

Examples:

1) Consider \(y^{(3)}(t)-y''(t)+y'(t)-y(t)=e^{-t}\). The characteristic equation is: \(r^3-r^2+r-1=0\) or equivalently \((r-1)(r^2+1)=0\). So the general solution of the homogeneous differential equation is: \(y_h(t)=c_1e^t+c_2\cos(t)+c_3\sin(t)\). Before we have already seen using the method of undetermined coefficients that \(y_p(t)=-\frac{1}{4}e^{-t}\) is a particular solution.

Now we apply the method of variation of parameters: suppose that \(y(t)=u_1(t)e^t+u_2(t)\cos(t)+u_3(t)\sin(t)\), then we have

\[y'(t)=\underbrace{u_1'(t)e^t+u_2'(t)\cos(t)+u_3'(t)\sin(t)}_{\text{set this equal to zero}}+u_1(t)e^t-u_2(t)\sin(t)+u_3(t)\cos(t),\] \[y''(t)=\underbrace{u_1'(t)e^t-u_2'(t)\sin(t)+u_3'(t)\cos(t)}_{\text{set this equal to zero}}+u_1(t)e^t-u_2(t)\cos(t)-u_3(t)\sin(t)\]

and

\[y^{(3)}(t)=u_1'(t)e^t-u_2'(t)\cos(t)-u_3'(t)\sin(t)+u_1(t)e^t+u_2(t)\sin(t)-u_3(t)\cos(t).\]

Substitution then gives (the terms with \(u_1(t)\), \(u_2(t)\) and \(u_3(t)\) cancel): \(u_1'(t)e^t-u_2'(t)\cos(t)-u_3'(t)\sin(t)=e^{-t}\). Hence:

\[\left\{\begin{array}{rcl}u_1'(t)e^t+u_2'(t)\cos(t)+u_3'(t)\sin(t)&=&0\\u_1'(t)e^t-u_2'(t)\sin(t)+u_3'(t)\cos(t)&=&0\\ u_1'(t)e^t-u_2'(t)\cos(t)-u_3'(t)\sin(t)&=&e^{-t}\end{array}\right.\quad\Longleftrightarrow\quad \begin{pmatrix}e^t&\cos(t)&\sin(t)\\e^t&-\sin(t)&\cos(t)\\e^t&-\cos(t)&-\sin(t)\end{pmatrix} \begin{pmatrix}u_1'(t)\\u_2'(t)\\u_3'(t)\end{pmatrix}=\begin{pmatrix}0\\0\\e^{-t}\end{pmatrix}.\]

Finally this implies: \(u_1'(t)=\frac{1}{2}e^{-2t}\), \(u_2'(t)=-\frac{1}{2}e^{-t}\left(\cos(t)-\sin(t)\right)\) and \(u_3'(t)=-\frac{1}{2}e^{-t}\left(\cos(t)+\sin(t)\right)\). Integration then gives: \(u_1(t)=-\frac{1}{4}e^{-2t}+c_1\), \(u_2(t)=-\frac{1}{2}e^{-t}\sin(t)+c_2\) and \(u_3(t)=\frac{1}{2}e^{-t}\cos(t)+c_3\). Finally we then obtain:

\begin{align*} y(t)=u_1(t)e^t+u_2(t)\cos(t)+u_3(t)\sin(t)&=-\tfrac{1}{4}e^{-t}-\tfrac{1}{2}e^{-t}\sin(t)\cos(t)+\tfrac{1}{2}e^{-t}\sin(t)\cos(t) +c_1e^t+c_2\cos(t)+c_3\sin(t)\\[2.5mm] &=\underbrace{-\tfrac{1}{4}e^{-t}}_{y_p(t)}+\underbrace{c_1e^t+c_2\cos(t)+c_3\sin(t)}_{y_h(t)}. \end{align*}

2) Consider \(y^{(4)}(t)-5y''(t)+4y(t)=1\). The characteristic equation is: \(r^4-5r^2+4=0\) or equivalently \((r^2-1)(r^2-4)=0\). So the general solution of the homogeneous differential equation is: \(y_h(t)=c_1e^t+c_2^{-t}+c_3e^{2t}+c_4e^{-2t}\). It is easily seen that \(y_p(t)=\frac{1}{4}\) is a particular solution. Variation of parameters would work as follows: suppose that \(y(t)=u_1(t)e^t+u_2(t)e^{-t}+u_3(t)e^{2t}+u_4(t)e^{-2t}\), then we finally obtain that

\[\begin{pmatrix}e^t&e^{-t}&e^{2t}&e^{-2t}\\e^t&-e^{-t}&2e^{2t}&-2e^{-2t}\\e^t&e^{-t}&4e^{2t}&4e^{-2t}\\ e^t&-e^{-t}&8e^{2t}&-8e^{-2t}\end{pmatrix}\begin{pmatrix}u_1'(t)\\u_2'(t)\\u_3'(t)\\u_4'(t)\end{pmatrix} =\begin{pmatrix}0\\0\\0\\1\end{pmatrix}.\]

Finally this implies: \(u_1'(t)=-\frac{1}{6}e^{-t}\), \(u_2'(t)=\frac{1}{6}e^t\), \(u_3'(t)=\frac{1}{12}e^{-2t}\) and \(u_4'(t)=-\frac{1}{12}e^{2t}\). Integration then gives: \(u_1(t)=\frac{1}{6}e^{-t}+c_1\), \(u_2(t)=\frac{1}{6}e^t+c_2\), \(u_3(t)=-\frac{1}{24}e^{-2t}+c_3\) and \(u_4(t)=-\frac{1}{24}e^{2t}+c_4\). Finally we obtain:

\begin{align*} y(t)=u_1(t)e^t+u_2(t)e^{-t}+u_3(t)e^{2t}+u_4(t)e^{-2t}&=\tfrac{1}{6}+\tfrac{1}{6}-\tfrac{1}{24}-\tfrac{1}{24} +c_1e^t+c_2e^{-t}+c_3e^{2t}+c_4e^{-2t}\\[2.5mm] &=\underbrace{\tfrac{1}{4}}_{y_p(t)}+\underbrace{c_1e^t+c_2e^{-t}+c_3e^{2t}+c_4e^{-2t}}_{y_h(t)}. \end{align*}

3) Consider \(y^{(3)}(t)-3y''(t)+3y'(t)-y(t)=\displaystyle\frac{e^t}{t^3}\) for \(t>0\). The characteristic equation is \(r^3-3r^2+3r-1=0\) or equivalently \((r-1)^3=0\). So the general solution of the homogeneous differential equation is \(y_h(t)=c_1e^t+c_2te^t+c_3t^2e^t\). Now suppose that \(y(t)=u_1(t)e^t+u_2(t)te^t+u_3(t)t^2e^t\), then we have:

\[\begin{pmatrix}e^t&te^t&t^2e^t\\e^t&(t+1)e^t&(t^2+2t)e^t\\e^t&(t+2)e^t&(t^2+4t+2)e^t\end{pmatrix} \begin{pmatrix}u_1'(t)\\u_2'(t)\\u_3'(t)\end{pmatrix}=\begin{pmatrix}0\\0\\\frac{e^t}{t^3}\end{pmatrix}.\]

Finally this implies that \(u_1'(t)=\displaystyle\frac{1}{2t}\), \(u_2'(t)=-\displaystyle\frac{1}{t^2}\) and \(u_3'(t)=\displaystyle\frac{1}{2t^3}\). Integration then gives: \(u_1(t)=\frac{1}{2}\ln(t)+k_1\), \(u_2(t)=\displaystyle\frac{1}{t}+k_2\) and \(u_3(t)=-\displaystyle\frac{1}{4t^2}+k_3\). Finally we then obtain:

\[y(t)=u_1(t)e^t+u_2(t)te^t+u_3(t)t^2e^t=\tfrac{1}{2}e^t\ln(t)+e^t-\tfrac{1}{4}e^t+k_1e^t+k_2te^t+k_3t^2e^t.\]

This can be simplified to: \(y(t)=\frac{1}{2}e^t\ln(t)+c_1e^t+c_2te^t+c_3t^2e^t\).

It might be clear that the method of variation of parameters should only be applied when the method of undetermined coefficients does not work!

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Last modified on April 21, 2021