Linear Nonhomogeneous ODEs : Undetermined Coefficients
Consider the linear nonhomogeneous second order ODE
\[y''+p(x)y'+q(x)y=g(x). \;\;\;\;(38)\]
Theorem.
The general solution of the nonhomogeneous ODE (38) is
\[y=c_1y_1+c_2y_2+y_p,\]
for arbitrary constants \(c_1\) and \(c_2\) where \(y_1\) and \(y_2\) are fundamental solutions of the corresponding homogeneous ODE \(y''+p(x)y'+q(x)y=0\) and \(y_p\) is a particular solution of (38).
Let \(y=\phi(x)\) be a solution of (38). Then
\[\phi''+p \phi'+q\phi=g.\]
Since \(y_p\) is a particular solution of (38), we have
\[y_p''+p y_p'+qy_p=g.\]
Using preceding two equations we get
\[(\phi-y_p)''+p (\phi-y_p)'+q(\phi-y_p)=g-g=0.\]
Thus \(\phi-y_p\) is solution of the homogeneous ODE \(y''+p(x)y'+q(x)y=0\). So \(\phi-y_p\) can be written as a linear combination of fundamental solutions \(y_1\) and \(y_2\). Thus
\[\begin{align*}
\phi-y_p&=c_1y_1+c_2y_2\\
\phi&=c_1y_1+c_2y_2+y_p.\; \blacksquare\\
\end{align*}\]
Theorem.
The general solution of the nonhomogeneous ODE
\[y''+p(x)y'+q(x)y=g_1(x)+g_2(x)+\cdots+g_n(x). \;\;\;\;(39)\]
is
\[y=c_1y_1+c_2y_2+y_{p_1}+y_{p_2}+\cdots+y_{p_n},\]
for arbitrary constants \(c_1\) and \(c_2\) where \(y_1\) and \(y_2\) are fundamental solutions of the corresponding homogeneous ODE \(y''+p(x)y'+q(x)y=0\) and \(y_{p_i}\) is a particular solutions of
\[y''+p(x)y'+q(x)y=g_i(x),\; i=1,\ldots,n.\]
Since \(y_{p_i}''+p(x)y_{p_i}'+q(x)y_{p_i}=g_i(x),\; i=1,\ldots,n\), we have
\[(y_{p_1}+\cdots+y_{p_n})''+p(x)(y_{p_1}+\cdots+y_{p_n})'+q(x)(y_{p_1}+\cdots+y_{p_n})=g_1(x)+\cdots+g_n(x)=g(x).\]
Thus \(y_{p_1}+\cdots+y_{p_n}\) is a particular solution of (39). By the preceding theorem, the general solution of the nonhomogeneous ODE (39) is
\[y=c_1y_1+c_2y_2+y_{p_1}+y_{p_2}+\cdots+y_{p_n},\]
for arbitrary constants \(c_1\) and \(c_2\). \(\blacksquare\)
Example.
Find the general solution.
\(y''-2y'-3y=3e^{2x}\)
\(y''-2y'-3y=-65\cos(2x)\)
\(y''-2y'-3y=9x^2+1\)
Solution. The characteristic equation is
\[\begin{align*}
r^2-2r-3&=0\\
(r+1)(r-3)&=0\\
r&=-1,3.
\end{align*}\]
So the homogeneous solution is \[y_h=c_1e^{-x}+c_2e^{3x},\]
for arbitrary constants \(c_1\) and \(c_2\). Now we will find a particular solution of the nonhomogeneous ODE.
Let \(y_p=ae^{2x}\) be a particular solution for some constant \(a\).
Then \(y_p'=2ae^{2x}\) and \(y_p''=4ae^{2x}\). Plugging these into \(y_p''-2y_p'-3y_p=3e^{2x}\), we get
\[\begin{align*}
4ae^{2x}-2\cdot 2ae^{2x}-3\cdot ae^{2x}&=3e^{2x}\\
(4a-4a-3a)e^{2x}&=3e^{2x}\\
-3a&=3\\
a&=-1.
\end{align*}\]
So \(y_p=-e^{2x}\) is a particular solution. Thus the general solution is
\[y=y_h+y_p=c_1e^{-x}+c_2e^{3x}-e^{2x}.\]
Let \(y_p=a\cos(2x)+b\sin(2x)\) be a particular solution for some constants \(a,\; b\).
Then \(y_p'=-2a\sin(2x)+2b\cos(2x)\) and \(y_p''=-4a\cos(2x)-4b\sin(2x)\). Plugging these into \(y_p''-2y_p'-3y_p=-65\cos(2x)\), we get
\[\begin{align*}
(-4a\cos(2x)-4b\sin(2x))-2\cdot(-2a\sin(2x)+2b\cos(2x)) -3\cdot(a\cos(2x)+b\sin(2x)) &=-65\cos(2x)\\
(-4a-4b-3a)\cos(2x)+(-4b+4a-3b)\sin(2x)&=-65\cos(2x)\\
(-7a-4b)\cos(2x)+(4a-7b)\sin(2x)&=-65\cos(2x)
\end{align*}\]
Comparing the coefficients of \(\cos(2x)\) and \(\sin(2x)\) in LHS and RHS, we get
\[\begin{align*}
-7a-4b&=-65 & 4a-7b&=0
\end{align*}\]
Solving we get \(a=7,\; b=4\). So the particular solution is \(y_p=7\cos(2x)+4\sin(2x)\). Thus the general solution is
\[y=y_h+y_p=c_1e^{-x}+c_2e^{3x}+7\cos(2x)+4\sin(2x).\]
Let \(y_p=ax^2+bx+c\) be a particular solution for some constants \(a,b,c\).
Then \(y_p'=2ax+b\) and \(y_p''=2a\). Plugging these into \(y_p''-2y_p'-3y_p=9x^2+1\), we get
\[\begin{align*}
2a-2\cdot (2ax+b)-3\cdot (ax^2+bx+c)&=9x^2+1\\
-3ax^2+(-4a-3b)x+(2a-2b-3c)&=9x^2+1
\end{align*}\]
Comparing the coefficients of \(x^2,x\) and the constant term in LHS and RHS, we get
\[\begin{align*}
-3a&=9 & -4a-3b&=0 & 2a-2b-3c&=1
\end{align*}\]
Solving we get \(a=-3,\; b=4,\; c=-5\). So \(y_p=-3x^2+4x-5\) is a particular solution. Thus the general solution is
\[y=y_h+y_p=c_1e^{-x}+c_2e^{3x}-3x^2+4x-5.\]
To find a particular solution of the ODE \(ay''+by'+cy=g(x)\) use the following table:
\[\begin{array}{|c|c|}
\hline g(x)&y_p\\
\hline c_0x^n+c_1x^{n-1}+\cdots+c_n & x^s(a_0x^n+a_1x^{n-1}+\cdots+a_n)\\
\hline (c_0x^n+c_1x^{n-1}+\cdots+c_n)e^{rx} & x^s(a_0x^n+a_1x^{n-1}+\cdots+a_n)e^{rx}\\
\hline (c_0x^n+c_1x^{n-1}+\cdots+c_n)\cos(\theta x), & x^s[(a_0x^n+a_1x^{n-1}+\cdots+a_n)\cos(\theta x)+\\
(c_0x^n+c_1x^{n-1}+\cdots+c_n)\sin(\theta x)\; & \;(b_0x^n+b_1x^{n-1}+\cdots+b_n)\sin(\theta x)]\\
\hline (c_0x^n+c_1x^{n-1}+\cdots+c_n)\cos(\theta x)e^{rx}, & x^s[(a_0x^n+a_1x^{n-1}+\cdots+a_n)\cos(\theta x)+\\
(c_0x^n+c_1x^{n-1}+\cdots+c_n)\sin(\theta x)e^{rx}\; & (b_0x^n+b_1x^{n-1}+\cdots+b_n)\sin(\theta x)]e^{rx}\\
\hline
\end{array}\]
where \(s\) is the smallest nonnegative integer \((s = 0, 1, 2)\) that will ensure that no term in \(y_p\) is a solution of the corresponding homogeneous equation.
Example.
Find the general solution.
\(y''+2y'-3y=16xe^{x}\)
\(y''+2y'-3y=-10x\sin x\)
\(y''+2y'-3y=16xe^{x}-10x\sin x\)
Solution. The characteristic equation is
\[\begin{align*}
r^2+2r-3&=0\\
(r-1)(r+3)&=0\\
r&=-3,1.
\end{align*}\]
So the homogeneous solution is \[y_h=c_1e^{-3x}+c_2e^{x},\]
for arbitrary constants \(c_1\) and \(c_2\). Now we will find a particular solution of the nonhomogeneous ODE.
Let \(y_p=(ax+b)e^{x}\) be a particular solution for some constants \(a,b\). Note that \(e^x\) is already in
the homogeneous solution. So let \(y_p=x(ax+b)e^{x}=(ax^2+bx)e^{x}\) be a particular solution for some constants \(a,b\).
Then \(y_p'=[ax^2+(2a+b)x+b]e^{x}\) and \(y_p''=[ax^2+(4a+b)x+2a+2b]e^{x}\). Plugging these into \(y_p''+2y_p'-3y_p=16e^{x}\), we get
\[\begin{align*}
[ax^2+(4a+b)x+2a+2b]e^{x}+2\cdot [ax^2+(2a+b)x+b]e^{x}-3\cdot (ax^2+bx)e^{x}&=16xe^{x}\\
[8ax+2a+4b]e^{x}&=16xe^{x}\\
8ax+2a+4b&=16x.
\end{align*}\]
Comparing the coefficient \(x\) and the constant term in LHS and RHS, we get
\[\begin{align*}
8a&=16 & 2a+4b&=0
\end{align*}\]
Solving we get \(a=2,\; b=-1\). So \(y_p=(2x^2-x)e^{x}\) is a particular solution. Thus the general solution is
\[y=y_h+y_p=c_1e^{-3x}+c_2e^{x}+(2x^2-x)e^{x}.\]
Let \(y_p=(ax+b)\cos x+(cx+d)\sin x\) be a particular solution for some constants \(a,b,c,d\). Then
\[y_p'=(cx+a+d)\cos x+(-ax-b+c)\sin x,\]
\[y_p''=(-ax-b+2c)\cos x+(-cx-2a-d)\sin x.\]
Plugging these into \(y_p''+2y_p'-3y_p=-10x\sin x\), we get
\[\begin{align*}
[(-ax-b+2c)\cos x+(-cx-2a-d)\sin x]+2[(cx+a+d)\cos x+(-ax-b+c)\sin x]&\\
-3[(ax+b)\cos x+(cx+d)\sin x]&=-10x\sin x\\
(-2a-4c)x\sin x+(-2a-d-2b+2c-3d)\sin x+(-ax-b+2c+2cx+2a+2d-3ax-3b)\cos x&=-10x\sin x\\
(-2a-4c)x\sin x+(-2a-2b+2c-4d)\sin x+(-4a+2c)x\cos x+(2a-4b+2c+2d)\cos x&=-10x\sin x
\end{align*}\]
Comparing the coefficients of \(x\sin x\), \(\sin x\), \(x\cos x\), and \(\cos x\) in LHS and RHS, we get
\[\begin{align*}
-2a-4c&=-10, & -2a-2b+2c-4d&=0, & -4a+2c&=0,& 2a-4b+2c+2d&=0
\end{align*}\]
Solving we get \(a=1,\; b=7/5,\; c=2,\; d=-1/5\). So
\[y_p=(x+7/5)\cos x+(2x-1/5)\sin x\] is a particular solution. Thus the general solution is
\[y=y_h+y_p=c_1e^{-3x}+c_2e^{x}+(x+7/5)\cos x+(2x-1/5)\sin x.\]
By preceding two parts and a theorem, \((2x^2-x)e^{x}+(x+7/5)\cos x+(2x-1/5)\sin x\) is a particular solution. Thus the general solution is
\[y=c_1e^{-3x}+c_2e^{x}+(2x^2-x)e^{x}+(x+7/5)\cos x+(2x-1/5)\sin x.\]