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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.

  1. \(y''-2y'-3y=3e^{2x}\)

  2. \(y''-2y'-3y=-65\cos(2x)\)

  3. \(y''-2y'-3y=9x^2+1\)

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.

  1. 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 \[y=y_h+y_p=c_1e^{-x}+c_2e^{3x}-e^{2x}.\]

  2. 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 \[y=y_h+y_p=c_1e^{-x}+c_2e^{3x}+7\cos(2x)+4\sin(2x).\]

  3. 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.

  1. \(y''+2y'-3y=16xe^{x}\)

  2. \(y''+2y'-3y=-10x\sin x\)

  3. \(y''+2y'-3y=16xe^{x}-10x\sin x\)

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.

  1. 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 \[y=y_h+y_p=c_1e^{-3x}+c_2e^{x}+(2x^2-x)e^{x}.\]

  2. 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 \[y=y_h+y_p=c_1e^{-3x}+c_2e^{x}+(x+7/5)\cos x+(2x-1/5)\sin x.\]

  3. 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 \[y=c_1e^{-3x}+c_2e^{x}+(2x^2-x)e^{x}+(x+7/5)\cos x+(2x-1/5)\sin x.\]


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