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Nonhomogeneous Constant Coefficient Linear Differential Equations

April 5th, 2009 |

Author: Merrin

We have already seen how to find the general solution for certain constant coefficient equations. Now we will add a monkey wrench into the mix and see what more we can solve. We add f(x) to the right hand side making the equation nonhomogeneous.

$a\cfrac{d^{2} y}{dx^{2} } +b\cfrac{dy}{dx} + cy=f(x)$

I will only present solutions to the equation for f(x) as a specific group of functions. The general strategy to solving nonhomogeneous differential
equations is to first imagine that f(x) is zero and solve that part.

$a\cfrac{d^{2} y}{dx^{2} } +b\cfrac{dy}{dx} +cy=0$

We will call the solution to this equation y_c for the complimentary solution. After we have solved that half we put back f(x) and find a particular solution, y_p to the differential equation. Now if we add the two solutions together we cover all our bases. The general solution is

$y=y_{c}(x)+y_{p}(x)$

I can show some good trial functions to find y_p only if f(x) is sufficiently simple. In this section, we will take f(x) to be a subset of the elementary functions. We will consider cases where f(x) is a polynomial, cosine or sine function, an exponential, or some combination of all three as a product. Finding the particular solution is one part educated guessing and one part educated guessing.

In the following examples we will use the same complimentary solutions and try solving different particular solutions.

Example 1. Find the general solution to

$\cfrac{d^{2} y}{dx^{2} } +2\cfrac{dy}{dx} +y = 4x^{2} +9x$

Solution 1. The form of the general solution will have a piece from the homogeneous equation and a piece from the nonhomogeneous equation. All these equations have the same homogeneous part. We can find that solution by looking at the characteristic equation

${\cfrac{d^{2} y}{dx^{2} } +2\cfrac{dy}{dx} +1=0}$
${\lambda ^{2} +2\lambda +1=0}$

This part is kind of tricky because there is a repeated root, but we remember the form of the general solution.

$y_{c} =c_{1} e^{-x} +c_{2} xe^{-x}$

There is a different trial solution to pick for the particular solution depending on the form of the nonhomogeneous term. With these trial solutions it is a simple matter to substitute into the equations and solve for the coefficients.

If f(x)=4x^2 +9x then try y_p =Ax^2+Bx+C. This substitution gives

${\cfrac{d^{2} }{dx^{2} } (Ax^{2} +Bx+C)+2\cfrac{d}{dx} (Ax^{2} +Bx+C)+(Ax^{2} +Bx+C)=4x^{2} +9x}$
${2A+4Ax+2B+Ax^{2} +Bx+C=4x^{2} +9x}$
${A=4} \qquad{(4A+B)=9}\qquad {B=-7} \qquad {2A+2B+C=0}$
${8-14+C=0} \qquad {C=6_{_{}^{} }^{} }$

Thus, the general solution is

$y=c_{1} e^{-x} +c_{2} xe^{-x} +4x^{2} -7x+6$

Example 2. Find the general solution.

$\cfrac{d^{2} y}{dx^{2} } +2\cfrac{dy}{dx} +y = \sin (3x)$

Solution 2. For sines and cosines of the same frequency in the nonhomogeneous terms, we try a trial solution of the form

$y_p(x)=A\sin(3x)+B\cos(3x)$
${\cfrac{d^{2} }{dx^{2} } (A\sin (3x)+B\cos (3x))+2\cfrac{d}{dx} (A\sin (3x)+B(\cos (3x))+....}$
$A\sin (3x)+B\cos (3x)=\sin (3x)$
${(-9A-6B+A)(\sin x)=1(\sin x)}$
${(-9B+6A+B)(\cos x)=0}$
${A=-2/25}$
${B=-3/50}$

Thus, the general solution is

$y_{c} =c_{1} e^{-x} +c_{2} xe^{-x} -\cfrac{2}{25} \sin (3x)-\cfrac{3}{50} \cos (3x)$

Example 3. Find the general solution.

$\cfrac{d^{2} y}{dx^{2} } +2\cfrac{dy}{dx} +y = 3e^{x}$

Solution 3. With an exponential in the nonhomogeneous term we use a trial solution of the form.

$y_{p}(x) =Ae^{x}$
${\cfrac{d^{2} }{dx^{2} } Ae^{x} +2\cfrac{d}{dx} Ae^{x} +Ae^{x} =3e^{x} }$
${A+2A+A=3}$
${A=3/4}$

Thus, the general solution is

$y=c_{1} e^{-x} +c_{2} xe^{-x} +\cfrac{3}{4} e^{x}$

Example 4. Find the general solution.

$\cfrac{d^2y}{dx^2} +2\cfrac{dy}{dx} +y= 2e^{-x}$

Solution 4. In this equation, there is a trick since the complimentary solution already contains a term of the form

$c_{1} e^{-x}$

The particular solution can’t share this form in order for it to be unique. As we have seen before when there is a conflict in constant coefficient equations with double root the predicted solution is multiplied by \textit{x} to the power of the number of conflicts. Let us try a new trial solution of the form

$y_p(x)=Ax^{2} e^{-x}$
${\cfrac{d^{2} }{dx^{2} } Ax^{2} e^{-x} +2\cfrac{d}{dx} Ax^{2} e^{-x} +Ax^{2} e^{-x} =2e^{x} }$
${(2A-2Ax+Ax^{2} )e^{-x} +(4Ax-2Ax^{2} )e^{-x} +Ax^{2} e^{-x} =2e^{-x} }$
${Ax^{2} -2Ax^{2} +Ax^{2} -2Ax+2Ax+2A=2}$
${A=1}$

Thus, the general solution is

$y_{c} =c_{1} e^{-x} +c_{2} xe^{-x} +x^{2} e^{-x}$

Example 5. Generalize the trial solutions for an arbitrary nonhomogeneous f(x) term.

Solution 5. For polynomials if f(x) is of the form

${f(x)=a_{1} +a_{2} x+...+a_{n} x^{n} }$
${y_{p} (x)=A_{1} +A_{2} x+...+A_{n} x^{n} }$

For sines and cosines if f(x) is of the form

${f(x)=a_{1} \cos (k_{1} x)+b_{1} \sin (k_{1} x)+a_{2} \cos (k_{2} x)+b_{2} (\sin (k_{2} x)+...}$
${y_{p} =\sum _{i=1}^{n}A_{i} \cos (k_{i} x)+B_{i} \sin (k_{i} x) }$

For exponentials if f(x) is of the form

${f(x)=a_{1} e^{-b_{1} x} +a_{2} e^{-b_{2} x} +...}$
${y_{p} (x)=\sum _{i=1}^{n}A_{i} e^{-b_{i} x}}$

Now, if f(x) is the product of the two of these classes of functions then write the most general form of the trial functions that spans the product. If

${f(x)=(a_{1} +a_{2} x+a_{3} x^{2} )e^{-kx} }$
${y_{p} (x)=(A+Bx+Cx^{2} )e^{-kx}}$

If f(x) is given as f(x)=9x cos (2x) then the trial solution should be

$y_p =(A+Bx)(C\cos (2x)+D\sin (2x))$

If f(x) is given as

$(2+9x^{2} )(\sin (9x))e^{-4x}$

Then the trial solution that is general enough to solve it should be

$y_{p} (x)=(A+Bx+Cx^{2} )(D\cos (9x)+E\sin (9x))e^{-4x}$

From these examples, we know how to find particular solutions for the limited cases presented. And after all, a trial solution is just a guess, plug it in and see if it solves the equation. I am not aware of a simple way of solving these equations for the particular solution directly so, good guessing!

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