Nonhomogeneous linear equations with constant coefficients

Nonhomogeneous linear equations with constant coefficients
A second order nonhomogeneous equation with constant coefficients is written as
ay^ +by^ +cy=R(x) ?(1)
where a, b and c are constant a?0 ,R(x)?0 .
I- Undetermined Coefficients
We use the method of undetermined coefficients for finding particular solutions y_p when R(x) is one of the forms shown in table.
y_p Form of Notes R(x)
A_n x^n+A_(n-1) x^(n-1)+?+A_2 x^2+A_1 x+A_0 n?0 ax^n
Ae^rx r?m_1?m_2 ae^rx
Axe^rx r=m_1 or r=m_2
Ax^2 e^rx r=m_1=m_2
A sin?kx+B cos?kx k?? a sin?kx
or
a cos?kx
x(A sin?kx+B cos?kx ) k=?
We write the general solution of (1) as the sum of the homogeneous y_h and particular solutions y_p. y=y_h+y_p
Example 1: Solve y^ -y^ -12y=36x-15
Solution: First, we solve the homogeneous equation. The characteristic equation is
m^2-m-12=0 ?(m+3)(m-4)=0 ? m_1=-3 ,m_2=4
y_h=c_1 e^(-3x)+c_2 e^4x
Second,let y_p=Ax+B then y_p^ =A and y_p^ =0
Upon substitution into the differential equation, we have
0-A-12(Ax+B)=36x-15 ? ?(-12Ax) ?(?(-A-12B) )=?36x ?(?(-15) )
-12A=36 ?A=-3 and -A-12B=-15 ? B=1
So y_p=-3x+1
Then y=y_h+y_p=c_1 e^(-3x)+c_2 e^4x-3x+1
Example 2: Solve y^ -2y^ +5y=10x^2+7x
Solution: First, we solve the homogeneous equation. The characteristic equation is
m^2-2m+5=0
m=(-b??(b^2-4ac))/2a=(-(-2)??((2)^2-4×1×5))/(2×1)=1?2i
y_h=e^x (c_1 sin?2x+c_2 cos?2x )
Second,let y_p=Ax^2+Bx+C then y_p^ =2Ax+B and y_p^ =2A
Upon substitution into the differential equation, we have
2A-4Ax-2B+5Ax^2+5Bx+5C=10x^2+7x
?(?(5Ax^2 ))+?((-4A+5B)x)+(2A-2B+5C)=?(?(10x^2 ))+?7x+0
Equating coefficient x^2on the left side with the coefficient x^2 in the right side we get
5A=10 ? A=2
By the same way equate coefficient x : - 4A+5B=7 ? B=3
And 2A-2B+5C=0 ? C=2?5
So y_p=2x^2-3x+2?5
? y=y_h+y_p=e^x (c_1 sin?2x+c_2 cos?2x )+2x^2+3x+2?5
Example 3: Solve y^ -y^ -6y=8e^2x
Solution: First, we solve the homogeneous equation. The characteristic equation is
m^2-m-6=0 ?(m-3)(m+2)=0 ? m_1=3 ,m_2=-2
y_h=c_1 e^3x+c_2 e^(-2x)
Second, we find a particular solution of the nonhomogeneous equation. The form of the particular solution is chosen such that the exponential will cancel out of both sides of the differential equation. We choose
y_p=Ae^2x ? y_p^ =2Ae^2x ? y_p^ =4Ae^2x
4A-2A-6A=8 ? A=-2
So y_p=-2e^2x
Then y=y_h+y_p=c_1 e^3x+c_2 e^(-2x)-2e^2x
Example 4: Solve y^ -?5y?^ +6y=2e^3x
Solution: m^2-5m+6=0 ?(m-3)(m-2)=0 ? m_1=3 ,m_2=2
y_h=c_1 e^3x+c_2 e^2x
m_1=r=3 so y_p=Axe^3x
y_p^ =3Axe^3x+Ae^3x
y_p^ =9Axe^3x+3Ae^3x+3Ae^3x =9Axe^3x+6Ae^3x
9Axe^3x+6Ae^3x-15Axe^3x-5Ae^3x+6Axe^3x=2e^3x
A=2
So y_p=2e^3x
Then y=y_h+y_p=c_1 e^3x+c_2 e^2x+2e^3x