## - Non - Homogeneous Differential Equations

Non – homogeneous first – order and second – order differential equation are of the form

$$\displaystyle a {dy\over dx}+by=f(x)$$
$$ay''+by'+cy=f(x), f(x)≠0$$
The particular integral is any solution of these types of differential equations.

$$f(x)$$ is a Polynomial
LESSON 1
Solve the equation
$$y''-5y'+6y=x$$
SOLUTION
• Determine the complementary function.
Homogeneous equation
$$y''-5y'+6y=0$$
Auxiliary equation
$$u^2-5u+6=0$$
$$u=2, 3$$
Complementary function is $$y=Ae^{2x}+Be^{3x}$$
Once we have the complementary function we now need to determine the PARTICULAR INTEGRAL
Since $$f(x)=x$$ is a linear function it seems sensible to think that the PARTICULAR INTEGRAL is also a linear function i.e. of the form $$y=Cx+D$$
$$\displaystyle y=Cx+D → {dy\over dx}=C → {d^2 y\over dx^2}=0$$
• Substitute into original equation and determine the values of $$C$$ and $$D$$.
$$y''-5y'+6y=x$$
$$⟹0-5C+6(Cx+D)=x$$
$$⟹6Cx-5C+6D=x$$
• Equate coefficients of $$x$$
$$⟹6C=1$$
$$\displaystyle ⟹C={1\over6}$$
• Equate constants
$$⟹-5C+6D=0$$
$$\displaystyle ⟹D={5\over36}$$
Therefore, the particular integral is
$$\displaystyle y={1\over6} x+{5\over36}$$
Thus the complete solution is
$$\displaystyle y=Ae^{2x}+Be^{3x}+{1\over6} x+{5\over36}$$
which is the combination of the complementary function and the particular integral.

LESSON 2
Solve the differential equation
$$y''-4y=3x^2-2x+1$$
SOLUTION
Homogeneous equation
$$y''-4y=0$$
Auxiliary equation
$$u^2-4=0$$
$$u=±2$$
Complementary function is
$$y=Ae^{-2x}+Be^{2x}$$
PARTICULAR INTEGRAL:
Since $$f(x)$$ is a quadratic expression we assume that the particular integral is of the same form.
$$y=Cx^2+Dx+E$$
$$⟹y'=2Cx+D$$
$$⟹y''=2C$$
• Substitute into original equation
$$2C-4(Cx^2+Dx+E)=3x^2-2x+1$$
$$-4Cx^2-4Dx+2C-4E=3x^2-2x+1$$
• Equate coefficients of $$x^2$$
$$-4C=3$$
$$\displaystyle ⟹C=-{3\over4}$$
Equate coefficients of $$x$$
$$-4D=-2$$
$$\displaystyle ⟹D={1\over2}$$
• Equate constants
$$2C-4E=1$$
$$\displaystyle ⟹E=-{5\over8}$$
Particular Integral is
$$\displaystyle y=-{3\over4} x^2+{1\over2} x-{5\over8}$$
General solution is
$$\displaystyle y=Ae^{-2x}+Be^{2x}-{3\over4} x^2+{1\over2} x-{5\over8}$$

$$f(x)$$ is a Trigonometric Function
LESSON 3
Find the complete solution of the differential equation
$$4y''-5y'+y=\cos ⁡x-\sin ⁡x$$
SOLUTION
Homogeneous Equation
$$4y''-5y'+y=0$$
Auxiliary equation
$$4u^2-5u+1=0$$
$$\displaystyle u={1\over4}, 1$$
Complementary Function is
$$\displaystyle y=Ae^{{1\over4}x}+Be^x$$
PARTICULAR INTEGRAL
$$f(x)$$ is a trigonometric function in sine and cosine so we assume that the particular integral is of the form $$m \sin ⁡qx+n \cos ⁡qx$$.
$$y=C \cos⁡ x+D \sin ⁡x$$
$$⟹y'=-C \sin ⁡x+D \cos⁡ x$$
$$⟹y''=-C \cos ⁡x-D \sin ⁡x$$
• Substitute into original equation
$$4y''-5y'+y=\cos ⁡x-\sin ⁡x$$
$$4(-C \cos ⁡x-D \sin ⁡x )-5(-C \sin ⁡x+D \cos ⁡x)+C \cos⁡ x+D \sin ⁡x=\cos ⁡x-\sin ⁡x$$
• Group terms with sine and cosine individually.
$$(-4C-5D+C) \cos ⁡x+(-4D+5C+D) \sin ⁡x=\cos ⁡x-\sin ⁡x$$
• Equate coefficients of $$\cos ⁡x$$ and $$\sin ⁡x$$ to get a pair of simultaneous equations.
$$-3C-5D=1$$
$$-3D+5C=-1$$
$$\displaystyle C=-{4\over17}$$
$$\displaystyle D=-{1\over17}$$
Particular Integral is
$$\displaystyle y=-{4\over17} \cos ⁡x-{1\over17}\sin ⁡x$$
The complete solution is
$$\displaystyle y=-{4\over17} \cos ⁡x-{1\over17}\sin ⁡x+Ae^{{1\over4}x}+Be^x$$

LESSON 4
(i) Solve the differential equation $$y''+4y'+3y=65 \sin ⁡2x$$
(ii) Hence, find the particular solution for which
$$y(0)=3, y' (0)=7$$
SOLUTION
Homogenous equation
$$y''+4y'+3y=0$$
Auxiliary equation
$$u^2+4u+3=0$$
$$u=-1, -3$$
Complementary function is
$$y=Ae^{-x}+Be^{-3x}$$
PARTICULAR INTEGRAL
Though $$f(x)$$ is only a function of sine we still use $$m \sin⁡ qx+n \cos ⁡qx$$ as the particular integral
$$y=C \sin ⁡2x+D \cos⁡ 2x$$
$$y'=2C \cos ⁡2x-2D \sin ⁡2x$$
$$y''=-4C \sin⁡ 2x-4D \cos⁡ 2x$$

• Substitute into original equation
$$y''+4y'+3=65 \sin ⁡2x$$
$$-4C \sin⁡ 2x-4D \cos ⁡2x+4(2C \cos ⁡2x-2D \sin ⁡2x)+3(C \sin ⁡2x+D \cos ⁡2x)=65 \sin ⁡2x$$
• Group terms in sine and cosine individually
$$(-4C-8D+3C) \sin ⁡2x+(-4D+8C+3D) \cos⁡ 2x=65 \sin⁡ 2x$$
$$(-C-8D) \sin ⁡2x+(8C-D) \cos⁡ 2x=65 \sin ⁡2x$$
• Equate coefficients of $$\sin ⁡2x$$ and $$\cos ⁡2x$$ to get a pair of simultaneous equations.
$$-C-8D=65$$
$$8C-D=0$$
$$C=-1$$
$$D=-8$$
Particular Integral is
$$y=-\sin ⁡2x-8 \cos⁡ 2x$$
General solution is
$$y=Ae^{-x}+Be^{-3x}-\sin⁡ 2x-8 \cos ⁡2x$$
PARTICULAR SOLUTION
• Determine the values of $$A$$ and $$B$$ using the initial boundary conditions $$y(0)=3$$ and $$y'(0)=7$$
$$y(x)=Ae^{-x}+Be^{-3x}-\sin⁡ 2x-8 \cos ⁡2x$$
• Using $$y(0)=3$$
$$⟹3=Ae^{-0}+Be^{-3(0)}-\sin⁡(2(0))-8 \cos⁡(2(0))$$
$$⟹3=A+B-8$$
$$⟹A+B=11\:\:\:\:\:(1)$$

$$y'(x)=-Ae^{-x}-3Be^{-3x}-2 \cos ⁡2x+16 \sin ⁡2x$$
Using $$y'(0)=7$$
$$⟹7=-Ae^0-3Be^0-2 \cos⁡(0)+16 \sin⁡(0)$$
$$⟹7=-A-3B-2$$
$$⟹A+3B=-9\:\:\:\:\:(2)$$
• Solving (1) and (2) simultaneously
$$A+B=11$$
$$A+3B=-9$$
$$A=21$$
$$B=-10$$
Particular Solution is
$$y=21e^{-x}-10e^{-3x}-\sin⁡ 2x-8 \cos⁡ 2x$$

VIDEO LESSON Part (i) Part (ii)

$$f(x)$$ is an Exponential Function
LESSON 5
Solve the differential equation
$$y''+2y'+10y=26e^x$$
given that $$y(0)=5$$ and $$y'(0)=11$$. Give your answer in the form $$y=f(x)$$.
SOLUTION
Homogeneous equation
$$y''+2y'+10y=0$$
Auxiliary equation:
$$u^2+2u+10=0$$
$$\displaystyle u={-2\pm \sqrt{2^2-4(1)(10)}\over2(1)}$$
$$u=-1±3i$$
Complementary function:
$$y=e^{-x}(A \cos ⁡3x+B\sin ⁡3x)$$

PARTICULAR INTEGRAL
Since $$f(x)$$ is an exponential function we use $$Ce^x$$ as the particular integral.
$$y=Ce^x$$
$$⟹y'=Ce^x$$
$$⟹y''=Ce^x$$
• Substitute into original equation
$$y''+2y'+10y=e^x$$
$$Ce^x+2Ce^x+10Ce^x=26e^x$$
$$13Ce^x=26e^x$$
$$C=2$$
Particular Integral is
$$y=2e^x$$
GENERAL SOLUTION:
$$y=e^{-x}(A \cos ⁡3x+B \sin ⁡3x)+2e^x$$
PARTICULAR SOLUTION
Determine the Particular solution using the initial boundary conditions $$y(0)=5$$ and $$y'(0)=11$$.
$$y(x)=e^{-x}(A \cos⁡3x+B \sin⁡3x)+2e^x$$
$$y(0)=5$$
$$5=e^0 (A \cos⁡0+B \sin⁡0 )+2e^0$$
$$3=A$$
$$y'(x)=-e^{-x} (A \cos⁡3x+B \sin⁡3x )+e^{-x} (-3A \sin⁡3x+3B \cos⁡3x )+2e^x$$
$$11=-e^0 (A \cos⁡0+B \sin⁡0 )+e^0 (-3A \sin⁡0+3B \cos⁡0 )+2e^0$$
$$11=-1(3)+1(3B)+2$$
$$12=3B$$
$$B=4$$
$$y=e^{-x} (3 \cos⁡3x+4 \sin⁡3x )+2e^x$$

LESSON 6
Solve the differential equation
$$y''-2y'-3y=2e^{-x}$$
given that $$y→0$$ as $$x→\infty$$ and that $$y'(0)=-3$$.
SOLUTION
Homogeneous equation
$$y''-2y'-3y=0$$
Auxiliary equation:
$$u^2-2u-3=0$$
$$u=-1, 3$$
COMPLEMENTARY FUNCTION
$$y=Ae^{-x}+Be^{3x}$$
PARTICULAR INTEGRAL
In this case, the particular integral would be of the form $$y=Ae^{-x}$$ but since this is already included in the complementary function we have to use $$y=Cxe^{-x}$$.
$$y=Cxe^{-x}$$
$$⟹y'=Ce^{-x}-Cxe^{-x}$$
$$⟹y''=-Ce^{-x}-Ce^{-x}+Cxe^{-x}$$
$$=-2Ce^{-x}+Cxe^{-x}$$
• Substitute into original equation
$$y''-2y'-3y=2e^{-x}$$
$$(-2Ce^{-x}+Cxe^{-x})-2(Ce^{-x}-Cxe^{-x})-3Cxe^{-x}=2e^{-x}$$
$$-4Ce^{-x}=2e^{-x}$$
$$-4C=2$$
$$\displaystyle C=-{1\over2}$$
PARTICULAR INTEGRAL
$$\displaystyle y=-{1\over2} xe^{-x}$$
GENERAL SOLUTION
$$\displaystyle y=Ae^{-x}+Be^{3x}-{1\over2} xe^{-x}$$
PARTICULAR SOLUTION
$$\displaystyle y(x)=Ae^{-x}+Be^{3x}-{1\over2} xe^{-x}$$
$$y→0$$ as $$x→\infty$$
As $$x→\infty, e^{-x}→0$$
Therefore, we have
$$y=Be^{3x}$$
As $$y→0, B=0$$

Since $$B=0$$
$$\displaystyle y(x)=Ae^{-x}-{1\over2} xe^{-x}$$
$$\displaystyle y'(x)=-Ae^{-x}-{1\over2} e^{-x}+{1\over2} xe^{-x}$$
$$y'(0)=-3$$
$$\displaystyle -3=-Ae^0-{1\over2} e^0+{1\over2} (0) e^0$$
$$\displaystyle A={5\over2}$$
PARTICULAR SOLUTION:
$$\displaystyle y={5\over2} e^{-x}-{1\over2} xe^{-x}$$