- 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.

VIDEO SOLUTION


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
\(\displaystyle y=Ae^{-2x}+Be^{2x}-{3\over4} x^2+{1\over2} x-{5\over8}\)

VIDEO LESSON


\(f(x)\) 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
\(\displaystyle y=-{4\over17} \cos ⁡x-{1\over17}\sin ⁡x+Ae^{{1\over4}x}+Be^x\)

VIDEO LESSON



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 \(y'(0)=11\).
\(y(x)=e^{-x}(A \cos⁡3x+B \sin⁡3x)+2e^x\)
\(y(0)=5\)
\(5=e^0 (A \cos⁡0+B \sin⁡0 )+2e^0\)
\(3=A\)
\(y'(x)=-e^{-x} (A \cos⁡3x+B \sin⁡3x )+e^{-x} (-3A \sin⁡3x+3B \cos⁡3x )+2e^x\)
\(11=-e^0 (A \cos⁡0+B \sin⁡0 )+e^0 (-3A \sin⁡0+3B \cos⁡0 )+2e^0\)
\(11=-1(3)+1(3B)+2\)
\(12=3B\)
\(B=4\)
\(y=e^{-x} (3 \cos⁡3x+4 \sin⁡3x )+2e^x\)

VIDEO LESSON



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:
\(\displaystyle y={5\over2} e^{-x}-{1\over2} xe^{-x}\)

VIDEO LESSON