Rules for Finding C.F.
Consider the homogeneous equation \( f(D) y = 0 \). The auxiliary equation (A.E.) is:
\[ f(m) = m^n + a_1 m^{n-1} + a_2 m^{n-2} + \dots + a_{n-1} m + a_n = 0 \]
The roots of the auxiliary equation determine the form of the complementary function (C.F.):
1. Real and Distinct Roots
If the roots \( m_1, m_2, \dots, m_n \) are real and distinct, then:
\[ \text{C.F.} = c_1 e^{m_1 x} + c_2 e^{m_2 x} + \dots + c_n e^{m_n x} \]
2. Real and Repeated Roots
If some roots are repeated, for example, \( m_1 \) is repeated twice, then:
\[ \text{C.F.} = (c_1 + c_2 x) e^{m_1 x} + c_3 e^{m_3 x} + \dots + c_n e^{m_n x} \]
If \( m_1 \) is repeated three times, then:
\[ \text{C.F.} = (c_1 + c_2 x + c_3 x^2) e^{m_1 x} + c_4 e^{m_4 x} + \dots + c_n e^{m_n x} \]
3. Complex Roots
If the roots are complex, say \( \alpha \pm i\beta \), then:
\[ \text{C.F.} = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x) + c_3 e^{m_3 x} + \dots + c_n e^{m_n x} \]
If the complex roots are repeated, for example, \( \alpha \pm i\beta \) is repeated twice, then:
\[ \text{C.F.} = e^{\alpha x} \left[ (c_1 + c_2 x) \cos \beta x + (c_3 + c_4 x) \sin \beta x \right] + c_5 e^{m_5 x} + \dots + c_n e^{m_n x} \]
Summary
- The general solution of a linear differential equation with constant coefficients is the sum of the complementary function (C.F.) and the particular integral (P.I.).
- The C.F. is found by solving the auxiliary equation and using the roots to construct the solution.
- The P.I. is a specific solution to the non-homogeneous equation and can be found using methods such as undetermined coefficients or variation of parameters.
This approach provides a systematic way to solve higher-order linear differential equations with constant coefficients.
1. Solve the differential equation: \( y'' + 2y' = 0 \)
Solution:
The given equation can be written as:
\[ \left( D^2 + 2D \right) y = 0 \quad \text{where} \quad D = \frac{d}{dx} \]
Here, \( f(D) = D^2 + 2D \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + 2m = 0 \Rightarrow m(m + 2) = 0 \]
This gives the roots:
\[ m = 0, -2 \quad (\text{Real and Distinct Roots}) \]
The complete solution is:
\[ y = c_1 e^{0x} + c_2 e^{-2x} = c_1 + c_2 e^{-2x} \]
2. Solve the differential equation: \( \frac{d^2 y}{dx^2} - 4 \frac{dy}{dx} + y = 0 \)
Solution:
The given equation can be written as:
\[ \left( D^2 - 4D + 1 \right) y = 0 \quad \text{where} \quad D = \frac{d}{dx} \]
Here, \( f(D) = D^2 - 4D + 1 \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 - 4m + 1 = 0 \]
Solving the quadratic equation, we get:
\[ m = 2 \pm \sqrt{3} \quad (\text{Real and Distinct Roots}) \]
The complete solution is:
\[ y = c_1 e^{(2 - \sqrt{3})x} + c_2 e^{(2 + \sqrt{3})x} \]
3. Solve the differential equation: \( (4D^2 + 4D + 1)y = 0 \)
Solution:
The given equation is in the form \( f(D)y = 0 \), where \( D = \frac{d}{dx} \).
Here, \( f(D) = 4D^2 + 4D + 1 \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow 4m^2 + 4m + 1 = 0 \Rightarrow (2m + 1)^2 = 0 \]
This gives the roots:
\[ m = -\frac{1}{2}, -\frac{1}{2} \quad (\text{Repeated Real Roots}) \]
The complete solution is:
\[ y = (c_1 + c_2 x) e^{-\frac{x}{2}} \]
4. Solve the differential equation: \( \frac{d^3 y}{dx^3} - 3 \frac{d^2 y}{dx^2} + 4y = 0 \)
Solution:
The given equation can be written as:
\[ (D^3 - 3D^2 + 4)y = 0 \quad \text{where} \quad D = \frac{d}{dx} \]
Here, \( f(D) = D^3 - 3D^2 + 4 \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^3 - 3m^2 + 4 = 0 \Rightarrow (m + 1)(m^2 - 4m + 4) = 0 \]
This gives the roots:
\[ m = -1, 2, 2 \quad (\text{Repeated Real Roots}) \]
The complete solution is:
\[ y = (c_1 + c_2 x) e^{2x} + c_3 e^{-x} \]
5. Solve the differential equation: \( y'' + y' + y = 0 \)
Solution:
The given equation can be written as:
\[ (D^2 + D + 1)y = 0 \quad \text{where} \quad D = \frac{d}{dx} \]
Here, \( f(D) = D^2 + D + 1 \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + m + 1 = 0 \]
This gives the roots:
\[ m = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \quad (\text{Complex Conjugate Roots}) \]
The complete solution is:
\[ y = e^{-\frac{x}{2}} \left( c_1 \cos \frac{\sqrt{3}}{2} x + c_2 \sin \frac{\sqrt{3}}{2} x \right) \]
6. Solve the differential equation: \( (D^2 + 1)^2 (D - 1) = 0 \)
Solution:
The given equation is in the form \( f(D)y = 0 \), where \( D = \frac{d}{dx} \).
Here, \( f(D) = (D^2 + 1)^2 (D - 1) \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow (m^2 + 1)^2 (m - 1) = 0 \]
This gives the roots:
\[ m = \pm i, \pm i, 1 \quad (\text{Repeated Complex and Real Roots}) \]
The complete solution is:
\[ y = (c_1 + c_2 x) \cos x + (c_3 + c_4 x) \sin x + c_5 e^x \]
7. Solve the differential equation: \( \frac{d^2 x}{dt^2} - 4 \frac{dx}{dt} + 13x = 0 \quad \text{with} \quad x(0) = 0, \quad \frac{dx}{dt}(0) = 2 \)
Solution:
The given equation can be written as:
\[ (D^2 - 4D + 13)x = 0 \quad \text{where} \quad D = \frac{d}{dt} \]
Here, \( f(D) = D^2 - 4D + 13 \).
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 - 4m + 13 = 0 \]
This gives the roots:
\[ m = 2 \pm 3i \quad (\text{Complex Conjugate Roots}) \]
The complete solution is:
\[ x = e^{2t} (c_1 \cos 3t + c_2 \sin 3t) \]
Differentiating with respect to \( t \):
\[ \frac{dx}{dt} = 2e^{2t} (c_1 \cos 3t + c_2 \sin 3t) + e^{2t} (-3c_1 \sin 3t + 3c_2 \cos 3t) \]
Applying the initial conditions:
\[ x(0) = 0 \Rightarrow c_1 = 0 \]
\[ \frac{dx}{dt}(0) = 2 \Rightarrow c_2 = \frac{2}{3} \]
The particular solution is:
\[ x = \frac{2}{3} e^{2t} \sin 3t \]
Model Problems
Solve the following differential equations:
-
Solve: \(y'' - 3y' + 2y = 0\)
Answer: \(y = c_1 e^x + c_2 e^{2x}\) -
Solve: \(\frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - y = 0\)
Answer: \(y = (c_1 + c_2 x + c_3 x^2) e^x\) -
Solve: \(\frac{d^4y}{dx^4} + 8\frac{d^2y}{dx^2} + 16y = 0\)
Answer: \(y = (c_1 + c_2 x) \cos 2x + (c_3 + c_4 x) \sin 2x\) -
Solve: \((D^2 + D + 1)^2 y = 0\)
Answer: \(y = e^{-\frac{x}{2}} \left[ (c_1 + c_2 x) \cos \left( \frac{\sqrt{3}}{2} x \right) + (c_3 + c_4 x) \sin \left( \frac{\sqrt{3}}{2} x \right) \right]\) -
Solve: \((D^2 + 5D + 3) y = 0\)
Answer: \(y = c_1 e^{\left( \frac{-5 + \sqrt{13}}{2} \right) x} + c_2 e^{\left( \frac{-5 - \sqrt{13}}{2} \right) x}\) -
Solve: \(\frac{d^4x}{dt^4} + 4x = 0\)
Answer: \(x = e^{-t} (c_1 \cos t + c_2 \sin t) + e^t (c_3 \cos t + c_4 \sin t)\) -
Solve: \(\frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 9x = 0\)
Answer: \(x = e^{-3t} (c_1 + c_2 t)\) -
Solve: \(y'' + 6y' + 9y = 0\) with initial conditions \(y(0) = -4\) and \(y'(0) = 14\)
Answer: \(y = (2x - 4) e^{-3x}\)