An equation of the form: \[ x^n \frac{d^n y}{dx^n} + a_1 x^{n-1} \frac{d^{n-1} y}{dx^{n-1}} + a_2 x^{n-2} \frac{d^{n-2} y}{dx^{n-2}} + \dots + a_{n-1} x \frac{dy}{dx} + a_n y = Q(x) \quad \text{(1)} \] where \( a_1, a_2, \dots, a_n \) are constants and \( Q(x) \) is a function of \( x \), is called Cauchy’s Homogeneous Linear Equation of the \( n^{th} \) order.
Reduction to Linear Differential Equation with Constant Coefficients
Equation (1) can be reduced to a linear differential equation with constant coefficients by changing the independent variable. Let: \[ x = e^z \quad \text{(i.e., } z = \log x \text{)}. \] Then: \[ \frac{dz}{dx} = \frac{1}{x}. \]
Now, compute the derivatives of \( y \) with respect to \( x \) in terms of \( z \): \[ \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \frac{1}{x} \frac{dy}{dz} \quad \Rightarrow \quad x \frac{dy}{dx} = \frac{dy}{dz}. \] Let: \[ \frac{d}{dx} = D \quad \text{and} \quad \frac{d}{dz} = \theta. \] Then: \[ x D y = \theta y. \]
Compute the second derivative: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{1}{x} \frac{dy}{dz} \right). \] Using the product rule: \[ \frac{d^2 y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x} \frac{d}{dx} \left( \frac{dy}{dz} \right). \] Substitute \( \frac{d}{dx} = \frac{1}{x} \frac{d}{dz} \): \[ \frac{d^2 y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x^2} \frac{d^2 y}{dz^2}. \] Multiply through by \( x^2 \): \[ x^2 \frac{d^2 y}{dx^2} = \frac{d^2 y}{dz^2} - \frac{dy}{dz}. \] In operator notation: \[ x^2 D^2 y = \theta^2 y - \theta y = \theta (\theta - 1) y. \]
Similarly, for higher-order derivatives: \[ x^3 D^3 y = \theta (\theta - 1)(\theta - 2) y, \] and in general: \[ x^n D^n y = \theta (\theta - 1)(\theta - 2) \dots [\theta - (n - 1)] y. \]
Substituting these results into Equation (1), we obtain a linear differential equation with constant coefficients in terms of \( \theta \). This equation can then be solved using standard methods for linear differential equations with constant coefficients.
1. Solve the differential equation: \( x^2 \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} + 2y = e^x. \) by the method of variation of parameters.
Solution The given equation is: \[ \left( x^2 D^2 + 4x D + 2 \right) y = e^x \quad \text{(1)}, \] where \( D = \frac{d}{dx} \).
To solve this, we use the substitution \( x = e^z \) (i.e., \( z = \log x \)), so that: \[ x D y = \theta y \quad \text{and} \quad x^2 D^2 y = \theta (\theta - 1) y, \] where \( \theta = \frac{d}{dz} \).
Substituting into Equation (1), we get: \[ \left[ \theta (\theta - 1) + 4 \theta + 2 \right] y = e^{e^z}. \] Simplifying: \[ \left( \theta^2 + 3 \theta + 2 \right) y = e^{e^z}. \]
The auxiliary equation is: \[ m^2 + 3m + 2 = 0 \implies (m + 1)(m + 2) = 0 \implies m = -1, -2. \] Therefore, the complementary function (C.F.) is: \[ \text{C.F.} = c_1 e^{-z} + c_2 e^{-2z} = c_1 x^{-1} + c_2 x^{-2}. \]
To find the particular integral (P.I.), we solve: \[ \text{P.I.} = \frac{1}{\theta^2 + 3 \theta + 2} e^{e^z}. \] Using partial fractions: \[ \frac{1}{\theta^2 + 3 \theta + 2} = \frac{1}{(\theta + 1)(\theta + 2)} = \frac{1}{\theta + 1} - \frac{1}{\theta + 2}. \] Thus: \[ \text{P.I.} = \frac{1}{\theta + 1} e^{e^z} - \frac{1}{\theta + 2} e^{e^z}. \]
Simplify each term: \[ \frac{1}{\theta + 1} e^{e^z} = e^{-z} \frac{1}{\theta} \left( e^z e^{e^z} \right) = e^{-z} \int e^{e^z} e^z \, dz, \] and \[ \frac{1}{\theta + 2} e^{e^z} = e^{-2z} \frac{1}{\theta} \left( e^{2z} e^{e^z} \right) = e^{-2z} \int e^{e^z} e^{2z} \, dz. \]
Let \( t = e^z \), so \( dt = e^z \, dz \). Substituting: \[ \int e^{e^z} e^z \, dz = \int e^t \, dt = e^t = e^{e^z}, \] and \[ \int e^{e^z} e^{2z} \, dz = \int t e^t \, dt = t e^t - e^t = e^{e^z} (e^z - 1). \]
Substituting back: \[ \text{P.I.} = e^{-z} e^{e^z} - e^{-2z} e^{e^z} (e^z - 1). \] Simplifying: \[ \text{P.I.} = x^{-1} e^x - x^{-2} e^x (x - 1) = x^{-2} e^x. \]
The complete solution is: \[ y = \text{C.F.} + \text{P.I.} = c_1 x^{-1} + c_2 x^{-2} + x^{-2} e^x. \]
2. Solve the differential equation: \( x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} - 12y = x^3 \log x. \)
Solution The given equation is: \[ \left( x^2 D^2 + 2x D - 12 \right) y = x^3 \log x \quad \text{(1)}, \] where \( D = \frac{d}{dx} \).
To solve this, we use the substitution \( x = e^z \) (i.e., \( z = \log x \)), so that: \[ x D y = \theta y \quad \text{and} \quad x^2 D^2 y = \theta (\theta - 1) y, \] where \( \theta = \frac{d}{dz} \).
Substituting into Equation (1), we get: \[ \left[ \theta (\theta - 1) + 2 \theta - 12 \right] y = e^{3z} z. \] Simplifying: \[ \left( \theta^2 + \theta - 12 \right) y = e^{3z} z. \]
The auxiliary equation is: \[ m^2 + m - 12 = 0 \implies (m + 4)(m - 3) = 0 \implies m = 3, -4. \] Therefore, the complementary function (C.F.) is: \[ \text{C.F.} = c_1 e^{3z} + c_2 e^{-4z} = c_1 x^3 + c_2 x^{-4}. \]
To find the particular integral (P.I.), we solve: \[ \text{P.I.} = \frac{1}{\theta^2 + \theta - 12} e^{3z} z. \] Using the shift theorem: \[ \text{P.I.} = e^{3z} \frac{1}{(\theta + 3)^2 + (\theta + 3) - 12} z = e^{3z} \frac{1}{\theta^2 + 7 \theta} z. \] Simplify the operator: \[ \frac{1}{\theta^2 + 7 \theta} = \frac{1}{7 \theta} \left( 1 + \frac{\theta}{7} \right)^{-1}. \] Expand using the binomial theorem: \[ \left( 1 + \frac{\theta}{7} \right)^{-1} = 1 - \frac{\theta}{7} + \left( \frac{\theta}{7} \right)^2 - \dots. \] Retain terms up to first order: \[ \text{P.I.} = \frac{e^{3z}}{7} \frac{1}{\theta} \left( 1 - \frac{\theta}{7} \right) z. \] Apply the operator: \[ \text{P.I.} = \frac{e^{3z}}{7} \frac{1}{\theta} \left( z - \frac{1}{7} \right) = \frac{e^{3z}}{7} \left( \frac{z^2}{2} - \frac{z}{7} \right). \] Substitute \( z = \log x \): \[ \text{P.I.} = \frac{x^3}{7} \left( \frac{(\log x)^2}{2} - \frac{\log x}{7} \right). \]
The complete solution is: \[ y = \text{C.F.} + \text{P.I.} = c_1 x^3 + c_2 x^{-4} + \frac{x^3}{7} \left( \frac{(\log x)^2}{2} - \frac{\log x}{7} \right). \]
Model Problems
Solve the following differential equations:
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Solve: \(x^2 \frac{d^2y}{dx^2} - 3x \frac{dy}{dx} + 4y = 1 + x^2\)
Answer: \(y = (c_1 + c_2 \log x) x^2 + \frac{1}{4} \left[ 1 + 2 (\log x)^2 x^2 \right]\) -
Solve: \(x^3 \frac{d^3y}{dx^3} + 3x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = x + \log x\)
Answer: \(y = x^{1/2} \left[ c_1 \cos \left( \frac{\sqrt{3}}{2} \log x \right) + c_2 \sin \left( \frac{\sqrt{3}}{2} \log x \right) \right] + c_3 x^{-1} + \frac{x}{2} + \log x\) -
Solve: \(x^2 \frac{d^2y}{dx^2} - 3x \frac{dy}{dx} + 5y = x^2 \sin (\log x)\)
Answer: \(y = x^2 \left[ c_1 \cos (\log x) + c_2 \sin (\log x) \right] - \frac{x^2 \log x \cos (\log x)}{2}\) -
Solve: \(x^2 \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} - 12y = x^3 \log x\)
Answer: \(y = c_1 x^3 + c_2 x^{-4} + \frac{x^3}{98} \left[ 7 (\log x)^2 - 2 \log x \right]\)