Complimentary function of the differential equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = {e^{{x^2}}}\)
c 1x -1 + c 2x -2
The problem asks us to find the complimentary function of the given differential equation: $$ {x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = {e^{{x^2}}} $$
To determine the complimentary function (\(y_c\)), we need to solve the associated homogeneous differential equation. The homogeneous part of the given equation is obtained by setting the right-hand side to zero: $$ {x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = 0 $$
Differential Equation Type
This is a second-order linear homogeneous differential equation with variable coefficients. Specifically, it is a Cauchy-Euler (or Euler-Cauchy) equation, which has the general form: $$ ax^2\frac{d^2y}{dx^2} + bx\frac{dy}{dx} + cy = 0 $$ For Cauchy-Euler equations, we assume a solution of the form \(y = x^m\), where \(m\) is a constant to be determined.
Solution Derivation for Complimentary Function
Let's find the derivatives of our assumed solution \(y = x^m\):
- First derivative: $$ \frac{dy}{dx} = m x^{m-1} $$
- Second derivative: $$ \frac{d^2y}{dx^2} = m(m-1) x^{m-2} $$
Now, substitute these derivatives back into the homogeneous differential equation:
$$ {x^2}\left(m(m-1) x^{m-2}\right) + 4x\left(m x^{m-1}\right) + 2\left(x^m\right) = 0 $$Simplify each term:
$$ m(m-1) x^m + 4m x^m + 2x^m = 0 $$Factor out \(x^m\) (assuming \(x \ne 0\)):
$$ x^m \left(m(m-1) + 4m + 2\right) = 0 $$Since \(x^m \ne 0\), the expression in the parenthesis must be zero. This gives us the characteristic equation (also known as the auxiliary equation):
$$ m(m-1) + 4m + 2 = 0 $$ $$ m^2 - m + 4m + 2 = 0 $$ $$ m^2 + 3m + 2 = 0 $$Characteristic Equation Roots
Now, we need to solve this quadratic characteristic equation for \(m\). We can factor the quadratic equation:
$$ (m+1)(m+2) = 0 $$This yields two distinct real roots for \(m\):
- \(m_1 = -1\)
- \(m_2 = -2\)
Complimentary Function Formulation
Since we have two distinct real roots, \(m_1\) and \(m_2\), the general form of the complimentary function for a Cauchy-Euler equation is:
$$ y_c = c_1 x^{m_1} + c_2 x^{m_2} $$Substitute the values of \(m_1\) and \(m_2\) we found:
$$ y_c = c_1 x^{-1} + c_2 x^{-2} $$This represents the complimentary function of the given differential equation.
Conclusion
Comparing our derived complimentary function with the given options:
- Option 1: \(c_1 x^{-1} + c_2 x^{-2}\)
- Option 2: \(c_1 x + c_2 x^2\)
- Option 3: \(c_1 x^{-1} + c_2 x^2\)
- Option 4: \(c_1 x + c_2 x^{-2}\)
Our result matches Option 1.
Important Questions from Differential Equations
The Green's function for the differential equation \(\rm\frac{d^2x}{dt^2}\) + x = f(t), satisfying the initial conditions x(0) = \(\rm\frac{dx}{dt}\) (0) = 0, is
G(t, τ) = \(\begin{cases}0 & \text { for } \quad 0<\rm t<\tau \\ \sin (\rm t−\tau) & \text { for } \quad \rm t>\tau\end{cases}\)
The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
- \(\smallint \frac{{dx}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}}\) = ?
A differential equation is given as x (t + 2) + 3x (t + 1) + 2x (t) =0; x(0) = 0, x(1) = 1. The solution of this equation will be:
If y = \(\rm\left(\frac{1}{x}\right)^x \), then value of \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\) is:
If \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\) and f(f(x)) = x then
