The solution of differential equation \(x^2 \frac {d^2y}{dx^2} + 4x\frac{dy}{dx}+2y=0\) will be ________, where c1 and c2 are constants.
Mathematical Step-by-Step Derivation
This expression is a standard second-order homogeneous Cauchy-Euler differential equation, which takes the form:
$$x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} + 2y = 0$$
To solve a Cauchy-Euler equation, we assume a solution of the form $y = x^m$. Finding its derivatives yields:
$\frac{dy}{dx} = m x^{m-1} \implies x \frac{dy}{dx} = m x^m$
$\frac{d^2y}{dx^2} = m(m-1) x^{m-2} \implies x^2 \frac{d^2y}{dx^2} = m(m-1) x^m$
Substituting these values back into the primary differential equation gives:
$$[m(m-1) + 4(m) + 2]\, x^m = 0$$
Since $x^m \neq 0$, we can solve for the auxiliary characteristic equation:
$$m^2 - m + 4m + 2 = 0$$
$$m^2 + 3m + 2 = 0$$
Factoring the quadratic polynomial:
$$(m+1)(m+2) = 0 \implies m_1 = -1, \;\; m_2 = -2$$
Substituting the roots back into our assumed format yields the general solution:
$$y = c_1 x^{-1} + c_2 x^{-2} \implies y = \frac{c_1}{x} + \frac{c_2}{x^2}$$
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
