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.
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:
$$y = c_1 x^{-1} + c_2 x^{-2} \implies y = \frac{c_1}{x} + \frac{c_2}{x^2}$$
What is the order of the differential equation ?
What is the degree of the differential equation ?
A solution of the differential equation
\(\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \) is
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:
The general solution of the differential equation ydx - xdy = 0