Question

The general solution of the differential equation \( x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 4 \) is

• A. \( y = c_2 x^2 + c_1 x + 2 \)
• B. \( y = c_2 x + \frac{c_1}{x^2} + 2 \)
• C. \( y = c_1 x^2 + c_2 x + 4 \)
• D. \( y = c_2 x + \frac{c_1}{x^2} + 4 \)

Hint:

Cauchy-Euler Equation. For \( ax^2 y'' + bxy' + cy = g(x) \), solve the homogeneous part using \(y=x^m\) to get the auxiliary equation in \(m\). Find the complementary function \(y_c\). Find a particular integral \(y_p\) using methods like undetermined coefficients or variation of parameters. General solution \(y = y_c + y_p\). For constant RHS, try constant \(y_p\).

Answer 1

The Correct Option is A

This is a Cauchy-Euler equation: \( x^2 y'' - 2xy' + 2y = 4 \)
First, find the complementary function (\(y_c\)) by solving the homogeneous equation: \( x^2 y'' - 2xy' + 2y = 0 \)
Assume a solution \(y = x^m\)
Then \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\)
Substitute into the homogeneous equation: $$ x^2 [m(m-1)x^{m-2}] - 2x [mx^{m-1}] + 2[x^m] = 0 $$ $$ m(m-1)x^m - 2mx^m + 2x^m = 0 $$ $$ [m^2 - m - 2m + 2] x^m = 0 $$ $$ [m^2 - 3m + 2] = 0 $$ This is the auxiliary equation
Factorize it: $$ (m-1)(m-2) = 0 $$ The roots are \(m_1 = 1\) and \(m_2 = 2\)
Since the roots are real and distinct, the complementary function is: $$ y_c = c_1 x^{m_1} + c_2 x^{m_2} = c_1 x + c_2 x^2 $$ Next, find a particular integral (\(y_p\)) for the non-homogeneous equation \( x^2 y'' - 2xy' + 2y = 4 \)
Since the right-hand side is a constant (4), let's try a constant particular solution \(y_p = A\)
Then \(y_p' = 0\) and \(y_p'' = 0\)
Substitute into the full equation: $$ x^2(0) - 2x(0) + 2(A) = 4 $$ $$ 2A = 4 \implies A = 2 $$ So, the particular integral is \(y_p = 2\)
The general solution is \(y = y_c + y_p\): $$ y = c_1 x + c_2 x^2 + 2 $$ This matches the form of option (1)