Question

Given that y(x) is a solution of the deferential equation
\(x^2y''+xy'-4y-x^2\) 
on the interval (0,∞) such that the \(lim_{x→0^+}\) y(x) exists and y(1)=1. the value y'(1) is qual to____.(Rounded off to two decimal places).

Answer 1

Correct Answer: 2.24 - 2.26

1. Solve the homogeneous equation

\[x^{2}y'' + xy' - 4y = 0\]

Try (y = x^{m}):

\[m(m-1) + m - 4 = 0\]

\[m^{2} - 4 = 0\]

\[m = \pm 2\]

So,
\[y_h = C_{1}x^{2} + C_{2}x^{-2}\]

To keep (y(x)) finite as (x\to 0^{+}), we must set (C_{2} = 0).

Thus,

\[y_h = C_{1}x^{2}\]

2. Find a particular solution

Right-hand side is (x^{2}), which duplicates a homogeneous term ⇒ resonance.
So try:

\[y_p = A x^{2} \ln x\]

Compute derivatives:

\[y_p' = A(2x\ln x + x)\]

\[y_p'' = A(2\ln x + 3)\]

Substitute into the differential equation:

\[x^{2}(A(2\ln x + 3)) + x(A(2x\ln x + x)) - 4(Ax^{2}\ln x)\]

Simplify:

\[2Ax^{2}\ln x + 3Ax^{2} + 2Ax^{2}\ln x + Ax^{2} - 4Ax^{2}\ln x = x^{2}\]

The log terms cancel:

\[4Ax^{2} = x^{2}\]

\[A = \frac14\]

Thus,

\[y_p = \frac14 x^{2}\ln x\]

3. General solution

\[y = C_{1}x^{2} + \frac14 x^{2}\ln x\]

Apply condition (y(1)=1):

\[1 = C_{1}(1)^{2} + \frac14(1)^{2}\ln 1\]

\[C_{1} = 1\]

Final solution:

\[y = x^{2} + \frac14 x^{2}\ln x\]

4. Compute derivative

\[y' = 2x + \frac14(2x\ln x + x)\]

\[y' = 2x + \frac12 x\ln x + \frac14 x\]

\[y' = x\left( \frac94 + \frac12 \ln x \right)\]

Evaluate at (x=1):

\[y'(1) = 1\left( \frac94 + \frac12\ln 1 \right)\]

\[y'(1) = \frac94\]

\[y'(1) = 2.25\]

Final Answer

\[y'(1) = 2.25\]