Question

If \(y_{1}=x^{2}\) is a known solution of \[ x^{2}y^{\prime\prime}-3xy^{\prime}+4y=0, \] then the second independent solution \(y_{2}\) is

• A. \(\dfrac{1}{x}\)
• B. \(x^{2}\)
• C. \(\log x\)
• D. \(x^{2}\log x\)

Hint:

For a second-order linear homogeneous differential equation, \[ y''+P(x)y'+Q(x)y=0, \] if one solution \(y_{1}\) is known, immediately use \[ y_{2}=y_{1}\int \frac{e^{-\int P(x)\,dx}} {(y_{1})^{2}}\,dx. \] This formula is frequently used in university examinations and competitive exams for finding the second independent solution without solving the entire differential equation from scratch.

Answer 1

The Correct Option is D

Concept: A second-order linear homogeneous differential equation is of the form \[ y''+P(x)y'+Q(x)y=0. \] If one non-trivial solution \(y_{1}\) is already known, then the second linearly independent solution can be obtained by the Method of Reduction of Order. The formula is \[ y_{2}=y_{1}\int \frac{e^{-\int P(x)\,dx}} {\left(y_{1}\right)^{2}}\,dx. \] This method guarantees that \(y_{1}\) and \(y_{2}\) are linearly independent and together form the complete complementary solution.

Step 1: Convert the given differential equation into standard form.
The given equation is \[ x^{2}y''-3xy'+4y=0. \] Divide the entire equation by \(x^{2}\): \[ y''-\frac{3}{x}y' +\frac{4}{x^{2}}y=0. \] Comparing with \[ y''+P(x)y'+Q(x)y=0, \] we obtain \[ P(x)=-\frac{3}{x}. \] Also, \[ y_{1}=x^{2}. \]

Step 2: Evaluate the exponential term appearing in the reduction of order formula.
First, \[ \int P(x)\,dx = \int -\frac{3}{x}\,dx = -3\log x. \] Hence, \[ e^{-\int P(x)\,dx} = e^{-(-3\log x)} = e^{3\log x} = x^{3}. \]

Step 3: Substitute into the reduction of order formula.
Using \[ y_{2} = y_{1} \int \frac{e^{-\int Pdx}} {y_{1}^{2}}\,dx, \] we get \[ y_{2} = x^{2} \int \frac{x^{3}} {(x^{2})^{2}} \,dx. \] Since \[ (x^{2})^{2}=x^{4}, \] therefore, \[ y_{2} = x^{2} \int \frac{x^{3}}{x^{4}} \,dx = x^{2} \int \frac{1}{x}\,dx. \]

Step 4: Evaluate the integral.
We know that \[ \int\frac{1}{x}\,dx = \log x. \] Hence, \[ y_{2} = x^{2}\log x. \] Thus, the second linearly independent solution is \[ \boxed{y_{2}=x^{2}\log x.} \]

Step 5: Write the final answer.
Therefore, \[ \boxed{y_{2}=x^{2}\log x.} \] Hence, the correct option is \[ \boxed{\text{(D) }x^{2}\log x.} \]