Question

If y = sin (ln x), then which one of the following is correct?
1. \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\)
2. \(\frac{{{d^2}y}}{{d{x^2}}} = 0\)
3. \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + y = 0\)
4. \({x^2}\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + y = 0\)
5. None of these

Answer 1

Correct Answer - Option 3 : \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + y = 0\)

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( x \right)\;{\rm{g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}{{\rm{f}}^{{\rm{'}}\left( {\rm{x}} \right)}}{\rm{g}}\left( {\rm{x}} \right) + {\rm{f}}\left( x \right){\rm{\;}}{{\rm{g}}^{{\rm{'}}\left( {\rm{x}} \right)}}\)
• Quotient Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {\frac{{{\rm{f}}\left( x \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right] = \frac{{{\rm{g}}\left( {\rm{x}} \right){\rm{\;f'}}\left( {\rm{x}} \right) - {\rm{f}}\left( x \right){\rm{\;g'}}\left( {\rm{x}} \right)}}{{{{\left[ {{\rm{g}}\left( {\rm{x}} \right)} \right]}^2}}}\)

Calculation:

Given:

y = sin (ln x)    ….(1)

Now differentiating both sides, we get

\(\Rightarrow {\rm{\;}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{\;}}\cos \left( {\ln {\rm{x}}} \right){\rm{\;}} \times {\rm{\;}}\frac{1}{{\rm{x}}}\)

\(\Rightarrow {\rm{\;x}}\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{\;}}\cos \left( {\ln {\rm{x}}} \right){\rm{\;}}\)

Again differentiating both sides, we get

\(\Rightarrow \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) + {\rm{x\;}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - \sin \left( {\ln {\rm{x}}} \right) \times {\rm{\;}}\frac{1}{{\rm{x}}}\)

From equation 1^st,

\(\Rightarrow \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) + {\rm{x\;}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {\rm{y}} \times {\rm{\;}}\frac{1}{{\rm{x}}}\)

\(\Rightarrow {{\rm{x}}^2}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0\)

Hence Option 3 is correct.