pls help me with this three questions thank you. I will upvote for it

Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/4Given:The given statement is \( \mathrm{\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} \) can be expressed as \( \mathrm{{D}{y}} \) whereas \( \mathrm{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}} \) can be expressed as \( \mathrm{{D}{y}^{{2}}} \).Objective:To establish if the assertion is accurate or not.Explanation:Please refer to solution in this step.Step2/4\( \mathrm{{1}{)}} \) Consider the statement, \( \mathrm{\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} \) can be expressed as \( \mathrm{{D}{y}} \) whereas \( \mathrm{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}} \) can be expressed as \( \mathrm{{D}{y}^{{2}}} \).Since differential operator is \( \mathrm{{D}=\frac{{d}}{{\left.{d}{x}\right.}}} \).Hence \( \mathrm{\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} \) can be expressed as \( \mathrm{{D}{y}} \).But \( \mathrm{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}} \) cannot be expressed as \( \mathrm{{D}{y}^{{2}}} \).It can be written as \( \mathrm{{D}^{{2}}{y}} \). Here \( \mathrm{{D}{y}^{{2}}} \) gives \( \mathrm{\frac{{{\left.{d}{y}\right.}^{{2}}}}{{{\left.{d}{x}\right.}}}} \) hence it is not same as \( \mathrm{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}} \).Therefore the statement given is false.ExplanationEvery group of derivatives used on a function is known as a differential operator.Explanation:Please refer to solution in this step.Step3/4\( \mathrm{{2}{)}} \) Consider the differential equation \( \mathrm{{y}{''}-{y}'-{12}{y}={0}} \).To write it by using differential operator.Here assume \( \mathrm{{D}=\frac{{d}}{{\left.{d}{x}\right.}}} \) and \( \mathrm{{D}^{{2}}=\frac{{d}^{{2}}}{{\left.{d}{x}\right.}^{{2}}}} \).Consider the differential equation as \( \mathrm{{\left({f{{\left({D}\right)}}}\right)}{y}={0}} \) with \( \mathrm{{D}^{{2}}-{ ... See the full answer