Home / Mathematics /     pls help me with this three questions thank you. I will upvote for it \( \frac{d y}{d x} \) can be expressed as \( D y \) whereas \( \frac{d^{2} y}{d x^{2}} \) can be expressed as \( D y^{2} \) True False QUESTION 2 \( y^{\prime \prime}-y^{\prime}-12 y=0 \) can be written using differential operator. Which of the following is incorrectly written? \[ \begin{array}{l} D^{2} y-D y-12=0 \\ D^{2}-D-12=0 \\ \left(D^{2}-D-12\right) y=0 \end{array} \] none of the above A linear operator \( L \) can be used as short hand notation for an n-order linear differential equation. Given \( 4 y^{\prime \prime}-4 y^{\prime}+y=0 \). Which is incorrectly expressed in terms of the linear operator? \[ \begin{array}{l} L[y]=0 \\ L(y)=0 \\ L y=0 \\ L=4 D^{2}-4 D+1=0 \end{array} \]

Question     pls help me with this three questions thank you. I will upvote for it \( \frac{d y}{d x} \) can be expressed as \( D y \) whereas \( \frac{d^{2} y}{d x^{2}} \) can be expressed as \( D y^{2} \) True False QUESTION 2 \( y^{\prime \prime}-y^{\prime}-12 y=0 \) can be written using differential operator. Which of the following is incorrectly written? \[ \begin{array}{l} D^{2} y-D y-12=0 \\ D^{2}-D-12=0 \\ \left(D^{2}-D-12\right) y=0 \end{array} \] none of the above A linear operator \( L \) can be used as short hand notation for an n-order linear differential equation. Given \( 4 y^{\prime \prime}-4 y^{\prime}+y=0 \). Which is incorrectly expressed in terms of the linear operator? \[ \begin{array}{l} L[y]=0 \\ L(y)=0 \\ L y=0 \\ L=4 D^{2}-4 D+1=0 \end{array} \]

FMFJZD The Asker · Advanced Mathematics

 

 

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

Transcribed Image Text: \( \frac{d y}{d x} \) can be expressed as \( D y \) whereas \( \frac{d^{2} y}{d x^{2}} \) can be expressed as \( D y^{2} \) True False QUESTION 2 \( y^{\prime \prime}-y^{\prime}-12 y=0 \) can be written using differential operator. Which of the following is incorrectly written? \[ \begin{array}{l} D^{2} y-D y-12=0 \\ D^{2}-D-12=0 \\ \left(D^{2}-D-12\right) y=0 \end{array} \] none of the above A linear operator \( L \) can be used as short hand notation for an n-order linear differential equation. Given \( 4 y^{\prime \prime}-4 y^{\prime}+y=0 \). Which is incorrectly expressed in terms of the linear operator? \[ \begin{array}{l} L[y]=0 \\ L(y)=0 \\ L y=0 \\ L=4 D^{2}-4 D+1=0 \end{array} \]
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Transcribed Image Text: \( \frac{d y}{d x} \) can be expressed as \( D y \) whereas \( \frac{d^{2} y}{d x^{2}} \) can be expressed as \( D y^{2} \) True False QUESTION 2 \( y^{\prime \prime}-y^{\prime}-12 y=0 \) can be written using differential operator. Which of the following is incorrectly written? \[ \begin{array}{l} D^{2} y-D y-12=0 \\ D^{2}-D-12=0 \\ \left(D^{2}-D-12\right) y=0 \end{array} \] none of the above A linear operator \( L \) can be used as short hand notation for an n-order linear differential equation. Given \( 4 y^{\prime \prime}-4 y^{\prime}+y=0 \). Which is incorrectly expressed in terms of the linear operator? \[ \begin{array}{l} L[y]=0 \\ L(y)=0 \\ L y=0 \\ L=4 D^{2}-4 D+1=0 \end{array} \]
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【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

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