1

Please Help me in writing this equation

enter image description here

2
  • 1
    Could you please show us what you've got so far. This is not a we-do-it-for-you-site, show some own effort and everybody here will be glad to help. You might take a look at the the LaTeX-Wikibook if you have no idea how to start.
    – Skillmon
    Commented Jun 29, 2017 at 11:46
  • Slashed letter can be obtained with the cancel package.
    – Bernard
    Commented Jun 29, 2017 at 12:12

3 Answers 3

3

Fixing also various glitches in the original (due to indiscriminate usage of \left and \right, mainly):

\documentclass{article}
\usepackage{amsmath}
\usepackage{slashed}

\begin{document}

\begin{equation*}
\left(
  \begin{aligned}
  -\slashed{g}
    &= y''\Bigl(\frac{dx}{ds}\Bigr)^{\!2}v^2
     = \frac{y''v^2}{(ds/dx)^2}
  \\
    &= \frac{y''v^2}{1+(y')^2}
     = \frac{y''2\slashed{g}(y_0-y)}{1+(y')^2}
  \end{aligned}
\right)
\implies
2y''(y-y_0) = 1+(y')^2
\end{equation*}

\end{document}

enter image description here

1

Using array

  \documentclass[a4paper]{article}
  \usepackage[left=1cm,right=1cm]{geometry}
  \usepackage{amsmath,cancel}

  \begin{document}
  \begin{center}

  $\left(
  \begin{array}{llll}
  -\cancel{g} & =  y''\left( \dfrac{dx}{ds} \right)^{2}v^{2}&=\dfrac{y''v_{2}}{(ds/dx)^{2}}\\
  &&&\\
                 & =  \dfrac{y''v_{2}}{1+(y')^{2}}&=\dfrac{y''2\cancel{g}(y_{0}-y)}{1+(y')^{2}}\\
  \end{array}
  \right)$$\implies 2y''(y-y_{0})=1+(y')^{2}$.

  \end{center}

  \end{document}

enter image description here

0

Use the array from amsmath for alignment at = and the cancel package to cancel out some terms.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,cancel}

\begin{document}
\[\left(
  \begin{array}{@{}r@{{}={}}l@{}}
  -\cancel{g}& y''\Big(\dfrac{dx}{ds}\Big)^2 v^2=\dfrac{y''v_2}{(ds/dx)^2}\\
   & \dfrac{y'' v^2}{1+(y^\prime)^2}=\dfrac{y''2\cancel{g}(y_0-y)}{1+(y')^2}
  \end{array}\right) \Longrightarrow 2y''(y-y_0)=1+(y')^2.
\]

\end{document}

enter image description here

This can also be represented in a slightly different way, since it is a single equation divided over two lines and not a matrix.

\[\left.
  \begin{array}{r@{{}={}}l}
  -\cancel{g} & y''\Big(\dfrac{dx}{ds}\Big)^2 v^2=\dfrac{y''v_2}{(ds/dx)^2}\\
   & \dfrac{y'' v^2}{1+(y')^2}=\dfrac{y''2\cancel{g}(y_0-y)}{1+(y')^2}
  \end{array}\right\} \Longrightarrow 2y''(y-y_0)=1+(y')^2.
\]

enter image description here

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