Test #1 Review, Webwork First Order Linear Problem 1

6 Comments

1. 

   Oscar

   September 25, 2020 at 6:51 pm

   I’ve been cheating and taking a screen capture and then uploading an image. Just type “image” in the search block after you click the + sign.

   there’s always latex too:

   replace * * with [ ] and this should post as the pretty math symbols:

   *latextext*

   $frac{dy}{dt}=y+4$
   $y’=y+4$
   $\frac{y’}{y+4}=\frac{y+4}{y+4}
   $\frac{y’}{y+4}=1$
   $\int \frac{y’dt}{y+4}=\int 1 dt$

   let u= y+4
   du= y’dt

   $\int \frac{du}{u}= \int 1 dt$
   $ln\lvert u \rvert +c = t+c$
   $ln\lvert y+4 \rvert +c = t+c$
   $e^{ln\lvert y+4 \rvert +c}=e^{t+c}$
   $y+4= e^{t+4}$
   $y=e^{t+c}-4$

   Should look like this (fingers crossed, no preview for comments):

   [latextext]

   $frac{dy}{dt}=y+4$
   $y’=y+4$
   $\frac{y’}{y+4}=\frac{y+4}{y+4}
   $\frac{y’}{y+4}=1$
   $\int \frac{y’dt}{y+4}=\int 1 dt$

   let u= y+4
   du= y’dt

   $\int \frac{du}{u}= \int 1 dt$
   $ln\lvert u \rvert +c = t+c$
   $ln\lvert y+4 \rvert +c = t+c$
   $e^{ln\lvert y+4 \rvert +c}=e^{t+c}$
   $y+4= e^{t+4}$
   $y=e^{t+c}-4$

   Reply
   • 

     Oscar

     September 25, 2020 at 6:53 pm

     ops *latexpage*

     [latexpage]

     $frac{dy}{dt}=y+4$
     $y’=y+4$
     $\frac{y’}{y+4}=\frac{y+4}{y+4}
     $\frac{y’}{y+4}=1$
     $\int \frac{y’dt}{y+4}=\int 1 dt$

     let u= y+4
     du= y’dt

     $\int \frac{du}{u}= \int 1 dt$
     $ln\lvert u \rvert +c = t+c$
     $ln\lvert y+4 \rvert +c = t+c$
     $e^{ln\lvert y+4 \rvert +c}=e^{t+c}$
     $y+4= e^{t+4}$
     $y=e^{t+c}-4$

     Reply
     • 

       Oscar

       September 25, 2020 at 6:56 pm

       hahaha where is the edit button!

       $\frac{dy}{dt}=y+4$
       $y’=y+4$
       $\frac{y’}{y+4}=\frac{y+4}{y+4}$
       $\frac{y’}{y+4}=1$
       $\int \frac{y’dt}{y+4}=\int 1 dt$

       let u= y+4
       du= y’dt

       $\int \frac{du}{u}= \int 1 dt$
       $ln\lvert u \rvert +c = t+c$
       $ln\lvert y+4 \rvert +c = t+c$
       $e^{ln\lvert y+4 \rvert +c}=e^{t+c}$
       $y+4= e^{t+4}$
       $y=e^{t+c}-4$

       Reply
       • 

         Oscar

         September 25, 2020 at 6:57 pm

         just 5 more attempts and then I’m out
         [latexpage]

         $\frac{dy}{dt}=y+4$
         $y’=y+4$
         $\frac{y’}{y+4}=\frac{y+4}{y+4}$
         $\frac{y’}{y+4}=1$
         $\int \frac{y’dt}{y+4}=\int 1 dt$

         let u= y+4
         du= y’dt

         $\int \frac{du}{u}= \int 1 dt$
         $ln\lvert u \rvert +c = t+c$
         $ln\lvert y+4 \rvert +c = t+c$
         $e^{ln\lvert y+4 \rvert +c}=e^{t+c}$
         $y+4= e^{t+4}$
         $y=e^{t+c}-4$

         Reply
         • 

           Oscar

           September 25, 2020 at 7:01 pm

           ok fix this and it’s golden… :

           $y+4= e^{t+4}$

           should be:

           $y+4=e^{t+c}
         • 

           Kate Poirier

           September 28, 2020 at 4:23 pm

           Oscar, I absolutely love that you’re all in on LaTeX! If you wanna get real nerdy, add a backslash to the beginning of your natural logarithm (so \ln instead of just ln). Since the natural log is a well-known function, people usually typeset it in roman characters instead of italics (which is what we see for letters in math mode), so there’s already this built-in macro to do this. Same goes for \sin, \cos, etc…