$y$ be the solution of $y'+y=|x|,xin mathbb{R}$, find $y(1)$ [duplicate]












0












$begingroup$



This question already has an answer here:




  • Find the value of $y(1)$ of the ODE $y'+y=|x|$.

    3 answers




Let $y$ be the solution of $$y'+y=|x|,xin mathbb{R}$$
$$y(-1)=0$$
Then what is the value of $y(1)$:



Efforts:



I know how to solve the linear equations of general form $$y'+Py=Q$$ where $P$ and $Q$ are continuous function of $x$.



In my case $P=1$ and $Q=|x|$ and both are continuous.



Now if I apply the formula I get



$$ye^x=int e^x |x|dx+C$$



Now if I put the initial value I get $$0=int e^{-1}dx+C$$



Now what should I do? I mean what should be the range of integration and how I proceed from here.



Thanks.






ordinary-differential-equations







share|cite|improve this question









$endgroup$



marked as duplicate by StammeringMathematician, Community♦ Jan 31 at 3:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
    $endgroup$
    – StammeringMathematician
    Jan 31 at 3:28








  • 1




    $begingroup$
    "Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
    $endgroup$
    – Did
    Jan 31 at 7:22










  • $begingroup$
    @Did Can you explain? I did not get it
    $endgroup$
    – StammeringMathematician
    Jan 31 at 10:12






  • 1




    $begingroup$
    Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
    $endgroup$
    – Did
    Jan 31 at 10:56
















0












$begingroup$



This question already has an answer here:




  • Find the value of $y(1)$ of the ODE $y'+y=|x|$.

    3 answers




Let $y$ be the solution of $$y'+y=|x|,xin mathbb{R}$$
$$y(-1)=0$$
Then what is the value of $y(1)$:



Efforts:



I know how to solve the linear equations of general form $$y'+Py=Q$$ where $P$ and $Q$ are continuous function of $x$.



In my case $P=1$ and $Q=|x|$ and both are continuous.



Now if I apply the formula I get



$$ye^x=int e^x |x|dx+C$$



Now if I put the initial value I get $$0=int e^{-1}dx+C$$



Now what should I do? I mean what should be the range of integration and how I proceed from here.



Thanks.






ordinary-differential-equations







share|cite|improve this question









$endgroup$



marked as duplicate by StammeringMathematician, Community♦ Jan 31 at 3:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
    $endgroup$
    – StammeringMathematician
    Jan 31 at 3:28








  • 1




    $begingroup$
    "Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
    $endgroup$
    – Did
    Jan 31 at 7:22










  • $begingroup$
    @Did Can you explain? I did not get it
    $endgroup$
    – StammeringMathematician
    Jan 31 at 10:12






  • 1




    $begingroup$
    Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
    $endgroup$
    – Did
    Jan 31 at 10:56














0












0








0


1



$begingroup$



This question already has an answer here:




  • Find the value of $y(1)$ of the ODE $y'+y=|x|$.

    3 answers




Let $y$ be the solution of $$y'+y=|x|,xin mathbb{R}$$
$$y(-1)=0$$
Then what is the value of $y(1)$:



Efforts:



I know how to solve the linear equations of general form $$y'+Py=Q$$ where $P$ and $Q$ are continuous function of $x$.



In my case $P=1$ and $Q=|x|$ and both are continuous.



Now if I apply the formula I get



$$ye^x=int e^x |x|dx+C$$



Now if I put the initial value I get $$0=int e^{-1}dx+C$$



Now what should I do? I mean what should be the range of integration and how I proceed from here.



Thanks.






ordinary-differential-equations







share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Find the value of $y(1)$ of the ODE $y'+y=|x|$.

    3 answers




Let $y$ be the solution of $$y'+y=|x|,xin mathbb{R}$$
$$y(-1)=0$$
Then what is the value of $y(1)$:



Efforts:



I know how to solve the linear equations of general form $$y'+Py=Q$$ where $P$ and $Q$ are continuous function of $x$.



In my case $P=1$ and $Q=|x|$ and both are continuous.



Now if I apply the formula I get



$$ye^x=int e^x |x|dx+C$$



Now if I put the initial value I get $$0=int e^{-1}dx+C$$



Now what should I do? I mean what should be the range of integration and how I proceed from here.



Thanks.





This question already has an answer here:




  • Find the value of $y(1)$ of the ODE $y'+y=|x|$.

    3 answers







ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 31 at 1:27









StammeringMathematicianStammeringMathematician

2,7311324




2,7311324




marked as duplicate by StammeringMathematician, Community♦ Jan 31 at 3:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by StammeringMathematician, Community♦ Jan 31 at 3:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
    $endgroup$
    – StammeringMathematician
    Jan 31 at 3:28








  • 1




    $begingroup$
    "Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
    $endgroup$
    – Did
    Jan 31 at 7:22










  • $begingroup$
    @Did Can you explain? I did not get it
    $endgroup$
    – StammeringMathematician
    Jan 31 at 10:12






  • 1




    $begingroup$
    Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
    $endgroup$
    – Did
    Jan 31 at 10:56


















  • $begingroup$
    I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
    $endgroup$
    – StammeringMathematician
    Jan 31 at 3:28








  • 1




    $begingroup$
    "Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
    $endgroup$
    – Did
    Jan 31 at 7:22










  • $begingroup$
    @Did Can you explain? I did not get it
    $endgroup$
    – StammeringMathematician
    Jan 31 at 10:12






  • 1




    $begingroup$
    Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
    $endgroup$
    – Did
    Jan 31 at 10:56
















$begingroup$
I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
$endgroup$
– StammeringMathematician
Jan 31 at 3:28






$begingroup$
I have voted my question as duplicate. I have found the solution here math.stackexchange.com/questions/1850959/… I am not deleting it as it has received two upvotes and one favorites. Thanks
$endgroup$
– StammeringMathematician
Jan 31 at 3:28






1




1




$begingroup$
"Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
$endgroup$
– Did
Jan 31 at 7:22




$begingroup$
"Now if I put the initial value I get" What? Come again? What do you think the formula just after that even means?
$endgroup$
– Did
Jan 31 at 7:22












$begingroup$
@Did Can you explain? I did not get it
$endgroup$
– StammeringMathematician
Jan 31 at 10:12




$begingroup$
@Did Can you explain? I did not get it
$endgroup$
– StammeringMathematician
Jan 31 at 10:12




1




1




$begingroup$
Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
$endgroup$
– Did
Jan 31 at 10:56




$begingroup$
Again, what do you mean by "Now if I put the initial value I get $0=int e^{-1}dx+C$"? Can you explain? I do not get it.
$endgroup$
– Did
Jan 31 at 10:56










1 Answer
1






active

oldest

votes


















0












$begingroup$

In general terms, the solution of



$y' + Py = Q, ; y(-1) = 0, tag 1$



may be derived as follows:



$exp left (displaystyle int_{x_0}^x P(s); ds right ) y' + exp left (displaystyle int_{x_0}^x P(s); ds right ) Py = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 2$



$left ( exp left (displaystyle int_{x_0}^x P(s); ds right )y right )' = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 3$



$exp left (displaystyle int_{x_0}^x P(s); ds right )y(x) - y(x_0) =displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du; tag 4$



$y(x) = exp left (-displaystyle int_{x_0}^x P(s); ds right )left ( y(x_0) + displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du right ); tag 5$



in the present case,



$y' + y = vert x vert, ; y(-1) = 0; tag 6$



$P(x) = 1, ; Q(x) = vert x vert; tag 7$



$displaystyle int_{x_0}^x P(s) ; ds = x - x_0; tag 8$



$exp left ( displaystyle int_{x_0}^x P(s); ds right ) = e^{x - x_0}; tag 9$



with



$x_0 = -1, ; y(x_0) = y(-1) = 0 tag{10}$



we find



$exp left ( displaystyle int_{-1}^x P(s); ds right ) = e^{x + 1}, tag{11}$



$exp left (- displaystyle int_{-1}^x P(s); ds right ) = e^{-(x + 1)}, tag{12}$



and from (5),



$y(x) = e^{-(x + 1)} displaystyle int_{-1}^x e^{u + 1} vert u vert ; du = e^{-x} displaystyle int_{-1}^x e^u vert u vert ; du ; tag{13}$



with $x ge 0$,



$displaystyle int_{-1}^x e^u vert u vert ; du = int_{-1}^0 (-ue^u) ; du + int_0^x ue^u ; du = int_0^x ue^u ; du - int_{-1}^0 ue^u ; du; tag{14}$



the integrals of the form occurring in (14) are well-known; indeed the anti-derivative of $ue^u$ is found from



$(e^u(u - 1))' = e^u(u - 1) + e^u = ue^u; tag{15}$



it then follows that



$displaystyle int_a^b ue^e ; u = (e^u(u - 1)]_a^b =e^b(b - 1) - e^a(a - 1) = be^b - ae^a -(e^b - e^a); tag{16}$



we thus may evaluate (14):



$displaystyle int_0^x ue^u ; du - int_{-1}^0 ue^u ; du$
$= (xe^x - e^x + 1) - (e^{-1} -1 + e^{-1}) = xe^x - e^x + 2 - 2e^{-1}; tag{17}$



thus, from (13),



$y(x) = x - 1 + 2e^{-x} - 2e^{-1 - x}; tag{18}$



finally we find



$y(1) = 2e^{-1} - 2e^{-2}. tag{19}$






share|cite|improve this answer











$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    In general terms, the solution of



    $y' + Py = Q, ; y(-1) = 0, tag 1$



    may be derived as follows:



    $exp left (displaystyle int_{x_0}^x P(s); ds right ) y' + exp left (displaystyle int_{x_0}^x P(s); ds right ) Py = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 2$



    $left ( exp left (displaystyle int_{x_0}^x P(s); ds right )y right )' = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 3$



    $exp left (displaystyle int_{x_0}^x P(s); ds right )y(x) - y(x_0) =displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du; tag 4$



    $y(x) = exp left (-displaystyle int_{x_0}^x P(s); ds right )left ( y(x_0) + displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du right ); tag 5$



    in the present case,



    $y' + y = vert x vert, ; y(-1) = 0; tag 6$



    $P(x) = 1, ; Q(x) = vert x vert; tag 7$



    $displaystyle int_{x_0}^x P(s) ; ds = x - x_0; tag 8$



    $exp left ( displaystyle int_{x_0}^x P(s); ds right ) = e^{x - x_0}; tag 9$



    with



    $x_0 = -1, ; y(x_0) = y(-1) = 0 tag{10}$



    we find



    $exp left ( displaystyle int_{-1}^x P(s); ds right ) = e^{x + 1}, tag{11}$



    $exp left (- displaystyle int_{-1}^x P(s); ds right ) = e^{-(x + 1)}, tag{12}$



    and from (5),



    $y(x) = e^{-(x + 1)} displaystyle int_{-1}^x e^{u + 1} vert u vert ; du = e^{-x} displaystyle int_{-1}^x e^u vert u vert ; du ; tag{13}$



    with $x ge 0$,



    $displaystyle int_{-1}^x e^u vert u vert ; du = int_{-1}^0 (-ue^u) ; du + int_0^x ue^u ; du = int_0^x ue^u ; du - int_{-1}^0 ue^u ; du; tag{14}$



    the integrals of the form occurring in (14) are well-known; indeed the anti-derivative of $ue^u$ is found from



    $(e^u(u - 1))' = e^u(u - 1) + e^u = ue^u; tag{15}$



    it then follows that



    $displaystyle int_a^b ue^e ; u = (e^u(u - 1)]_a^b =e^b(b - 1) - e^a(a - 1) = be^b - ae^a -(e^b - e^a); tag{16}$



    we thus may evaluate (14):



    $displaystyle int_0^x ue^u ; du - int_{-1}^0 ue^u ; du$
    $= (xe^x - e^x + 1) - (e^{-1} -1 + e^{-1}) = xe^x - e^x + 2 - 2e^{-1}; tag{17}$



    thus, from (13),



    $y(x) = x - 1 + 2e^{-x} - 2e^{-1 - x}; tag{18}$



    finally we find



    $y(1) = 2e^{-1} - 2e^{-2}. tag{19}$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      In general terms, the solution of



      $y' + Py = Q, ; y(-1) = 0, tag 1$



      may be derived as follows:



      $exp left (displaystyle int_{x_0}^x P(s); ds right ) y' + exp left (displaystyle int_{x_0}^x P(s); ds right ) Py = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 2$



      $left ( exp left (displaystyle int_{x_0}^x P(s); ds right )y right )' = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 3$



      $exp left (displaystyle int_{x_0}^x P(s); ds right )y(x) - y(x_0) =displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du; tag 4$



      $y(x) = exp left (-displaystyle int_{x_0}^x P(s); ds right )left ( y(x_0) + displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du right ); tag 5$



      in the present case,



      $y' + y = vert x vert, ; y(-1) = 0; tag 6$



      $P(x) = 1, ; Q(x) = vert x vert; tag 7$



      $displaystyle int_{x_0}^x P(s) ; ds = x - x_0; tag 8$



      $exp left ( displaystyle int_{x_0}^x P(s); ds right ) = e^{x - x_0}; tag 9$



      with



      $x_0 = -1, ; y(x_0) = y(-1) = 0 tag{10}$



      we find



      $exp left ( displaystyle int_{-1}^x P(s); ds right ) = e^{x + 1}, tag{11}$



      $exp left (- displaystyle int_{-1}^x P(s); ds right ) = e^{-(x + 1)}, tag{12}$



      and from (5),



      $y(x) = e^{-(x + 1)} displaystyle int_{-1}^x e^{u + 1} vert u vert ; du = e^{-x} displaystyle int_{-1}^x e^u vert u vert ; du ; tag{13}$



      with $x ge 0$,



      $displaystyle int_{-1}^x e^u vert u vert ; du = int_{-1}^0 (-ue^u) ; du + int_0^x ue^u ; du = int_0^x ue^u ; du - int_{-1}^0 ue^u ; du; tag{14}$



      the integrals of the form occurring in (14) are well-known; indeed the anti-derivative of $ue^u$ is found from



      $(e^u(u - 1))' = e^u(u - 1) + e^u = ue^u; tag{15}$



      it then follows that



      $displaystyle int_a^b ue^e ; u = (e^u(u - 1)]_a^b =e^b(b - 1) - e^a(a - 1) = be^b - ae^a -(e^b - e^a); tag{16}$



      we thus may evaluate (14):



      $displaystyle int_0^x ue^u ; du - int_{-1}^0 ue^u ; du$
      $= (xe^x - e^x + 1) - (e^{-1} -1 + e^{-1}) = xe^x - e^x + 2 - 2e^{-1}; tag{17}$



      thus, from (13),



      $y(x) = x - 1 + 2e^{-x} - 2e^{-1 - x}; tag{18}$



      finally we find



      $y(1) = 2e^{-1} - 2e^{-2}. tag{19}$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        In general terms, the solution of



        $y' + Py = Q, ; y(-1) = 0, tag 1$



        may be derived as follows:



        $exp left (displaystyle int_{x_0}^x P(s); ds right ) y' + exp left (displaystyle int_{x_0}^x P(s); ds right ) Py = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 2$



        $left ( exp left (displaystyle int_{x_0}^x P(s); ds right )y right )' = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 3$



        $exp left (displaystyle int_{x_0}^x P(s); ds right )y(x) - y(x_0) =displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du; tag 4$



        $y(x) = exp left (-displaystyle int_{x_0}^x P(s); ds right )left ( y(x_0) + displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du right ); tag 5$



        in the present case,



        $y' + y = vert x vert, ; y(-1) = 0; tag 6$



        $P(x) = 1, ; Q(x) = vert x vert; tag 7$



        $displaystyle int_{x_0}^x P(s) ; ds = x - x_0; tag 8$



        $exp left ( displaystyle int_{x_0}^x P(s); ds right ) = e^{x - x_0}; tag 9$



        with



        $x_0 = -1, ; y(x_0) = y(-1) = 0 tag{10}$



        we find



        $exp left ( displaystyle int_{-1}^x P(s); ds right ) = e^{x + 1}, tag{11}$



        $exp left (- displaystyle int_{-1}^x P(s); ds right ) = e^{-(x + 1)}, tag{12}$



        and from (5),



        $y(x) = e^{-(x + 1)} displaystyle int_{-1}^x e^{u + 1} vert u vert ; du = e^{-x} displaystyle int_{-1}^x e^u vert u vert ; du ; tag{13}$



        with $x ge 0$,



        $displaystyle int_{-1}^x e^u vert u vert ; du = int_{-1}^0 (-ue^u) ; du + int_0^x ue^u ; du = int_0^x ue^u ; du - int_{-1}^0 ue^u ; du; tag{14}$



        the integrals of the form occurring in (14) are well-known; indeed the anti-derivative of $ue^u$ is found from



        $(e^u(u - 1))' = e^u(u - 1) + e^u = ue^u; tag{15}$



        it then follows that



        $displaystyle int_a^b ue^e ; u = (e^u(u - 1)]_a^b =e^b(b - 1) - e^a(a - 1) = be^b - ae^a -(e^b - e^a); tag{16}$



        we thus may evaluate (14):



        $displaystyle int_0^x ue^u ; du - int_{-1}^0 ue^u ; du$
        $= (xe^x - e^x + 1) - (e^{-1} -1 + e^{-1}) = xe^x - e^x + 2 - 2e^{-1}; tag{17}$



        thus, from (13),



        $y(x) = x - 1 + 2e^{-x} - 2e^{-1 - x}; tag{18}$



        finally we find



        $y(1) = 2e^{-1} - 2e^{-2}. tag{19}$






        share|cite|improve this answer











        $endgroup$



        In general terms, the solution of



        $y' + Py = Q, ; y(-1) = 0, tag 1$



        may be derived as follows:



        $exp left (displaystyle int_{x_0}^x P(s); ds right ) y' + exp left (displaystyle int_{x_0}^x P(s); ds right ) Py = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 2$



        $left ( exp left (displaystyle int_{x_0}^x P(s); ds right )y right )' = exp left (displaystyle int_{x_0}^x P(s); ds right )Q; tag 3$



        $exp left (displaystyle int_{x_0}^x P(s); ds right )y(x) - y(x_0) =displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du; tag 4$



        $y(x) = exp left (-displaystyle int_{x_0}^x P(s); ds right )left ( y(x_0) + displaystyle int_{x_0}^x exp left (displaystyle int_{x_0}^u P(s); ds right )Q(u) ; du right ); tag 5$



        in the present case,



        $y' + y = vert x vert, ; y(-1) = 0; tag 6$



        $P(x) = 1, ; Q(x) = vert x vert; tag 7$



        $displaystyle int_{x_0}^x P(s) ; ds = x - x_0; tag 8$



        $exp left ( displaystyle int_{x_0}^x P(s); ds right ) = e^{x - x_0}; tag 9$



        with



        $x_0 = -1, ; y(x_0) = y(-1) = 0 tag{10}$



        we find



        $exp left ( displaystyle int_{-1}^x P(s); ds right ) = e^{x + 1}, tag{11}$



        $exp left (- displaystyle int_{-1}^x P(s); ds right ) = e^{-(x + 1)}, tag{12}$



        and from (5),



        $y(x) = e^{-(x + 1)} displaystyle int_{-1}^x e^{u + 1} vert u vert ; du = e^{-x} displaystyle int_{-1}^x e^u vert u vert ; du ; tag{13}$



        with $x ge 0$,



        $displaystyle int_{-1}^x e^u vert u vert ; du = int_{-1}^0 (-ue^u) ; du + int_0^x ue^u ; du = int_0^x ue^u ; du - int_{-1}^0 ue^u ; du; tag{14}$



        the integrals of the form occurring in (14) are well-known; indeed the anti-derivative of $ue^u$ is found from



        $(e^u(u - 1))' = e^u(u - 1) + e^u = ue^u; tag{15}$



        it then follows that



        $displaystyle int_a^b ue^e ; u = (e^u(u - 1)]_a^b =e^b(b - 1) - e^a(a - 1) = be^b - ae^a -(e^b - e^a); tag{16}$



        we thus may evaluate (14):



        $displaystyle int_0^x ue^u ; du - int_{-1}^0 ue^u ; du$
        $= (xe^x - e^x + 1) - (e^{-1} -1 + e^{-1}) = xe^x - e^x + 2 - 2e^{-1}; tag{17}$



        thus, from (13),



        $y(x) = x - 1 + 2e^{-x} - 2e^{-1 - x}; tag{18}$



        finally we find



        $y(1) = 2e^{-1} - 2e^{-2}. tag{19}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 31 at 5:51

























        answered Jan 31 at 5:03









        Robert LewisRobert Lewis

        48.6k23167




        48.6k23167















            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            How to fix TextFormField cause rebuild widget in Flutter

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more