Solving a differential equation with a linear solution and initial conditions












2












$begingroup$


The problem reads:



The solution of a certain differential equation is of the form



$$y(t)= aexp(5t) + bexp(8t)$$



where $a$ and $b$ are constants.



The solution has initial conditions $y(0)=5$ and $y'(0)=5$



Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................



What I did was solve using the initial conditions and I found that



$a + b = 5 $



and
$
5a + 8b =5. $



Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
    $endgroup$
    – mfl
    Sep 4 '16 at 0:09
















2












$begingroup$


The problem reads:



The solution of a certain differential equation is of the form



$$y(t)= aexp(5t) + bexp(8t)$$



where $a$ and $b$ are constants.



The solution has initial conditions $y(0)=5$ and $y'(0)=5$



Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................



What I did was solve using the initial conditions and I found that



$a + b = 5 $



and
$
5a + 8b =5. $



Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
    $endgroup$
    – mfl
    Sep 4 '16 at 0:09














2












2








2





$begingroup$


The problem reads:



The solution of a certain differential equation is of the form



$$y(t)= aexp(5t) + bexp(8t)$$



where $a$ and $b$ are constants.



The solution has initial conditions $y(0)=5$ and $y'(0)=5$



Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................



What I did was solve using the initial conditions and I found that



$a + b = 5 $



and
$
5a + 8b =5. $



Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.






ordinary-differential-equations







share|cite|improve this question











$endgroup$




The problem reads:



The solution of a certain differential equation is of the form



$$y(t)= aexp(5t) + bexp(8t)$$



where $a$ and $b$ are constants.



The solution has initial conditions $y(0)=5$ and $y'(0)=5$



Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................



What I did was solve using the initial conditions and I found that



$a + b = 5 $



and
$
5a + 8b =5. $



Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.






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








edited Sep 4 '16 at 0:36









ForgotALot

3,6271816




3,6271816










asked Sep 4 '16 at 0:04









J McDowellJ McDowell

192




192








  • 2




    $begingroup$
    You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
    $endgroup$
    – mfl
    Sep 4 '16 at 0:09














  • 2




    $begingroup$
    You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
    $endgroup$
    – mfl
    Sep 4 '16 at 0:09








2




2




$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09




$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09










1 Answer
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1












$begingroup$

You're absolutely right.



If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.



Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.



To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.



Personally I would use matrix algebra, but it's up to you.



$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$



The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives



$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$



Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}



Your final solution is then



$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$






share|cite|improve this answer









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    1












    $begingroup$

    You're absolutely right.



    If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.



    Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.



    To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.



    Personally I would use matrix algebra, but it's up to you.



    $$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
    left(begin{array}{c} a \ b end{array}right)=
    left(begin{array}{c} 5 \ 5 end{array}right) $$



    The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives



    $$
    left(begin{array}{c} a \ b end{array}right)=
    frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
    left(begin{array}{c} 5 \ 5 end{array}right) $$



    Expanding the right hand side gives
    begin{eqnarray*}
    a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
    b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
    end{eqnarray*}



    Your final solution is then



    $$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You're absolutely right.



      If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.



      Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.



      To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.



      Personally I would use matrix algebra, but it's up to you.



      $$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
      left(begin{array}{c} a \ b end{array}right)=
      left(begin{array}{c} 5 \ 5 end{array}right) $$



      The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives



      $$
      left(begin{array}{c} a \ b end{array}right)=
      frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
      left(begin{array}{c} 5 \ 5 end{array}right) $$



      Expanding the right hand side gives
      begin{eqnarray*}
      a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
      b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
      end{eqnarray*}



      Your final solution is then



      $$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You're absolutely right.



        If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.



        Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.



        To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.



        Personally I would use matrix algebra, but it's up to you.



        $$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
        left(begin{array}{c} a \ b end{array}right)=
        left(begin{array}{c} 5 \ 5 end{array}right) $$



        The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives



        $$
        left(begin{array}{c} a \ b end{array}right)=
        frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
        left(begin{array}{c} 5 \ 5 end{array}right) $$



        Expanding the right hand side gives
        begin{eqnarray*}
        a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
        b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
        end{eqnarray*}



        Your final solution is then



        $$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$






        share|cite|improve this answer









        $endgroup$



        You're absolutely right.



        If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.



        Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.



        To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.



        Personally I would use matrix algebra, but it's up to you.



        $$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
        left(begin{array}{c} a \ b end{array}right)=
        left(begin{array}{c} 5 \ 5 end{array}right) $$



        The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives



        $$
        left(begin{array}{c} a \ b end{array}right)=
        frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
        left(begin{array}{c} 5 \ 5 end{array}right) $$



        Expanding the right hand side gives
        begin{eqnarray*}
        a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
        b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
        end{eqnarray*}



        Your final solution is then



        $$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 4 '16 at 1:02









        Fly by NightFly by Night

        26.1k32978




        26.1k32978






























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